How do I find a file whose name has five digits, in a row, somewhere in the file name?

Normally, I would ask a professor or classmate this, but as it's a Saturday and everyone is gone. I have to find a file in a directory that has five digits in a row somewhere in the file name. They don't have to be ascending or descending digits.

edited Jan 19 at 16:03

Jeff Schaller

42.9k1159136

asked Jan 19 at 15:58

Irish Jet

1

Related: unix.stackexchange.com/questions/134684/…

– Jeff Schaller
Jan 19 at 16:06

Also: unix.stackexchange.com/questions/234201/…

– Jeff Schaller
Jan 19 at 16:06

This command line should do it. find . -name "*[0-9][0-9][0-9][0-9][0-9]*". It can probably be simplified. If there must be exactly 5 digits, that expression will be more complicated.

– sudodus
Jan 19 at 16:08

See also unix.stackexchange.com/q/414226/117549

– Jeff Schaller
Jan 19 at 16:18

If you want to search subdirectories as well: find -E directory/ -regex '.*[0-9]{5,}.*'

– Nick Garvey
Jan 19 at 17:46

|
show 1 more comment

edited Jan 19 at 16:03

Jeff Schaller

42.9k1159136

asked Jan 19 at 15:58

Irish Jet

1

Related: unix.stackexchange.com/questions/134684/…

– Jeff Schaller
Jan 19 at 16:06

Also: unix.stackexchange.com/questions/234201/…

– Jeff Schaller
Jan 19 at 16:06

This command line should do it. find . -name "*[0-9][0-9][0-9][0-9][0-9]*". It can probably be simplified. If there must be exactly 5 digits, that expression will be more complicated.

– sudodus
Jan 19 at 16:08

See also unix.stackexchange.com/q/414226/117549

– Jeff Schaller
Jan 19 at 16:18

If you want to search subdirectories as well: find -E directory/ -regex '.*[0-9]{5,}.*'

– Nick Garvey
Jan 19 at 17:46

|
show 1 more comment

edited Jan 19 at 16:03

Jeff Schaller

42.9k1159136

asked Jan 19 at 15:58

Irish Jet

linux command-line filenames

edited Jan 19 at 16:03

Jeff Schaller

42.9k1159136

asked Jan 19 at 15:58

Irish Jet

edited Jan 19 at 16:03

Jeff Schaller

42.9k1159136

asked Jan 19 at 15:58

Irish Jet

edited Jan 19 at 16:03

Jeff Schaller

42.9k1159136

edited Jan 19 at 16:03

Jeff Schaller

42.9k1159136

edited Jan 19 at 16:03

Jeff Schaller

42.9k1159136

asked Jan 19 at 15:58

Irish Jet

asked Jan 19 at 15:58

Irish Jet

asked Jan 19 at 15:58

Irish Jet

1

Related: unix.stackexchange.com/questions/134684/…

– Jeff Schaller
Jan 19 at 16:06

Also: unix.stackexchange.com/questions/234201/…

– Jeff Schaller
Jan 19 at 16:06

This command line should do it. find . -name "*[0-9][0-9][0-9][0-9][0-9]*". It can probably be simplified. If there must be exactly 5 digits, that expression will be more complicated.

– sudodus
Jan 19 at 16:08

See also unix.stackexchange.com/q/414226/117549

– Jeff Schaller
Jan 19 at 16:18

If you want to search subdirectories as well: find -E directory/ -regex '.*[0-9]{5,}.*'

– Nick Garvey
Jan 19 at 17:46

|
show 1 more comment

1

Related: unix.stackexchange.com/questions/134684/…

– Jeff Schaller
Jan 19 at 16:06

Also: unix.stackexchange.com/questions/234201/…

– Jeff Schaller
Jan 19 at 16:06

This command line should do it. find . -name "*[0-9][0-9][0-9][0-9][0-9]*". It can probably be simplified. If there must be exactly 5 digits, that expression will be more complicated.

– sudodus
Jan 19 at 16:08

See also unix.stackexchange.com/q/414226/117549

– Jeff Schaller
Jan 19 at 16:18

If you want to search subdirectories as well: find -E directory/ -regex '.*[0-9]{5,}.*'

– Nick Garvey
Jan 19 at 17:46

Related: unix.stackexchange.com/questions/134684/…

– Jeff Schaller
Jan 19 at 16:06

Also: unix.stackexchange.com/questions/234201/…

– Jeff Schaller
Jan 19 at 16:06

This command line should do it. find . -name "*[0-9][0-9][0-9][0-9][0-9]*". It can probably be simplified. If there must be exactly 5 digits, that expression will be more complicated.

– sudodus
Jan 19 at 16:08

See also unix.stackexchange.com/q/414226/117549

– Jeff Schaller
Jan 19 at 16:18

If you want to search subdirectories as well: find -E directory/ -regex '.*[0-9]{5,}.*'

– Nick Garvey
Jan 19 at 17:46

|
show 1 more comment

2 Answers
2

active

oldest

votes

Wildcards (or globs) can accomplish this, with a numeric range:

ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

This tells the shell to look in /path/to/directory for filenames that start with:

* -- anything (or nothing)

[0-9] -- a digit

(four more digits)

* -- and ending in anything (or nothing)

That list of filenames is then passed to ls to list them.

