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5

I have been trying to:

cp file.csv file.$(date +%D).csv

But it fails because the filenames is: file.03/27/19.csv with the slash of separate directories.

And I have been trying again to:

cp file.csv file.$(printf "%q" $(date +%D)).csv

But it still fails.

shell filenames date escape-characters slash

share|improve this question

edited 17 mins ago

Stéphane Chazelas

312k57589946

asked 16 hours ago

tres.14159tres.14159

3814

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…
  
  – tres.14159
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  the problem is your use of the date format using the / character. You said it yourself, the shell is seeing them as directory markers. Try one of the many other options available from date. You might be able to get the / escaped so the filename uses the character code (like putting a space in a filename), but that is often problematic.
  
  – 0xSheepdog
  16 hours ago
  
  
  

  
  
  
  

add a comment | 

5

I have been trying to:

cp file.csv file.$(date +%D).csv

But it fails because the filenames is: file.03/27/19.csv with the slash of separate directories.

And I have been trying again to:

cp file.csv file.$(printf "%q" $(date +%D)).csv

But it still fails.

shell filenames date escape-characters slash

share|improve this question

edited 17 mins ago

Stéphane Chazelas

312k57589946

asked 16 hours ago

tres.14159tres.14159

3814

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…
  
  – tres.14159
  16 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  the problem is your use of the date format using the / character. You said it yourself, the shell is seeing them as directory markers. Try one of the many other options available from date. You might be able to get the / escaped so the filename uses the character code (like putting a space in a filename), but that is often problematic.
  
  – 0xSheepdog
  16 hours ago
  
  
  

  
  
  
  

add a comment | 

I have been trying to:

cp file.csv file.$(date +%D).csv

But it fails because the filenames is: file.03/27/19.csv with the slash of separate directories.

And I have been trying again to:

cp file.csv file.$(printf "%q" $(date +%D)).csv

But it still fails.

shell filenames date escape-characters slash

share|improve this question

edited 17 mins ago

Stéphane Chazelas

312k57589946

asked 16 hours ago

tres.14159tres.14159

3814

cp file.csv file.$(date +%D).csv

cp file.csv file.$(printf "%q" $(date +%D)).csv

share|improve this question

edited 17 mins ago

Stéphane Chazelas

312k57589946

asked 16 hours ago

tres.14159tres.14159

3814

share|improve this question

edited 17 mins ago

Stéphane Chazelas

312k57589946

asked 16 hours ago

tres.14159tres.14159

3814

edited 17 mins ago

Stéphane Chazelas

312k57589946

edited 17 mins ago

Stéphane Chazelas

312k57589946

asked 16 hours ago

tres.14159tres.14159

3814

asked 16 hours ago

tres.14159tres.14159

3814

1 Answer
1

active

oldest

votes

16

You can't have / (byte 0x2F on ASCII-based systems) in a file name, period.

You can use characters that look like / like ⁄ (fraction slash), so you could do:

cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"

Or with some shells (zsh, bash at least):

cp file.csv "file.$(date +%D | sed $'s|/|u2044|g').csv"

(using sed instead of tr as some tr implementations including GNU tr still don't support multi-byte characters).

But you may run into problems like the file name being rendered differently in locales using a different charset from the one that was in use at the time you created the file (and of course the confusion of users when they see what looks like a slash in a file name).

My advice would be to use the standard non-ambiguous (for most people outside the US, 03/12/18 would be interpreted as the 3^rd of December 2018 for instance) YYYY-mm-dd format instead (which also helps wrt sorting):

cp file.csv "file.$(date +%Y-%m-%d).csv"

Which with many date implementations you can shorten to:

cp file.csv "file.$(date +%F).csv"

share|improve this answer

edited 19 mins ago

answered 16 hours ago

Stéphane ChazelasStéphane Chazelas

312k57589946

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Other characters similar to slash can also be cumbersome to type in terminal.
  
  – gronostaj
  55 mins ago
  

  
  
  
  

add a comment | 

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16

You can't have / (byte 0x2F on ASCII-based systems) in a file name, period.

