grep specific number from lines 5 but not 25 or 52 and so on












3















I need to find a user for a userid. The return from the authentication system is as follows. Now with a bash script I need to extract the line with the exact number. Not 5 and 25 ..



------------

ID  LOGIN   

------------

28  user1

25  user2 

5   user3


If I use grep 5 I do get 2 lines but I need the line with "5".



Any ideas ?






text-processing







share|improve this question





























    3















    I need to find a user for a userid. The return from the authentication system is as follows. Now with a bash script I need to extract the line with the exact number. Not 5 and 25 ..



    ------------
    
ID  LOGIN   
    
------------
    
28  user1
    
25  user2 
    
5   user3


    If I use grep 5 I do get 2 lines but I need the line with "5".



    Any ideas ?






    text-processing







    share|improve this question



























      3












      3








      3








      I need to find a user for a userid. The return from the authentication system is as follows. Now with a bash script I need to extract the line with the exact number. Not 5 and 25 ..



      ------------
      
ID  LOGIN   
      
------------
      
28  user1
      
25  user2 
      
5   user3


      If I use grep 5 I do get 2 lines but I need the line with "5".



      Any ideas ?






      text-processing







      share|improve this question
















      I need to find a user for a userid. The return from the authentication system is as follows. Now with a bash script I need to extract the line with the exact number. Not 5 and 25 ..



      ------------
      
ID  LOGIN   
      
------------
      
28  user1
      
25  user2 
      
5   user3


      If I use grep 5 I do get 2 lines but I need the line with "5".



      Any ideas ?






      text-processing




      text-processing






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 2 days ago









      don_crissti

      50.1k15132162




      50.1k15132162










      asked 2 days ago









      Dave.Dave.

      241




      241






















          3 Answers
          3






          active

          oldest

          votes


















          8














          There are several ways of doing it.



          IMHO the best way is to use awk, which is useful when dealing with fields.



          If you want agrep based solution, I would do:



          grep -w '^5'


          The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.



          The awk solution would look like:



          awk '$1 == 5'


          If you want only the username, which is the second column:



          awk '$1 == 5 {print $2}'


          If you're searching for a string and not a numeric value, enclose the string in double quotes:



          awk '$1 == "abc" {print $2}'





          share|improve this answer





















          • 1





            thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

            – Dave.
            2 days ago








          • 2





            @Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

            – bishop
            2 days ago



















          2














          You could try with a regex (first char in line) and including the space:



          grep -E "^5 "





          share|improve this answer








          New contributor




          rbrtflr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.
















          • 4





            This would be better as grep -E "^5s" in case there's a tab and not a space.

            – terdon♦
            2 days ago



















          0














          Try the following:



          query-auth-system | grep "^5\>"




          • ^: means "match at start of line


          • \>: matches a word-boundary. So it will match 5, but not 50.






          share|improve this answer








          New contributor




          Ralf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.
















          • 2





            That will also match lines starting with 5:, 5%, 5| etc.

            – terdon♦
            2 days ago











          Your Answer








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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          8














          There are several ways of doing it.



          IMHO the best way is to use awk, which is useful when dealing with fields.



          If you want agrep based solution, I would do:



          grep -w '^5'


          The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.



          The awk solution would look like:



          awk '$1 == 5'


          If you want only the username, which is the second column:



          awk '$1 == 5 {print $2}'


          If you're searching for a string and not a numeric value, enclose the string in double quotes:



          awk '$1 == "abc" {print $2}'





          share|improve this answer





















          • 1





            thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

            – Dave.
            2 days ago








          • 2





            @Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

            – bishop
            2 days ago
















          8














          There are several ways of doing it.



          IMHO the best way is to use awk, which is useful when dealing with fields.



          If you want agrep based solution, I would do:



          grep -w '^5'


          The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.



          The awk solution would look like:



          awk '$1 == 5'


          If you want only the username, which is the second column:



          awk '$1 == 5 {print $2}'


          If you're searching for a string and not a numeric value, enclose the string in double quotes:



          awk '$1 == "abc" {print $2}'





          share|improve this answer





















          • 1





            thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

            – Dave.
            2 days ago








          • 2





            @Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

            – bishop
            2 days ago














          8












          8








          8







          There are several ways of doing it.



          IMHO the best way is to use awk, which is useful when dealing with fields.



          If you want agrep based solution, I would do:



          grep -w '^5'


          The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.



          The awk solution would look like:



          awk '$1 == 5'


          If you want only the username, which is the second column:



          awk '$1 == 5 {print $2}'


          If you're searching for a string and not a numeric value, enclose the string in double quotes:



          awk '$1 == "abc" {print $2}'





          share|improve this answer















          There are several ways of doing it.



