“let” command is not found in my bash script

I have the following bash script,

#!/bin/bash

echo $SHELL



i=1

for img in ../img/*.png; do

  echo $img

  new=$(printf "../img/tmp/pyr%d.png" "$i")

  echo $new

  # cp "$img" "$new"

  let i=i+1

done

# avconv -r 2  -i ../img/tmp/pyr%d.png -b:v 1000k ../pop_pyr.mp4

In the ../img folder, there are files pyr1920.png, pyr1925.png, ...

When I run this script, I get this error

../img/pyr1920.png

../img/tmp/pyr1.png

make_video.sh: 10: make_video.sh: let: not found

../img/pyr1925.png

../img/tmp/pyr1.png

But I pasted the script in a terminal, it works perfectly fine.

../img/pyr1920.png

../img/tmp/pyr1.png

../img/pyr1925.png

../img/tmp/pyr2.png

My environment it ubuntu 14.04. I understand that default /bin/sh is dash. I make sure to use bash and check by echo $SHELL.

Why it doesn't work as a script but works fine from a terminal? I tried to find a solution but all I can find is about "dash" and "bash" setting.

edited Feb 22 '17 at 13:52

asked Feb 22 '17 at 13:41

user26767

4626

That's not a bash error message; you aren't running your script with bash.

– chepner
Feb 22 '17 at 13:52

Thank you for the info, I will check and try them out.

– user26767
Feb 22 '17 at 13:54

@chepner, Yes, now I realized it works with "bash myscript.sh". Doesn't it run bash if I write "#!/bin/bash" on the top of the script?

– user26767
Feb 22 '17 at 13:57

1

It should; how are you running it? Incidentally, $SHELL does not contain the name of the current shell; it's the name of your default shell.

– chepner
Feb 22 '17 at 14:00

Even my default shell is /bin/bash, I checked just open a terminal and type $SHELL. I am confused.

– user26767
Feb 22 '17 at 14:07

|
show 1 more comment

I have the following bash script,

#!/bin/bash

echo $SHELL



i=1

for img in ../img/*.png; do

  echo $img

  new=$(printf "../img/tmp/pyr%d.png" "$i")

  echo $new

  # cp "$img" "$new"

  let i=i+1

done

# avconv -r 2  -i ../img/tmp/pyr%d.png -b:v 1000k ../pop_pyr.mp4

In the ../img folder, there are files pyr1920.png, pyr1925.png, ...

When I run this script, I get this error

../img/pyr1920.png

../img/tmp/pyr1.png

make_video.sh: 10: make_video.sh: let: not found

../img/pyr1925.png

../img/tmp/pyr1.png

But I pasted the script in a terminal, it works perfectly fine.

../img/pyr1920.png

../img/tmp/pyr1.png

../img/pyr1925.png

../img/tmp/pyr2.png

My environment it ubuntu 14.04. I understand that default /bin/sh is dash. I make sure to use bash and check by echo $SHELL.

Why it doesn't work as a script but works fine from a terminal? I tried to find a solution but all I can find is about "dash" and "bash" setting.

edited Feb 22 '17 at 13:52

asked Feb 22 '17 at 13:41

user26767

4626

That's not a bash error message; you aren't running your script with bash.

– chepner
Feb 22 '17 at 13:52

Thank you for the info, I will check and try them out.

– user26767
Feb 22 '17 at 13:54

@chepner, Yes, now I realized it works with "bash myscript.sh". Doesn't it run bash if I write "#!/bin/bash" on the top of the script?

– user26767
Feb 22 '17 at 13:57

1

It should; how are you running it? Incidentally, $SHELL does not contain the name of the current shell; it's the name of your default shell.

– chepner
Feb 22 '17 at 14:00

Even my default shell is /bin/bash, I checked just open a terminal and type $SHELL. I am confused.

