Reading in File line by line w/ Bash

I'm creating a bash script to read a file in line by line, that is formatted later to be organised by name and then date. I cannot see why this code isn't working at this time though no errors show up even though I have tried with the input and output filename variables on their own, with a directory finder and export command.

export inputfilename="employees.txt"

export outputfilename="branch.txt"

directoryinput=$(find -name $inputfilename)

directoryoutput=$(find -name $outputfilename)

n=1



if [[ -f "$directoryinput" ]]; then

     while read line; do

         echo "$line"

         n=$((n+1))

     done < "$directoryoutput"

 else

    echo "Input file does not exist. Please create a employees.txt file"

 fi

All help is very much appreciated, thank you!
NOTE: As people noticed, I forgot to add in the $ sign on the data transfer to file, but it was just in copying my code, I do have the $ sign in my actual application and still no result

edited Nov 20 '18 at 10:15

asked Nov 19 '18 at 19:35

Sam Thompson

3

done < "directoryoutput" should probably be done < "$directoryoutput" (it was missing the dollar sign). You can toss these shell scripts into shellcheck.net and it's pretty good at helping to find issues like this. In this case it highlights that 4th line and notes that the variable is loaded, but never used.
– JNevill
Nov 19 '18 at 19:38

Much safer to use find . -name "$inputfilename" -print -quit, where you're guaranteed to only find one file, vs having multiple names concatenated together in your variable, if you must use find
– Charles Duffy
Nov 19 '18 at 20:30

1

BTW, the original indentation here is extremely confusing. Why does the do move back to outside the while loop? Why are the contents of that loop not indented? I'm taking the liberty of revising it to follow the code flow; feel free to roll back if that's too heavy-handed, but please do avoid outdenting symbols that are part of a loop to an indentation level preceding that loop.
– Charles Duffy
Nov 19 '18 at 20:35

(Also, the . in find . -name is suggested above for portability reasons: Being able to leave it out is a GNUism, but doesn't work in versions of find that hew more closely to the baseline POSIX specification).
– Charles Duffy
Nov 19 '18 at 20:38

add a comment |

export inputfilename="employees.txt"

export outputfilename="branch.txt"

directoryinput=$(find -name $inputfilename)

directoryoutput=$(find -name $outputfilename)

n=1



if [[ -f "$directoryinput" ]]; then

     while read line; do

         echo "$line"

         n=$((n+1))

     done < "$directoryoutput"

 else

    echo "Input file does not exist. Please create a employees.txt file"

 fi

edited Nov 20 '18 at 10:15

asked Nov 19 '18 at 19:35

Sam Thompson

3

done < "directoryoutput" should probably be done < "$directoryoutput" (it was missing the dollar sign). You can toss these shell scripts into shellcheck.net and it's pretty good at helping to find issues like this. In this case it highlights that 4th line and notes that the variable is loaded, but never used.
– JNevill
Nov 19 '18 at 19:38

Much safer to use find . -name "$inputfilename" -print -quit, where you're guaranteed to only find one file, vs having multiple names concatenated together in your variable, if you must use find
– Charles Duffy
Nov 19 '18 at 20:30

1

BTW, the original indentation here is extremely confusing. Why does the do move back to outside the while loop? Why are the contents of that loop not indented? I'm taking the liberty of revising it to follow the code flow; feel free to roll back if that's too heavy-handed, but please do avoid outdenting symbols that are part of a loop to an indentation level preceding that loop.
– Charles Duffy
Nov 19 '18 at 20:35

