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





.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}







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 Jan 9 at 12:15









      don_crissti

      51.8k15141168




      51.8k15141168










      asked Jan 9 at 12:13









      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 a grep 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.
            Jan 9 at 14:05








          • 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
            Jan 9 at 20:59



















          2














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



          grep -E "^5 "





          share|improve this answer



















          • 4





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

            – terdon♦
            Jan 9 at 13:50



















          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



















          • 2





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

            – terdon♦
            Jan 9 at 13:49












          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 a grep 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.
            Jan 9 at 14:05








          • 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
            Jan 9 at 20:59
















          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 a grep 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.
            Jan 9 at 14:05








          • 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
            Jan 9 at 20:59














          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 a grep 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 a grep 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 Jan 12 at 19:22









          Jeff Schaller♦

          44.6k1162145




          44.6k1162145










          answered Jan 9 at 12:24









          wurtelwurtel

          11.1k11628




          11.1k11628








          • 1





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

            – Dave.
            Jan 9 at 14:05








          • 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
            Jan 9 at 20:59














          • 1





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

            – Dave.
            Jan 9 at 14:05








          • 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
            Jan 9 at 20:59








          1




          1





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

          – Dave.
          Jan 9 at 14:05







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

          – Dave.
          Jan 9 at 14:05






          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
          Jan 9 at 20:59





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

          – bishop
          Jan 9 at 20:59













          2














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



          grep -E "^5 "





          share|improve this answer



















          • 4





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

            – terdon♦
            Jan 9 at 13:50
















          2














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



          grep -E "^5 "





          share|improve this answer



















          • 4





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

            – terdon♦
            Jan 9 at 13:50














          2












          2








          2







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



          grep -E "^5 "





          share|improve this answer













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



          grep -E "^5 "






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 9 at 12:21









          rbrtflrrbrtflr

          613




          613








          • 4





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

            – terdon♦
            Jan 9 at 13:50














          • 4





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

            – terdon♦
            Jan 9 at 13:50








          4




          4





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

          – terdon♦
          Jan 9 at 13:50





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

          – terdon♦
          Jan 9 at 13:50











          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



















          • 2





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

            – terdon♦
            Jan 9 at 13:49
















          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



















          • 2





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

            – terdon♦
            Jan 9 at 13:49














          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













          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












          share|improve this answer



          share|improve this answer










          answered Jan 9 at 12:19









          RalfRalf

          36718




          36718








          • 2





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

            – terdon♦
            Jan 9 at 13:49














          • 2





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

            – terdon♦
            Jan 9 at 13:49








          2




          2





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

          – terdon♦
          Jan 9 at 13:49





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

          – terdon♦
          Jan 9 at 13:49


















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