– user26767
Feb 22 '17 at 14:07

|
show 1 more comment

I have the following bash script,

#!/bin/bash

echo $SHELL



i=1

for img in ../img/*.png; do

  echo $img

  new=$(printf "../img/tmp/pyr%d.png" "$i")

  echo $new

  # cp "$img" "$new"

  let i=i+1

done

# avconv -r 2  -i ../img/tmp/pyr%d.png -b:v 1000k ../pop_pyr.mp4

In the ../img folder, there are files pyr1920.png, pyr1925.png, ...

When I run this script, I get this error

../img/pyr1920.png

../img/tmp/pyr1.png

make_video.sh: 10: make_video.sh: let: not found

../img/pyr1925.png

../img/tmp/pyr1.png

But I pasted the script in a terminal, it works perfectly fine.

../img/pyr1920.png

../img/tmp/pyr1.png

../img/pyr1925.png

../img/tmp/pyr2.png

My environment it ubuntu 14.04. I understand that default /bin/sh is dash. I make sure to use bash and check by echo $SHELL.

Why it doesn't work as a script but works fine from a terminal? I tried to find a solution but all I can find is about "dash" and "bash" setting.

edited Feb 22 '17 at 13:52

asked Feb 22 '17 at 13:41

user26767

4626

I have the following bash script,

#!/bin/bash

echo $SHELL



i=1

for img in ../img/*.png; do

  echo $img

  new=$(printf "../img/tmp/pyr%d.png" "$i")

  echo $new

  # cp "$img" "$new"

  let i=i+1

done

# avconv -r 2  -i ../img/tmp/pyr%d.png -b:v 1000k ../pop_pyr.mp4

In the ../img folder, there are files pyr1920.png, pyr1925.png, ...

When I run this script, I get this error

../img/pyr1920.png

../img/tmp/pyr1.png

make_video.sh: 10: make_video.sh: let: not found

../img/pyr1925.png

../img/tmp/pyr1.png

But I pasted the script in a terminal, it works perfectly fine.

../img/pyr1920.png

../img/tmp/pyr1.png

../img/pyr1925.png

../img/tmp/pyr2.png

My environment it ubuntu 14.04. I understand that default /bin/sh is dash. I make sure to use bash and check by echo $SHELL.

Why it doesn't work as a script but works fine from a terminal? I tried to find a solution but all I can find is about "dash" and "bash" setting.

bash

edited Feb 22 '17 at 13:52

asked Feb 22 '17 at 13:41

user26767

4626

edited Feb 22 '17 at 13:52

asked Feb 22 '17 at 13:41

user26767

4626

edited Feb 22 '17 at 13:52

asked Feb 22 '17 at 13:41

user26767

4626

asked Feb 22 '17 at 13:41

user26767

4626

asked Feb 22 '17 at 13:41

user26767

4626

That's not a bash error message; you aren't running your script with bash.

– chepner
Feb 22 '17 at 13:52

Thank you for the info, I will check and try them out.

– user26767
Feb 22 '17 at 13:54

@chepner, Yes, now I realized it works with "bash myscript.sh". Doesn't it run bash if I write "#!/bin/bash" on the top of the script?

– user26767
Feb 22 '17 at 13:57

1

It should; how are you running it? Incidentally, $SHELL does not contain the name of the current shell; it's the name of your default shell.

– chepner
Feb 22 '17 at 14:00

Even my default shell is /bin/bash, I checked just open a terminal and type $SHELL. I am confused.

– user26767
Feb 22 '17 at 14:07

|
show 1 more comment

That's not a bash error message; you aren't running your script with bash.

– chepner
Feb 22 '17 at 13:52

Thank you for the info, I will check and try them out.

– user26767
Feb 22 '17 at 13:54

@chepner, Yes, now I realized it works with "bash myscript.sh". Doesn't it run bash if I write "#!/bin/bash" on the top of the script?

– user26767
Feb 22 '17 at 13:57

1

It should; how are you running it? Incidentally, $SHELL does not contain the name of the current shell; it's the name of your default shell.

– chepner
Feb 22 '17 at 14:00

Even my default shell is /bin/bash, I checked just open a terminal and type $SHELL. I am confused.