(Also, the . in find . -name is suggested above for portability reasons: Being able to leave it out is a GNUism, but doesn't work in versions of find that hew more closely to the baseline POSIX specification).
– Charles Duffy
Nov 19 '18 at 20:38

add a comment |

export inputfilename="employees.txt"

export outputfilename="branch.txt"

directoryinput=$(find -name $inputfilename)

directoryoutput=$(find -name $outputfilename)

n=1



if [[ -f "$directoryinput" ]]; then

     while read line; do

         echo "$line"

         n=$((n+1))

     done < "$directoryoutput"

 else

    echo "Input file does not exist. Please create a employees.txt file"

 fi

edited Nov 20 '18 at 10:15

asked Nov 19 '18 at 19:35

Sam Thompson

export inputfilename="employees.txt"

export outputfilename="branch.txt"

directoryinput=$(find -name $inputfilename)

directoryoutput=$(find -name $outputfilename)

n=1



if [[ -f "$directoryinput" ]]; then

     while read line; do

         echo "$line"

         n=$((n+1))

     done < "$directoryoutput"

 else

    echo "Input file does not exist. Please create a employees.txt file"

 fi

bash unix while-loop line-by-line

edited Nov 20 '18 at 10:15

asked Nov 19 '18 at 19:35

Sam Thompson

edited Nov 20 '18 at 10:15

asked Nov 19 '18 at 19:35

Sam Thompson

edited Nov 20 '18 at 10:15

asked Nov 19 '18 at 19:35

Sam Thompson

asked Nov 19 '18 at 19:35

Sam Thompson

asked Nov 19 '18 at 19:35

Sam Thompson

3

done < "directoryoutput" should probably be done < "$directoryoutput" (it was missing the dollar sign). You can toss these shell scripts into shellcheck.net and it's pretty good at helping to find issues like this. In this case it highlights that 4th line and notes that the variable is loaded, but never used.
– JNevill
Nov 19 '18 at 19:38

Much safer to use find . -name "$inputfilename" -print -quit, where you're guaranteed to only find one file, vs having multiple names concatenated together in your variable, if you must use find
– Charles Duffy
Nov 19 '18 at 20:30

1

BTW, the original indentation here is extremely confusing. Why does the do move back to outside the while loop? Why are the contents of that loop not indented? I'm taking the liberty of revising it to follow the code flow; feel free to roll back if that's too heavy-handed, but please do avoid outdenting symbols that are part of a loop to an indentation level preceding that loop.
– Charles Duffy
Nov 19 '18 at 20:35

(Also, the . in find . -name is suggested above for portability reasons: Being able to leave it out is a GNUism, but doesn't work in versions of find that hew more closely to the baseline POSIX specification).
– Charles Duffy
Nov 19 '18 at 20:38

add a comment |

3

done < "directoryoutput" should probably be done < "$directoryoutput" (it was missing the dollar sign). You can toss these shell scripts into shellcheck.net and it's pretty good at helping to find issues like this. In this case it highlights that 4th line and notes that the variable is loaded, but never used.
– JNevill
Nov 19 '18 at 19:38

Much safer to use find . -name "$inputfilename" -print -quit, where you're guaranteed to only find one file, vs having multiple names concatenated together in your variable, if you must use find
– Charles Duffy
Nov 19 '18 at 20:30

1

BTW, the original indentation here is extremely confusing. Why does the do move back to outside the while loop? Why are the contents of that loop not indented? I'm taking the liberty of revising it to follow the code flow; feel free to roll back if that's too heavy-handed, but please do avoid outdenting symbols that are part of a loop to an indentation level preceding that loop.
– Charles Duffy
Nov 19 '18 at 20:35

(Also, the . in find . -name is suggested above for portability reasons: Being able to leave it out is a GNUism, but doesn't work in versions of find that hew more closely to the baseline POSIX specification).
– Charles Duffy
Nov 19 '18 at 20:38

done < "directoryoutput" should probably be done < "$directoryoutput" (it was missing the dollar sign). You can toss these shell scripts into shellcheck.net and it's pretty good at helping to find issues like this. In this case it highlights that 4th line and notes that the variable is loaded, but never used.
– JNevill
Nov 19 '18 at 19:38

