Escape a backup date in a file name












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

























  • You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

    – tres.14159
    9 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
    9 hours ago


















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

























  • You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

    – tres.14159
    9 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
    9 hours ago
















3












3








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
















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




shell filenames date escape-characters slash






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 4 hours ago









Gilles

544k12811041621




544k12811041621










asked 9 hours ago









tres.14159tres.14159

2814




2814













  • You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

    – tres.14159
    9 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
    9 hours ago





















  • You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

    – tres.14159
    9 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
    9 hours ago



















You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

– tres.14159
9 hours ago





You can not set a filename with slash characters: stackoverflow.com/questions/9847288/…

– tres.14159
9 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
9 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
9 hours ago












1 Answer
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12














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 3rd 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"





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    12














    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 3rd 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






























      12














      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 3rd 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




























        12












        12








        12







        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 3rd 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















        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 3rd 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














        share|improve this answer



        share|improve this answer








        edited 9 hours ago

























        answered 9 hours ago









        Stéphane ChazelasStéphane Chazelas

        311k57589946




        311k57589946






























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