– user26767
Feb 22 '17 at 14:07

That's not a bash error message; you aren't running your script with bash.

– chepner
Feb 22 '17 at 13:52

Thank you for the info, I will check and try them out.

– user26767
Feb 22 '17 at 13:54

@chepner, Yes, now I realized it works with "bash myscript.sh". Doesn't it run bash if I write "#!/bin/bash" on the top of the script?

– user26767
Feb 22 '17 at 13:57

It should; how are you running it? Incidentally, $SHELL does not contain the name of the current shell; it's the name of your default shell.

– chepner
Feb 22 '17 at 14:00

Even my default shell is /bin/bash, I checked just open a terminal and type $SHELL. I am confused.

– user26767
Feb 22 '17 at 14:07

|
show 1 more comment

3 Answers
3

active

oldest

votes

do in this way:

i=$((i+1))

i=$(expr $i + 1)

answered Feb 22 '17 at 13:45

Kent

144k25153214

1

If $((...)) doesn't work, there's no reason to believe expr is available either.

– chepner
Feb 22 '17 at 13:51

add a comment |

I am sorry for answering my question.
I was running my script with sh script.sh not bash script.sh. bash script.sh worked fine.

But isn't that I can specify it by #!/bin/bash on the top of a script?

answered Feb 22 '17 at 14:03

user26767

4626

2

If you run the script by passing its name as an argument to a specific interpreter (like sh), the shebang is ignored. It is only used when you make the script executable and run it directly (chmod +x script.sh; ./script.sh)

– chepner
Feb 22 '17 at 14:04

1

Thank you very much for the comment! It explained clearly and saved me a lot of time. I was stucked in this for long time.

– user26767
Feb 22 '17 at 14:10

@user26767: please accept an answer to prevent others spending time on this.

– cdarke
Feb 22 '17 at 15:31

add a comment |

I also had the same error, checking I found that with this code you get what you need

i=`expr $i + 1`

answered Nov 19 '18 at 23:28

jhonxito

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

do in this way:

i=$((i+1))

i=$(expr $i + 1)

answered Feb 22 '17 at 13:45

Kent

144k25153214

1

If $((...)) doesn't work, there's no reason to believe expr is available either.

– chepner
Feb 22 '17 at 13:51

add a comment |

do in this way:

i=$((i+1))

i=$(expr $i + 1)

answered Feb 22 '17 at 13:45

Kent

144k25153214

1

If $((...)) doesn't work, there's no reason to believe expr is available either.

– chepner
Feb 22 '17 at 13:51

add a comment |

do in this way:

i=$((i+1))

i=$(expr $i + 1)

answered Feb 22 '17 at 13:45

Kent

144k25153214

do in this way:

i=$((i+1))

i=$(expr $i + 1)

answered Feb 22 '17 at 13:45

Kent

144k25153214

answered Feb 22 '17 at 13:45

Kent

144k25153214

answered Feb 22 '17 at 13:45

Kent

144k25153214

answered Feb 22 '17 at 13:45

Kent

144k25153214

1

If $((...)) doesn't work, there's no reason to believe expr is available either.

– chepner
Feb 22 '17 at 13:51

add a comment |

1

If $((...)) doesn't work, there's no reason to believe expr is available either.

– chepner
Feb 22 '17 at 13:51

If $((...)) doesn't work, there's no reason to believe expr is available either.

– chepner
Feb 22 '17 at 13:51

add a comment |

I am sorry for answering my question.
I was running my script with sh script.sh not bash script.sh. bash script.sh worked fine.

But isn't that I can specify it by #!/bin/bash on the top of a script?

answered Feb 22 '17 at 14:03

user26767

4626

2

If you run the script by passing its name as an argument to a specific interpreter (like sh), the shebang is ignored. It is only used when you make the script executable and run it directly (chmod +x script.sh; ./script.sh)

– chepner
Feb 22 '17 at 14:04

1

Thank you very much for the comment! It explained clearly and saved me a lot of time. I was stucked in this for long time.