Much safer to use find . -name "$inputfilename" -print -quit, where you're guaranteed to only find one file, vs having multiple names concatenated together in your variable, if you must use find
– Charles Duffy
Nov 19 '18 at 20:30

BTW, the original indentation here is extremely confusing. Why does the do move back to outside the while loop? Why are the contents of that loop not indented? I'm taking the liberty of revising it to follow the code flow; feel free to roll back if that's too heavy-handed, but please do avoid outdenting symbols that are part of a loop to an indentation level preceding that loop.
– Charles Duffy
Nov 19 '18 at 20:35

(Also, the . in find . -name is suggested above for portability reasons: Being able to leave it out is a GNUism, but doesn't work in versions of find that hew more closely to the baseline POSIX specification).
– Charles Duffy
Nov 19 '18 at 20:38

add a comment |

1 Answer
1

active

oldest

votes

Reading in File line by line w/ Bash

The best and idiomatic way to read file line by line is to do:

while IFS= read -r line; do

  // parse line

  printf "%s" "$line"

done < "file"

More on this topic can be found on bashfaq

However don't read files in bash line by line. You can (ok, almost) always not read a stream line by line in bash. Reading a file line by line in bash is extremely slow and shouldn't be done. For simple cases all the unix tools with the help of xargs or parallel can be used, for more complicated awk and datamesh are used.

done < "directoryoutput"

The code is not working, because you are passing to your while read loop as input to standard input the content of a file named directoryoutput. As such a file does not exists, your script fails.

directoryoutput=$(find -name $outputfilename)

One can simply append the variable value with newline appended to a read while loop using a HERE-string construction:

done <<< "$directoryoutput"

directoryinput=$(find -name $inputfilename)
if [[ -f "$directoryinput" ]]

This is ok as long as you have only one file named $inputfilename in your directory. Also it makes no sense to find a file and then check for it's existance. In case of more files, find return a newline separated list of names. However a small check if [ "$(printf "$directoryinput" | wc -l)" -eq 1 ] or using find -name $inputfilename | head -n1 I think would be better.

while read line;

   do

      echo "$line"

      n=$((n+1))

  done < "directoryoutput"

The intention is pretty clear here. This is just:

 n=$(<directoryoutput wc -l)

 cat "directoryoutput"

Except that while read line removed trailing and leading newlines and is IFS dependent.

Also always remember to quote your variables unless you have a reason not to.

Have a look at shellcheck which can find most common mistakes in scripts.

I would do it more like this:

inputfilename="employees.txt"

outputfilename="branch.txt"



directoryinput=$(find . -name "$inputfilename")

directoryinput_cnt=$(printf "%sn" "$directoryinput" | wc -l)

if [ "$directoryinput_cnt" -eq 0 ]; then

   echo "Input file does not exist. Please create a '$inputfilename' file" >&2

   exit 1

elif [ "$directoryinput_cnt" -gt 1 ]; then

   echo "Multiple file named '$inputfilename' exists in the current path" >&2

   exit 1

fi



directoryoutput=$(find . -name "$outputfilename")

directoryoutput_cnt=$(printf "%sn" "$directoryoutput" | wc -l)



if [ "$directoryoutput_cnt" -eq 0 ]; then 

    echo "Input file does not exist. Please create a '$outputfilename' file" >&2

    exit 1

elif [ "$directoryoutput_cnt" -gt 1 ]; then 

   echo "Multiple file named '$outputfilename' exists in the current path" >&2

    exit 1

fi



cat "$directoryoutput"

n=$(<"$directoryoutput" wc -l)

edited Nov 20 '18 at 15:17

answered Nov 19 '18 at 20:21

Kamil Cuk

9,2501524

printf "$directoryinput" would be better written as printf '%sn' "$directoryinput", so you aren't treating your filename (with arbitrary contents) as a format string. (Also, a trailing newline is needed to properly terminate a line of content on UNIX; POSIX explicitly directs wc -l to count the number of newlines, so content with no trailing newline is ignored).
– Charles Duffy
Nov 19 '18 at 20:32