You can use characters that look like / like ⁄ (fraction slash), so you could do:

cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"

Or with some shells (zsh, bash at least):

cp file.csv "file.$(date +%D | sed $'s|/|u2044|g').csv"

(using sed instead of tr as some tr implementations including GNU tr still don't support multi-byte characters).

But you may run into problems like the file name being rendered differently in locales using a different charset from the one that was in use at the time you created the file (and of course the confusion of users when they see what looks like a slash in a file name).

My advice would be to use the standard non-ambiguous (for most people outside the US, 03/12/18 would be interpreted as the 3^rd of December 2018 for instance) YYYY-mm-dd format instead (which also helps wrt sorting):

cp file.csv "file.$(date +%Y-%m-%d).csv"

Which with many date implementations you can shorten to:

cp file.csv "file.$(date +%F).csv"

share|improve this answer

edited 19 mins ago

answered 16 hours ago

Stéphane ChazelasStéphane Chazelas

312k57589946

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Other characters similar to slash can also be cumbersome to type in terminal.
  
  – gronostaj
  55 mins ago
  

  
  
  
  

add a comment | 

16

You can't have / (byte 0x2F on ASCII-based systems) in a file name, period.

You can use characters that look like / like ⁄ (fraction slash), so you could do:

cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"

Or with some shells (zsh, bash at least):

cp file.csv "file.$(date +%D | sed $'s|/|u2044|g').csv"

(using sed instead of tr as some tr implementations including GNU tr still don't support multi-byte characters).

But you may run into problems like the file name being rendered differently in locales using a different charset from the one that was in use at the time you created the file (and of course the confusion of users when they see what looks like a slash in a file name).

My advice would be to use the standard non-ambiguous (for most people outside the US, 03/12/18 would be interpreted as the 3^rd of December 2018 for instance) YYYY-mm-dd format instead (which also helps wrt sorting):

cp file.csv "file.$(date +%Y-%m-%d).csv"

Which with many date implementations you can shorten to:

cp file.csv "file.$(date +%F).csv"

share|improve this answer

edited 19 mins ago

answered 16 hours ago

Stéphane ChazelasStéphane Chazelas

312k57589946

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Other characters similar to slash can also be cumbersome to type in terminal.
  
  – gronostaj
  55 mins ago
  

  
  
  
  

add a comment | 

You can't have / (byte 0x2F on ASCII-based systems) in a file name, period.

You can use characters that look like / like ⁄ (fraction slash), so you could do:

cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"

Or with some shells (zsh, bash at least):

cp file.csv "file.$(date +%D | sed $'s|/|u2044|g').csv"

(using sed instead of tr as some tr implementations including GNU tr still don't support multi-byte characters).

But you may run into problems like the file name being rendered differently in locales using a different charset from the one that was in use at the time you created the file (and of course the confusion of users when they see what looks like a slash in a file name).

My advice would be to use the standard non-ambiguous (for most people outside the US, 03/12/18 would be interpreted as the 3^rd of December 2018 for instance) YYYY-mm-dd format instead (which also helps wrt sorting):

cp file.csv "file.$(date +%Y-%m-%d).csv"

Which with many date implementations you can shorten to:

cp file.csv "file.$(date +%F).csv"

share|improve this answer

edited 19 mins ago

answered 16 hours ago

Stéphane ChazelasStéphane Chazelas

312k57589946

cp file.csv "file.$(date +%D | sed 's|/|⁄|g').csv"

cp file.csv "file.$(date +%D | sed $'s|/|u2044|g').csv"

cp file.csv "file.$(date +%Y-%m-%d).csv"

cp file.csv "file.$(date +%F).csv"

share|improve this answer

edited 19 mins ago

answered 16 hours ago

Stéphane ChazelasStéphane Chazelas

312k57589946

answered 16 hours ago

Stéphane ChazelasStéphane Chazelas

312k57589946

answered 16 hours ago

Stéphane ChazelasStéphane Chazelas

312k57589946