Using grep and sed to find and replace a string

Question

I am using the following to search a directory recursively for specific string and replace it with another   grep -rl oldstr path   xargs sed -i  s oldstr newstr g    This works okay  The only problem is that if the string doesn t exist then sed fails because it doesn t get any arguments  This is a problem for me since i m running this automatically with ANT and the build fails since sed fails   Is there a way to make it fail-proof in case the string is not found   I m interested in a one line simple solution I can use  not necessarily with grep or sed but with common unix commands like these

User · Answer

Your solution is ok  only try it in this way   files   grep -rl oldstr path   amp  amp  echo  files   xargs sed       so execute the xargs only when grep return 0  e g  when found the string in some files

User · Answer

Not sure if this will be helpful but you can use this with a remote server like the example below ssh example server com  quot find  DIR NAME -type f -name  quot FILES LOOKING FOR quot  -exec sed -i  s LOOKINGFOR withThisString g       quot  replace the example server com with your server replace DIR NAME with your directory file locations replace FILES LOOKING FOR with files you are looking for replace LOOKINGFOR with what you are looking for replace withThisString with what your want to be replaced in the file

User · Answer

I think that without using -exec you can simply provide  dev null as at least one argument in case nothing is found   grep -rl oldstr path   xargs sed -i  s oldstr newstr g   dev null

User · Answer

I have taken Vlad s idea and changed it a little bit  Instead of   grep -rl oldstr path   xargs sed -i  s oldstr newstr g   dev null   Which yields   sed  couldn t edit  dev null  not a regular file   I m doing in 3 different connections to the remote server  touch deleteme grep -rl oldstr path   xargs sed -i  s oldstr newstr g    deleteme rm deleteme   Although this is less elegant and requires 2 more connections to the server  maybe there s a way to do it all in one line  it does the job efficiently as well

User · Answer

You can use find and -exec directly into sed rather than first locating oldstr with grep   It s maybe a bit less efficient  but that might not be important   This way  the sed replacement is executed over all files listed by find  but if oldstr isn t there it obviously won t operate on it   find  path -type f -exec sed -i  s oldstr newstr g

User · Answer

Standard xargs has no good way to do it  you re better off using find -exec as someone else suggested  or wrap the sed in a script which does nothing if there are no arguments   GNU xargs has the --no-run-if-empty option  and BSD   OS X xargs has the -L option which looks like it should do something similar

User · Answer

My use case was I wanted to replace  foo  Drive Letter with foo  bar baz xyz In my case I was able to do it with the following code  I was in the same directory location where there were bulk of files   find   -name    library  -print0   xargs -0 sed -i    -e  s foo   Drive Letter  foo   bar  baz  xyz g    hope that helped

User · Answer

If you are to replace a fixed string or some pattern  I would also like to add the bash builtin pattern string replacement variable substitution construct  Instead of describing it myself  I am quoting the section from the bash manual        parameter pattern string         The pattern is expanded to produce a pattern just as in pathname                 expansion    parameter is expanded and the longest match of pattern against its value is replaced  with  string    If  pattern                 begins  with    all matches of pattern are replaced with string                  Normally only the first match is replaced    If  pattern  begins                 with     it must match at the beginning of the expanded value of                 parameter   If pattern begins with    it must match at  the  end                 of  the expanded value of parameter   If string is null  matches                 of pattern are deleted and the   following pattern may be  omitted    If  parameter  is     or    the substitution operation is                 applied to each positional parameter in turn  and the  expansion                 is  the  resultant list   If parameter is an array variable subscripted with   or    the substitution operation is  applied  to                 each  member  of  the  array  in  turn  and the expansion is the                 resultant list

[linux] Using grep and sed to find and replace a string

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