Though the file exists, the code doesn't believe it does and returns the "Input file does not exist. Please create a '$inputfilename' file". I tried also putting dollar signs on the initialisation of the inputfilename and outputfilename variables. The rest of it looks good though, thanks!
– Sam Thompson
Nov 20 '18 at 10:51

@SamThompson you are right I made a typo. When counting the number of results I am missing one trailing newline, so we need to do like as Charles suggested, printf "%sn" "$...." | wc -l. Without the n it counts zero, even when one file exists :/ I forgot that command substitution $(...) removes trailing newlines. I've edited the answer. Alternatively with bash you can $(<<<"$..." wc -l) which will result in the same.
– Kamil Cuk
Nov 20 '18 at 15:17

add a comment |

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1 Answer
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1 Answer
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oldest

votes

Reading in File line by line w/ Bash

The best and idiomatic way to read file line by line is to do:

while IFS= read -r line; do

  // parse line

  printf "%s" "$line"

done < "file"

More on this topic can be found on bashfaq

done < "directoryoutput"

The code is not working, because you are passing to your while read loop as input to standard input the content of a file named directoryoutput. As such a file does not exists, your script fails.

directoryoutput=$(find -name $outputfilename)

One can simply append the variable value with newline appended to a read while loop using a HERE-string construction:

done <<< "$directoryoutput"

directoryinput=$(find -name $inputfilename)
if [[ -f "$directoryinput" ]]

while read line;

   do

      echo "$line"

      n=$((n+1))

  done < "directoryoutput"

The intention is pretty clear here. This is just:

 n=$(<directoryoutput wc -l)

 cat "directoryoutput"

Except that while read line removed trailing and leading newlines and is IFS dependent.

Also always remember to quote your variables unless you have a reason not to.

Have a look at shellcheck which can find most common mistakes in scripts.

I would do it more like this:

inputfilename="employees.txt"

outputfilename="branch.txt"



directoryinput=$(find . -name "$inputfilename")

directoryinput_cnt=$(printf "%sn" "$directoryinput" | wc -l)

if [ "$directoryinput_cnt" -eq 0 ]; then

   echo "Input file does not exist. Please create a '$inputfilename' file" >&2

   exit 1

elif [ "$directoryinput_cnt" -gt 1 ]; then

   echo "Multiple file named '$inputfilename' exists in the current path" >&2

   exit 1

fi



directoryoutput=$(find . -name "$outputfilename")

directoryoutput_cnt=$(printf "%sn" "$directoryoutput" | wc -l)



if [ "$directoryoutput_cnt" -eq 0 ]; then 

    echo "Input file does not exist. Please create a '$outputfilename' file" >&2

    exit 1

elif [ "$directoryoutput_cnt" -gt 1 ]; then 

   echo "Multiple file named '$outputfilename' exists in the current path" >&2

    exit 1

fi



cat "$directoryoutput"

n=$(<"$directoryoutput" wc -l)

edited Nov 20 '18 at 15:17

answered Nov 19 '18 at 20:21

Kamil Cuk

9,2501524

printf "$directoryinput" would be better written as printf '%sn' "$directoryinput", so you aren't treating your filename (with arbitrary contents) as a format string. (Also, a trailing newline is needed to properly terminate a line of content on UNIX; POSIX explicitly directs wc -l to count the number of newlines, so content with no trailing newline is ignored).
– Charles Duffy
Nov 19 '18 at 20:32

Though the file exists, the code doesn't believe it does and returns the "Input file does not exist. Please create a '$inputfilename' file". I tried also putting dollar signs on the initialisation of the inputfilename and outputfilename variables. The rest of it looks good though, thanks!
– Sam Thompson
Nov 20 '18 at 10:51

