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3

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 1 hour ago

Gilles

544k12811041621

asked 6 hours ago

tres.14159tres.14159

2612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…
  
  – tres.14159
  6 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
  6 hours ago
  
  

  
  
  
  

add a comment | 

3

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 1 hour ago

Gilles

544k12811041621

asked 6 hours ago

tres.14159tres.14159

2612

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…
  
  – tres.14159
  6 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
  6 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 1 hour ago

Gilles

544k12811041621

asked 6 hours ago

tres.14159tres.14159

2612

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

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

share|improve this question

edited 1 hour ago

Gilles

544k12811041621

asked 6 hours ago

tres.14159tres.14159

2612

share|improve this question

edited 1 hour ago

Gilles

544k12811041621

asked 6 hours ago

tres.14159tres.14159

2612

edited 1 hour ago

Gilles

544k12811041621

edited 1 hour ago

Gilles

544k12811041621

asked 6 hours ago

tres.14159tres.14159

2612

asked 6 hours ago

tres.14159tres.14159

2612

1 Answer
1

active

oldest

votes

11

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"

But you may run into problems like the file name being rendered differently in locales using a different charset.

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 5 hours ago

answered 6 hours ago

Stéphane ChazelasStéphane Chazelas

311k57588946

add a comment | 

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11

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"

But you may run into problems like the file name being rendered differently in locales using a different charset.

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 5 hours ago

answered 6 hours ago

Stéphane ChazelasStéphane Chazelas

311k57588946

add a comment | 

11

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"

But you may run into problems like the file name being rendered differently in locales using a different charset.

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 5 hours ago

answered 6 hours ago

Stéphane ChazelasStéphane Chazelas

311k57588946

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"

But you may run into problems like the file name being rendered differently in locales using a different charset.

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 5 hours ago

answered 6 hours ago

Stéphane ChazelasStéphane Chazelas

311k57588946

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

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

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

share|improve this answer

edited 5 hours ago

answered 6 hours ago

Stéphane ChazelasStéphane Chazelas

311k57588946

answered 6 hours ago

Stéphane ChazelasStéphane Chazelas

311k57588946

answered 6 hours ago

Stéphane ChazelasStéphane Chazelas

311k57588946