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Bash find files and filter by date and size

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0

1

I have a directory with a lot of files in it.
Each day, new files are added automatically.

The filenames are formatted like that :
[GROUP_ID]_[RANDOM_NUMBER].txt
Example : 012_1234.txt

For every day, for every GROUP_ID (032, 024, 044...etc), I want to keep only the biggest file of the day.

So for example, for the two days 27 and 28 march I have :

March 27 - 012_1234.txt - 12ko
March 27 - 012_0243.txt - 3000ko
March 27 - 016_5647.txt - 25ko
March 27 - 024_4354.txt - 20ko
March 27 - 032_8745.txt - 40ko

March 28 - 032_1254.txt - 16ko
March 28 - 036_0456.txt - 30ko
March 28 - 042_7645.txt - 500ko
March 28 - 042_2310.txt - 25ko
March 28 - 042_2125.txt - 34ko
March 28 - 044_4510.txt - 35ko

And I want to have :

March 27 - 012_0243.txt - 3000ko
March 27 - 016_5647.txt - 25ko
March 27 - 024_4354.txt - 20ko
March 27 - 032_8745.txt - 40ko

March 28 - 032_1254.txt - 16ko
March 28 - 036_0456.txt - 30ko
March 28 - 042_7645.txt - 500ko
March 28 - 044_4510.txt - 35ko

I don't find the right bash ls/find command to do that, somebody have an idea ?

With this command, I can display the biggest file for each day.

ls -l *.txt --time-style=+%s |
awk '$6 = int($6/86400); print' |
sort -nk6,6 -nrk5,5 | sort -sunk6,6

But I want the biggest file of each GROUP_ID file of each day.
So, if there is one file for "012" group_id file, of 10ko, I want to display it, even if there is bigger files for others group id...

bash shell find ls

share|improve this question

edited Mar 27 at 14:01

user2178964

asked Mar 27 at 12:34

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Stack Overflow is not a code writing service. What have you searched for before asking, and what did you find? Based on that, what did you try, and how did it fail?
  
  – tripleee
  Mar 27 at 12:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I edit my post. I just want some ideas, I don't ask for a clean and complete answer necessarly. Thanks
  
  – user2178964
  Mar 27 at 13:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And "ko" is your language's abbreviation for kB I guess?
  
  – tripleee
  Mar 27 at 13:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And *Z is actually a placeholder for *.txt or your files are actually named according to that wildcard?
  
  – tripleee
  Mar 27 at 13:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes, ko=kb , and I edit my post to replace *.Z by *.txt
  
  – user2178964
  Mar 27 at 14:02
  
  

  
  
  
  

 | 
show 5 more comments

0

1

I have a directory with a lot of files in it.
Each day, new files are added automatically.

The filenames are formatted like that :
[GROUP_ID]_[RANDOM_NUMBER].txt
Example : 012_1234.txt

For every day, for every GROUP_ID (032, 024, 044...etc), I want to keep only the biggest file of the day.

So for example, for the two days 27 and 28 march I have :

March 27 - 012_1234.txt - 12ko
March 27 - 012_0243.txt - 3000ko
March 27 - 016_5647.txt - 25ko
March 27 - 024_4354.txt - 20ko
March 27 - 032_8745.txt - 40ko

March 28 - 032_1254.txt - 16ko
March 28 - 036_0456.txt - 30ko
March 28 - 042_7645.txt - 500ko
March 28 - 042_2310.txt - 25ko
March 28 - 042_2125.txt - 34ko
March 28 - 044_4510.txt - 35ko

And I want to have :

March 27 - 012_0243.txt - 3000ko
March 27 - 016_5647.txt - 25ko
March 27 - 024_4354.txt - 20ko
March 27 - 032_8745.txt - 40ko

March 28 - 032_1254.txt - 16ko
March 28 - 036_0456.txt - 30ko
March 28 - 042_7645.txt - 500ko
March 28 - 044_4510.txt - 35ko

I don't find the right bash ls/find command to do that, somebody have an idea ?