@SamThompson you are right I made a typo. When counting the number of results I am missing one trailing newline, so we need to do like as Charles suggested, printf "%sn" "$...." | wc -l. Without the n it counts zero, even when one file exists :/ I forgot that command substitution $(...) removes trailing newlines. I've edited the answer. Alternatively with bash you can $(<<<"$..." wc -l) which will result in the same.
– Kamil Cuk
Nov 20 '18 at 15:17

add a comment |

Reading in File line by line w/ Bash

The best and idiomatic way to read file line by line is to do:

while IFS= read -r line; do

  // parse line

  printf "%s" "$line"

done < "file"

More on this topic can be found on bashfaq

done < "directoryoutput"

The code is not working, because you are passing to your while read loop as input to standard input the content of a file named directoryoutput. As such a file does not exists, your script fails.

directoryoutput=$(find -name $outputfilename)

One can simply append the variable value with newline appended to a read while loop using a HERE-string construction:

done <<< "$directoryoutput"

directoryinput=$(find -name $inputfilename)
if [[ -f "$directoryinput" ]]

while read line;

   do

      echo "$line"

      n=$((n+1))

  done < "directoryoutput"

The intention is pretty clear here. This is just:

 n=$(<directoryoutput wc -l)

 cat "directoryoutput"

Except that while read line removed trailing and leading newlines and is IFS dependent.

Also always remember to quote your variables unless you have a reason not to.

Have a look at shellcheck which can find most common mistakes in scripts.

I would do it more like this:

inputfilename="employees.txt"

outputfilename="branch.txt"



directoryinput=$(find . -name "$inputfilename")

directoryinput_cnt=$(printf "%sn" "$directoryinput" | wc -l)

if [ "$directoryinput_cnt" -eq 0 ]; then

   echo "Input file does not exist. Please create a '$inputfilename' file" >&2

   exit 1

elif [ "$directoryinput_cnt" -gt 1 ]; then

   echo "Multiple file named '$inputfilename' exists in the current path" >&2

   exit 1

fi



directoryoutput=$(find . -name "$outputfilename")

directoryoutput_cnt=$(printf "%sn" "$directoryoutput" | wc -l)



if [ "$directoryoutput_cnt" -eq 0 ]; then 

    echo "Input file does not exist. Please create a '$outputfilename' file" >&2

    exit 1

elif [ "$directoryoutput_cnt" -gt 1 ]; then 

   echo "Multiple file named '$outputfilename' exists in the current path" >&2

    exit 1

fi



cat "$directoryoutput"

n=$(<"$directoryoutput" wc -l)

edited Nov 20 '18 at 15:17

answered Nov 19 '18 at 20:21

Kamil Cuk

9,2501524

printf "$directoryinput" would be better written as printf '%sn' "$directoryinput", so you aren't treating your filename (with arbitrary contents) as a format string. (Also, a trailing newline is needed to properly terminate a line of content on UNIX; POSIX explicitly directs wc -l to count the number of newlines, so content with no trailing newline is ignored).
– Charles Duffy
Nov 19 '18 at 20:32

Though the file exists, the code doesn't believe it does and returns the "Input file does not exist. Please create a '$inputfilename' file". I tried also putting dollar signs on the initialisation of the inputfilename and outputfilename variables. The rest of it looks good though, thanks!
– Sam Thompson
Nov 20 '18 at 10:51

@SamThompson you are right I made a typo. When counting the number of results I am missing one trailing newline, so we need to do like as Charles suggested, printf "%sn" "$...." | wc -l. Without the n it counts zero, even when one file exists :/ I forgot that command substitution $(...) removes trailing newlines. I've edited the answer. Alternatively with bash you can $(<<<"$..." wc -l) which will result in the same.
– Kamil Cuk
Nov 20 '18 at 15:17

add a comment |

Reading in File line by line w/ Bash

The best and idiomatic way to read file line by line is to do:

while IFS= read -r line; do

  // parse line

  printf "%s" "$line"

done < "file"

More on this topic can be found on bashfaq

done < "directoryoutput"