More expansively, bash also allows character classes as wildcards, so if you have numbers in your language that aren't covered by [0-9], you could use:

ls -d *[[:digit:]][[:digit:]][[:digit:]][[:digit:]][[:digit:]]*

edited Jan 19 at 17:10

answered Jan 19 at 16:12

Jeff Schaller

42.9k1159136

1

Maybe add -d to avoid listing the contents of directories that match, i.e. ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

– bitinerant
Jan 19 at 17:06

add a comment |

-1

I have achieved by below method and i worked fine

find . -maxdepth 1  -type f| sed "s/.///g"| awk -F "." '{print $1}'|sed '/^$/d'| awk '/[0-9]/{print $0}'| awk '{print $1,gsub("[0-9]",$1)}'| awk '$2 == 5 {print $1}'

answered Jan 20 at 8:58

Praveen Kumar BS

1,5041310

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Wildcards (or globs) can accomplish this, with a numeric range:

ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

This tells the shell to look in /path/to/directory for filenames that start with:

* -- anything (or nothing)

[0-9] -- a digit

(four more digits)

* -- and ending in anything (or nothing)

That list of filenames is then passed to ls to list them.

More expansively, bash also allows character classes as wildcards, so if you have numbers in your language that aren't covered by [0-9], you could use:

ls -d *[[:digit:]][[:digit:]][[:digit:]][[:digit:]][[:digit:]]*

edited Jan 19 at 17:10

answered Jan 19 at 16:12

Jeff Schaller

42.9k1159136

1

Maybe add -d to avoid listing the contents of directories that match, i.e. ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

– bitinerant
Jan 19 at 17:06

add a comment |

Wildcards (or globs) can accomplish this, with a numeric range:

ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

This tells the shell to look in /path/to/directory for filenames that start with:

* -- anything (or nothing)

[0-9] -- a digit

(four more digits)

* -- and ending in anything (or nothing)

That list of filenames is then passed to ls to list them.

More expansively, bash also allows character classes as wildcards, so if you have numbers in your language that aren't covered by [0-9], you could use:

ls -d *[[:digit:]][[:digit:]][[:digit:]][[:digit:]][[:digit:]]*

edited Jan 19 at 17:10

answered Jan 19 at 16:12

Jeff Schaller

42.9k1159136

1

Maybe add -d to avoid listing the contents of directories that match, i.e. ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

– bitinerant
Jan 19 at 17:06

add a comment |

Wildcards (or globs) can accomplish this, with a numeric range:

ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

This tells the shell to look in /path/to/directory for filenames that start with:

* -- anything (or nothing)

[0-9] -- a digit

(four more digits)

* -- and ending in anything (or nothing)

That list of filenames is then passed to ls to list them.

More expansively, bash also allows character classes as wildcards, so if you have numbers in your language that aren't covered by [0-9], you could use:

ls -d *[[:digit:]][[:digit:]][[:digit:]][[:digit:]][[:digit:]]*

edited Jan 19 at 17:10

answered Jan 19 at 16:12

Jeff Schaller

42.9k1159136

Wildcards (or globs) can accomplish this, with a numeric range:

ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

This tells the shell to look in /path/to/directory for filenames that start with:

* -- anything (or nothing)

[0-9] -- a digit

(four more digits)

* -- and ending in anything (or nothing)

That list of filenames is then passed to ls to list them.

More expansively, bash also allows character classes as wildcards, so if you have numbers in your language that aren't covered by [0-9], you could use:

ls -d *[[:digit:]][[:digit:]][[:digit:]][[:digit:]][[:digit:]]*

edited Jan 19 at 17:10

answered Jan 19 at 16:12

Jeff Schaller

42.9k1159136

edited Jan 19 at 17:10

answered Jan 19 at 16:12

Jeff Schaller

42.9k1159136

answered Jan 19 at 16:12

Jeff Schaller

42.9k1159136

answered Jan 19 at 16:12

Jeff Schaller

42.9k1159136

1

Maybe add -d to avoid listing the contents of directories that match, i.e. ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

– bitinerant
Jan 19 at 17:06

add a comment |

1

Maybe add -d to avoid listing the contents of directories that match, i.e. ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

– bitinerant
Jan 19 at 17:06

Maybe add -d to avoid listing the contents of directories that match, i.e. ls -d /path/to/directory/*[0-9][0-9][0-9][0-9][0-9]*

– bitinerant
Jan 19 at 17:06

add a comment |

-1

I have achieved by below method and i worked fine

find . -maxdepth 1  -type f| sed "s/.///g"| awk -F "." '{print $1}'|sed '/^$/d'| awk '/[0-9]/{print $0}'| awk '{print $1,gsub("[0-9]",$1)}'| awk '$2 == 5 {print $1}'

answered Jan 20 at 8:58

Praveen Kumar BS

1,5041310

add a comment |

-1

I have achieved by below method and i worked fine

find . -maxdepth 1  -type f| sed "s/.///g"| awk -F "." '{print $1}'|sed '/^$/d'| awk '/[0-9]/{print $0}'| awk '{print $1,gsub("[0-9]",$1)}'| awk '$2 == 5 {print $1}'

answered Jan 20 at 8:58

Praveen Kumar BS

1,5041310

add a comment |

-1

I have achieved by below method and i worked fine

find . -maxdepth 1  -type f| sed "s/.///g"| awk -F "." '{print $1}'|sed '/^$/d'| awk '/[0-9]/{print $0}'| awk '{print $1,gsub("[0-9]",$1)}'| awk '$2 == 5 {print $1}'

answered Jan 20 at 8:58

Praveen Kumar BS

1,5041310

I have achieved by below method and i worked fine

find . -maxdepth 1  -type f| sed "s/.///g"| awk -F "." '{print $1}'|sed '/^$/d'| awk '/[0-9]/{print $0}'| awk '{print $1,gsub("[0-9]",$1)}'| awk '$2 == 5 {print $1}'

answered Jan 20 at 8:58

Praveen Kumar BS

1,5041310

answered Jan 20 at 8:58

Praveen Kumar BS

1,5041310

answered Jan 20 at 8:58

Praveen Kumar BS

1,5041310

answered Jan 20 at 8:58

Praveen Kumar BS

1,5041310

add a comment |

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