          IMHO the best way is to use awk, which is useful when dealing with fields.



          If you want agrep based solution, I would do:



          grep -w '^5'


          The -w tells grep to match the exact word, so this will not match "52". The "^" tells grep to search the 5 at the beginning of the line, which will fail if there are e.g. leading spaces.



          The awk solution would look like:



          awk '$1 == 5'


          If you want only the username, which is the second column:



          awk '$1 == 5 {print $2}'


          If you're searching for a string and not a numeric value, enclose the string in double quotes:



          awk '$1 == "abc" {print $2}'






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 days ago









          terdon♦

          129k32253428




          129k32253428










          answered 2 days ago









          wurtelwurtel

          10k11425




          10k11425








          • 1





            thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

            – Dave.
            2 days ago








          • 2





            @Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

            – bishop
            2 days ago














          • 1





            thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

            – Dave.
            2 days ago








          • 2





            @Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

            – bishop
            2 days ago








          1




          1





          thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

          – Dave.
          2 days ago







          thanks I have choosen the awk way. awk '$1 == 5 {print $2}' that worked.

          – Dave.
          2 days ago






          2




          2





          @Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

          – bishop
          2 days ago





          @Dave. Since this answer worked for you, please consider clicking the checkbox beside it to signal to future readers it answered your question.

          – bishop
          2 days ago













          2














          You could try with a regex (first char in line) and including the space:



          grep -E "^5 "





          share|improve this answer








          New contributor




          rbrtflr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.
















          • 4





            This would be better as grep -E "^5s" in case there's a tab and not a space.

            – terdon♦
            2 days ago
















          2














          You could try with a regex (first char in line) and including the space:



          grep -E "^5 "





          share|improve this answer








          New contributor




          rbrtflr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.
















          • 4





            This would be better as grep -E "^5s" in case there's a tab and not a space.

            – terdon♦
            2 days ago














          2












          2








          2







          You could try with a regex (first char in line) and including the space:



          grep -E "^5 "





          share|improve this answer








          New contributor




          rbrtflr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.










          You could try with a regex (first char in line) and including the space:



          grep -E "^5 "






          share|improve this answer








          New contributor




          rbrtflr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|improve this answer



          share|improve this answer






          New contributor




          rbrtflr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 2 days ago









          rbrtflrrbrtflr

          613




          613




          New contributor




          rbrtflr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          rbrtflr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          rbrtflr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.








          • 4





            This would be better as grep -E "^5s" in case there's a tab and not a space.

            – terdon♦
            2 days ago














          • 4





            This would be better as grep -E "^5s" in case there's a tab and not a space.

            – terdon♦
            2 days ago








          4




          4





          This would be better as grep -E "^5s" in case there's a tab and not a space.

          – terdon♦
          2 days ago





          This would be better as grep -E "^5s" in case there's a tab and not a space.

          – terdon♦
          2 days ago











          0














          Try the following:



          query-auth-system | grep "^5\>"




          • ^: means "match at start of line


          • \>: matches a word-boundary. So it will match 5, but not 50.






          share|improve this answer








          New contributor




          Ralf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.
















          • 2





            That will also match lines starting with 5:, 5%, 5| etc.

            – terdon♦
            2 days ago
















          0














          Try the following:



          query-auth-system | grep "^5\>"




          • ^: means "match at start of line


          • \>: matches a word-boundary. So it will match 5, but not 50.






          share|improve this answer








          New contributor




          Ralf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.
















          • 2





            That will also match lines starting with 5:, 5%, 5| etc.

            – terdon♦
            2 days ago














          0












          0








          0







          Try the following:



          query-auth-system | grep "^5\>"




          • ^: means "match at start of line


          • \>: matches a word-boundary. So it will match 5, but not 50.






          share|improve this answer








          New contributor




          Ralf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.










          Try the following:



          query-auth-system | grep "^5\>"




          • ^: means "match at start of line


          • \>: matches a word-boundary. So it will match 5, but not 50.







          share|improve this answer








          New contributor




          Ralf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|improve this answer



          share|improve this answer






          New contributor




          Ralf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 2 days ago









          RalfRalf

          3257




          3257




          New contributor




          Ralf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.





          New contributor





          Ralf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          Ralf is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.








          • 2





            That will also match lines starting with 5:, 5%, 5| etc.

            – terdon♦
            2 days ago














          • 2





            That will also match lines starting with 5:, 5%, 5| etc.

            – terdon♦
            2 days ago








          2




          2





          That will also match lines starting with 5:, 5%, 5| etc.

          – terdon♦
          2 days ago





          That will also match lines starting with 5:, 5%, 5| etc.

          – terdon♦
          2 days ago


















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