The code is not working, because you are passing to your while read loop as input to standard input the content of a file named directoryoutput. As such a file does not exists, your script fails.

directoryoutput=$(find -name $outputfilename)

One can simply append the variable value with newline appended to a read while loop using a HERE-string construction:

done <<< "$directoryoutput"

directoryinput=$(find -name $inputfilename)
if [[ -f "$directoryinput" ]]

while read line;

   do

      echo "$line"

      n=$((n+1))

  done < "directoryoutput"

The intention is pretty clear here. This is just:

 n=$(<directoryoutput wc -l)

 cat "directoryoutput"

Except that while read line removed trailing and leading newlines and is IFS dependent.

Also always remember to quote your variables unless you have a reason not to.

Have a look at shellcheck which can find most common mistakes in scripts.

I would do it more like this:

inputfilename="employees.txt"

outputfilename="branch.txt"



directoryinput=$(find . -name "$inputfilename")

directoryinput_cnt=$(printf "%sn" "$directoryinput" | wc -l)

if [ "$directoryinput_cnt" -eq 0 ]; then

   echo "Input file does not exist. Please create a '$inputfilename' file" >&2

   exit 1

elif [ "$directoryinput_cnt" -gt 1 ]; then

   echo "Multiple file named '$inputfilename' exists in the current path" >&2

   exit 1

fi



directoryoutput=$(find . -name "$outputfilename")

directoryoutput_cnt=$(printf "%sn" "$directoryoutput" | wc -l)



if [ "$directoryoutput_cnt" -eq 0 ]; then 

    echo "Input file does not exist. Please create a '$outputfilename' file" >&2

    exit 1

elif [ "$directoryoutput_cnt" -gt 1 ]; then 

   echo "Multiple file named '$outputfilename' exists in the current path" >&2

    exit 1

fi



cat "$directoryoutput"

n=$(<"$directoryoutput" wc -l)

edited Nov 20 '18 at 15:17

answered Nov 19 '18 at 20:21

Kamil Cuk

9,2501524

Reading in File line by line w/ Bash

The best and idiomatic way to read file line by line is to do:

while IFS= read -r line; do

  // parse line

  printf "%s" "$line"

done < "file"

More on this topic can be found on bashfaq

done < "directoryoutput"

The code is not working, because you are passing to your while read loop as input to standard input the content of a file named directoryoutput. As such a file does not exists, your script fails.

directoryoutput=$(find -name $outputfilename)

One can simply append the variable value with newline appended to a read while loop using a HERE-string construction:

done <<< "$directoryoutput"

directoryinput=$(find -name $inputfilename)
if [[ -f "$directoryinput" ]]

while read line;

   do

      echo "$line"

      n=$((n+1))

  done < "directoryoutput"

The intention is pretty clear here. This is just:

 n=$(<directoryoutput wc -l)

 cat "directoryoutput"

Except that while read line removed trailing and leading newlines and is IFS dependent.

Also always remember to quote your variables unless you have a reason not to.

Have a look at shellcheck which can find most common mistakes in scripts.

I would do it more like this:

inputfilename="employees.txt"

outputfilename="branch.txt"



directoryinput=$(find . -name "$inputfilename")

directoryinput_cnt=$(printf "%sn" "$directoryinput" | wc -l)

if [ "$directoryinput_cnt" -eq 0 ]; then

   echo "Input file does not exist. Please create a '$inputfilename' file" >&2

   exit 1

elif [ "$directoryinput_cnt" -gt 1 ]; then

   echo "Multiple file named '$inputfilename' exists in the current path" >&2

   exit 1

fi



directoryoutput=$(find . -name "$outputfilename")

directoryoutput_cnt=$(printf "%sn" "$directoryoutput" | wc -l)



if [ "$directoryoutput_cnt" -eq 0 ]; then 

    echo "Input file does not exist. Please create a '$outputfilename' file" >&2

    exit 1

elif [ "$directoryoutput_cnt" -gt 1 ]; then 

   echo "Multiple file named '$outputfilename' exists in the current path" >&2

    exit 1

fi



cat "$directoryoutput"