– user26767
Feb 22 '17 at 14:10

@user26767: please accept an answer to prevent others spending time on this.

– cdarke
Feb 22 '17 at 15:31

add a comment |

I am sorry for answering my question.
I was running my script with sh script.sh not bash script.sh. bash script.sh worked fine.

But isn't that I can specify it by #!/bin/bash on the top of a script?

answered Feb 22 '17 at 14:03

user26767

4626

2

If you run the script by passing its name as an argument to a specific interpreter (like sh), the shebang is ignored. It is only used when you make the script executable and run it directly (chmod +x script.sh; ./script.sh)

– chepner
Feb 22 '17 at 14:04

1

Thank you very much for the comment! It explained clearly and saved me a lot of time. I was stucked in this for long time.

– user26767
Feb 22 '17 at 14:10

@user26767: please accept an answer to prevent others spending time on this.

– cdarke
Feb 22 '17 at 15:31

add a comment |

I am sorry for answering my question.
I was running my script with sh script.sh not bash script.sh. bash script.sh worked fine.

But isn't that I can specify it by #!/bin/bash on the top of a script?

answered Feb 22 '17 at 14:03

user26767

4626

I am sorry for answering my question.
I was running my script with sh script.sh not bash script.sh. bash script.sh worked fine.

But isn't that I can specify it by #!/bin/bash on the top of a script?

answered Feb 22 '17 at 14:03

user26767

4626

answered Feb 22 '17 at 14:03

user26767

4626

answered Feb 22 '17 at 14:03

user26767

4626

answered Feb 22 '17 at 14:03

user26767

4626

2

If you run the script by passing its name as an argument to a specific interpreter (like sh), the shebang is ignored. It is only used when you make the script executable and run it directly (chmod +x script.sh; ./script.sh)

– chepner
Feb 22 '17 at 14:04

1

Thank you very much for the comment! It explained clearly and saved me a lot of time. I was stucked in this for long time.

– user26767
Feb 22 '17 at 14:10

@user26767: please accept an answer to prevent others spending time on this.

– cdarke
Feb 22 '17 at 15:31

add a comment |

2

If you run the script by passing its name as an argument to a specific interpreter (like sh), the shebang is ignored. It is only used when you make the script executable and run it directly (chmod +x script.sh; ./script.sh)

– chepner
Feb 22 '17 at 14:04

1

Thank you very much for the comment! It explained clearly and saved me a lot of time. I was stucked in this for long time.

– user26767
Feb 22 '17 at 14:10

@user26767: please accept an answer to prevent others spending time on this.

– cdarke
Feb 22 '17 at 15:31

If you run the script by passing its name as an argument to a specific interpreter (like sh), the shebang is ignored. It is only used when you make the script executable and run it directly (chmod +x script.sh; ./script.sh)

– chepner
Feb 22 '17 at 14:04

Thank you very much for the comment! It explained clearly and saved me a lot of time. I was stucked in this for long time.

– user26767
Feb 22 '17 at 14:10

@user26767: please accept an answer to prevent others spending time on this.

– cdarke
Feb 22 '17 at 15:31

add a comment |

I also had the same error, checking I found that with this code you get what you need

i=`expr $i + 1`

answered Nov 19 '18 at 23:28

jhonxito

add a comment |

I also had the same error, checking I found that with this code you get what you need

i=`expr $i + 1`

answered Nov 19 '18 at 23:28

jhonxito

add a comment |

I also had the same error, checking I found that with this code you get what you need

i=`expr $i + 1`

answered Nov 19 '18 at 23:28

jhonxito

I also had the same error, checking I found that with this code you get what you need

i=`expr $i + 1`

answered Nov 19 '18 at 23:28

jhonxito

answered Nov 19 '18 at 23:28

jhonxito

answered Nov 19 '18 at 23:28

jhonxito

answered Nov 19 '18 at 23:28

jhonxito

add a comment |

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