How to cut a string after a specific character in unix

Question

So I have this string     var server 10 200 200 20  home some directory file   I just want to extract the directory address meaning I only want the bit after the     character and get    home some directory file   thanks   I need a generic command so the cut command wont work as the  var variable doesn t have a fixed length

User · Answer

This might work for you   echo   var       See Example 10-10  Pattern matching in parameter substitution

User · Answer

This will also do   echo  var   cut -f2 -d

User · Answer

awk -F    print  2    lt  lt  lt   var

User · Answer

You don t say which shell you re using   If it s a POSIX-compatible one such as Bash  then parameter expansion can do what you want      Parameter Expansion                 parameter word       Remove Smallest Prefix Pattern     The word is expanded to produce a pattern   The parameter expansion then results in parameter  with the smallest portion of the prefix matched by the pattern deleted    In other words  you can write   var    var        which will remove anything matching    from  var  i e  everything up to and including the first      If you want to match up to the last    then you could use    in place of     This is all assuming that the part to remove does not contain    true for IPv4 addresses  but not for IPv6 addresses

User · Answer

For completeness  using cut  cut -d   -f 2  lt  lt  lt   var   And using only bash   IFS   read a b  lt  lt  lt   var   echo  b

User · Answer

This should do the trick     echo   var    awk -F      print  NF    home some directory file

User · Answer

Using sed     var server 10 200 200 20  home some directory file   echo  var   sed  s         home some directory file

[string] How to cut a string after a specific character in unix

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