n=$(<"$directoryoutput" wc -l)

edited Nov 20 '18 at 15:17

answered Nov 19 '18 at 20:21

Kamil Cuk

9,2501524

edited Nov 20 '18 at 15:17

answered Nov 19 '18 at 20:21

Kamil Cuk

9,2501524

answered Nov 19 '18 at 20:21

Kamil Cuk

9,2501524

answered Nov 19 '18 at 20:21

Kamil Cuk

9,2501524

printf "$directoryinput" would be better written as printf '%sn' "$directoryinput", so you aren't treating your filename (with arbitrary contents) as a format string. (Also, a trailing newline is needed to properly terminate a line of content on UNIX; POSIX explicitly directs wc -l to count the number of newlines, so content with no trailing newline is ignored).
– Charles Duffy
Nov 19 '18 at 20:32

Though the file exists, the code doesn't believe it does and returns the "Input file does not exist. Please create a '$inputfilename' file". I tried also putting dollar signs on the initialisation of the inputfilename and outputfilename variables. The rest of it looks good though, thanks!
– Sam Thompson
Nov 20 '18 at 10:51

@SamThompson you are right I made a typo. When counting the number of results I am missing one trailing newline, so we need to do like as Charles suggested, printf "%sn" "$...." | wc -l. Without the n it counts zero, even when one file exists :/ I forgot that command substitution $(...) removes trailing newlines. I've edited the answer. Alternatively with bash you can $(<<<"$..." wc -l) which will result in the same.
– Kamil Cuk
Nov 20 '18 at 15:17

add a comment |

printf "$directoryinput" would be better written as printf '%sn' "$directoryinput", so you aren't treating your filename (with arbitrary contents) as a format string. (Also, a trailing newline is needed to properly terminate a line of content on UNIX; POSIX explicitly directs wc -l to count the number of newlines, so content with no trailing newline is ignored).
– Charles Duffy
Nov 19 '18 at 20:32

Though the file exists, the code doesn't believe it does and returns the "Input file does not exist. Please create a '$inputfilename' file". I tried also putting dollar signs on the initialisation of the inputfilename and outputfilename variables. The rest of it looks good though, thanks!
– Sam Thompson
Nov 20 '18 at 10:51

@SamThompson you are right I made a typo. When counting the number of results I am missing one trailing newline, so we need to do like as Charles suggested, printf "%sn" "$...." | wc -l. Without the n it counts zero, even when one file exists :/ I forgot that command substitution $(...) removes trailing newlines. I've edited the answer. Alternatively with bash you can $(<<<"$..." wc -l) which will result in the same.
– Kamil Cuk
Nov 20 '18 at 15:17

printf "$directoryinput" would be better written as printf '%sn' "$directoryinput", so you aren't treating your filename (with arbitrary contents) as a format string. (Also, a trailing newline is needed to properly terminate a line of content on UNIX; POSIX explicitly directs wc -l to count the number of newlines, so content with no trailing newline is ignored).
– Charles Duffy
Nov 19 '18 at 20:32

Though the file exists, the code doesn't believe it does and returns the "Input file does not exist. Please create a '$inputfilename' file". I tried also putting dollar signs on the initialisation of the inputfilename and outputfilename variables. The rest of it looks good though, thanks!
– Sam Thompson
Nov 20 '18 at 10:51

@SamThompson you are right I made a typo. When counting the number of results I am missing one trailing newline, so we need to do like as Charles suggested, printf "%sn" "$...." | wc -l. Without the n it counts zero, even when one file exists :/ I forgot that command substitution $(...) removes trailing newlines. I've edited the answer. Alternatively with bash you can $(<<<"$..." wc -l) which will result in the same.
– Kamil Cuk
Nov 20 '18 at 15:17

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