With this command, I can display the biggest file for each day.

ls -l *.txt --time-style=+%s |
awk '$6 = int($6/86400); print' |
sort -nk6,6 -nrk5,5 | sort -sunk6,6

But I want the biggest file of each GROUP_ID file of each day.
So, if there is one file for "012" group_id file, of 10ko, I want to display it, even if there is bigger files for others group id...

bash shell find ls

share|improve this question

edited Mar 27 at 14:01

user2178964

asked Mar 27 at 12:34

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Stack Overflow is not a code writing service. What have you searched for before asking, and what did you find? Based on that, what did you try, and how did it fail?
  
  – tripleee
  Mar 27 at 12:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I edit my post. I just want some ideas, I don't ask for a clean and complete answer necessarly. Thanks
  
  – user2178964
  Mar 27 at 13:49
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And "ko" is your language's abbreviation for kB I guess?
  
  – tripleee
  Mar 27 at 13:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  And *Z is actually a placeholder for *.txt or your files are actually named according to that wildcard?
  
  – tripleee
  Mar 27 at 13:56
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  yes, ko=kb , and I edit my post to replace *.Z by *.txt
  
  – user2178964
  Mar 27 at 14:02
  
  

  
  
  
  

 | 
show 5 more comments

I have a directory with a lot of files in it.
Each day, new files are added automatically.

The filenames are formatted like that :
[GROUP_ID]_[RANDOM_NUMBER].txt
Example : 012_1234.txt

For every day, for every GROUP_ID (032, 024, 044...etc), I want to keep only the biggest file of the day.

So for example, for the two days 27 and 28 march I have :

March 27 - 012_1234.txt - 12ko
March 27 - 012_0243.txt - 3000ko
March 27 - 016_5647.txt - 25ko
March 27 - 024_4354.txt - 20ko
March 27 - 032_8745.txt - 40ko

March 28 - 032_1254.txt - 16ko
March 28 - 036_0456.txt - 30ko
March 28 - 042_7645.txt - 500ko
March 28 - 042_2310.txt - 25ko
March 28 - 042_2125.txt - 34ko
March 28 - 044_4510.txt - 35ko

And I want to have :

March 27 - 012_0243.txt - 3000ko
March 27 - 016_5647.txt - 25ko
March 27 - 024_4354.txt - 20ko
March 27 - 032_8745.txt - 40ko

March 28 - 032_1254.txt - 16ko
March 28 - 036_0456.txt - 30ko
March 28 - 042_7645.txt - 500ko
March 28 - 044_4510.txt - 35ko

I don't find the right bash ls/find command to do that, somebody have an idea ?

With this command, I can display the biggest file for each day.

ls -l *.txt --time-style=+%s |
awk '$6 = int($6/86400); print' |
sort -nk6,6 -nrk5,5 | sort -sunk6,6

But I want the biggest file of each GROUP_ID file of each day.
So, if there is one file for "012" group_id file, of 10ko, I want to display it, even if there is bigger files for others group id...

bash shell find ls

share|improve this question

edited Mar 27 at 14:01

user2178964

asked Mar 27 at 12:34

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

March 27 - 012_1234.txt - 12ko
March 27 - 012_0243.txt - 3000ko
March 27 - 016_5647.txt - 25ko
March 27 - 024_4354.txt - 20ko
March 27 - 032_8745.txt - 40ko

March 28 - 032_1254.txt - 16ko
March 28 - 036_0456.txt - 30ko
March 28 - 042_7645.txt - 500ko
March 28 - 042_2310.txt - 25ko
March 28 - 042_2125.txt - 34ko
March 28 - 044_4510.txt - 35ko

March 27 - 012_0243.txt - 3000ko
March 27 - 016_5647.txt - 25ko
March 27 - 024_4354.txt - 20ko
March 27 - 032_8745.txt - 40ko

March 28 - 032_1254.txt - 16ko
March 28 - 036_0456.txt - 30ko
March 28 - 042_7645.txt - 500ko
March 28 - 044_4510.txt - 35ko

ls -l *.txt --time-style=+%s |
awk '$6 = int($6/86400); print' |
sort -nk6,6 -nrk5,5 | sort -sunk6,6

share|improve this question

edited Mar 27 at 14:01

user2178964

asked Mar 27 at 12:34

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

share|improve this question

edited Mar 27 at 14:01

user2178964

asked Mar 27 at 12:34

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

asked Mar 27 at 12:34

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

asked Mar 27 at 12:34

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

1 Answer
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oldest

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0

I found myself the solution:

ls -l | tail -n+2 |
awk ' split($0,var,"_"); group_id=var[5]; print $0" "group_id ' |
sort -k9,9 -k5,5nr |
awk '$10 != x print x = $10 '

This gives me the biggest file for each group_id, so now I just add to handle the day part.

For information:

tail -n+2: hide the "total" part of the ls command's output

First awk: get the group_id part (012, 036...) and display it after the original line ($0)

Sort: sort on filename and size

Take the biggest size of each group_id (column 10 added by awk at beginning)

share|improve this answer

edited Mar 28 at 5:58

tripleee

102k1515 gold badges150150 silver badges204204 bronze badges

answered Mar 27 at 15:45

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

add a comment | 

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0

I found myself the solution:

ls -l | tail -n+2 |
awk ' split($0,var,"_"); group_id=var[5]; print $0" "group_id ' |
sort -k9,9 -k5,5nr |
awk '$10 != x print x = $10 '

This gives me the biggest file for each group_id, so now I just add to handle the day part.

For information:

tail -n+2: hide the "total" part of the ls command's output

First awk: get the group_id part (012, 036...) and display it after the original line ($0)

Sort: sort on filename and size

Take the biggest size of each group_id (column 10 added by awk at beginning)

share|improve this answer

edited Mar 28 at 5:58

tripleee

102k1515 gold badges150150 silver badges204204 bronze badges

answered Mar 27 at 15:45

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

add a comment | 

0

I found myself the solution:

ls -l | tail -n+2 |
awk ' split($0,var,"_"); group_id=var[5]; print $0" "group_id ' |
sort -k9,9 -k5,5nr |
awk '$10 != x print x = $10 '

This gives me the biggest file for each group_id, so now I just add to handle the day part.

For information:

tail -n+2: hide the "total" part of the ls command's output

First awk: get the group_id part (012, 036...) and display it after the original line ($0)

Sort: sort on filename and size

Take the biggest size of each group_id (column 10 added by awk at beginning)

share|improve this answer

edited Mar 28 at 5:58

tripleee

102k1515 gold badges150150 silver badges204204 bronze badges

answered Mar 27 at 15:45

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

add a comment | 

I found myself the solution:

ls -l | tail -n+2 |
awk ' split($0,var,"_"); group_id=var[5]; print $0" "group_id ' |
sort -k9,9 -k5,5nr |
awk '$10 != x print x = $10 '

This gives me the biggest file for each group_id, so now I just add to handle the day part.

For information:

tail -n+2: hide the "total" part of the ls command's output

First awk: get the group_id part (012, 036...) and display it after the original line ($0)

Sort: sort on filename and size

Take the biggest size of each group_id (column 10 added by awk at beginning)

share|improve this answer

edited Mar 28 at 5:58

tripleee

102k1515 gold badges150150 silver badges204204 bronze badges

answered Mar 27 at 15:45

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

ls -l | tail -n+2 |
awk ' split($0,var,"_"); group_id=var[5]; print $0" "group_id ' |
sort -k9,9 -k5,5nr |
awk '$10 != x print x = $10 '

share|improve this answer

edited Mar 28 at 5:58

tripleee

102k1515 gold badges150150 silver badges204204 bronze badges

answered Mar 27 at 15:45

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

edited Mar 28 at 5:58

tripleee

102k1515 gold badges150150 silver badges204204 bronze badges

edited Mar 28 at 5:58

tripleee

102k1515 gold badges150150 silver badges204204 bronze badges

answered Mar 27 at 15:45

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges

answered Mar 27 at 15:45

user2178964user2178964

4155 gold badges1414 silver badges2727 bronze badges