Replace a string in shell script using a variable

Question

I am using the below code for replacing a string inside a shell script   echo  LINE   sed -e  s 12345678   replace  g    but it s getting replaced with  replace instead of the value of that variable   Could anybody tell what went wrong

User · Answer

echo  LINE   sed -e  s 12345678   replace  g    you can still use single quotes  but you have to  open  them when you want the variable expanded at the right place  otherwise the string is taken  literally   as  paxdiablo correctly stated  his answer is correct as well

User · Answer

I had a similar requirement to this but my replace var contained an ampersand  Escaping the ampersand like this solved my problem   replace  salt  amp  pepper  echo  pass the salt    sed  s salt   replace  amp    amp   g

User · Answer

If you want to interpret  replace  you should not use single quotes since they prevent variable substitution  Try  echo  LINE   sed -e  quot s 12345678   replace  g quot   Transcript  pax gt  export replace 987654321 pax gt  echo X123456789X   sed  quot s 123456789   replace   quot  X987654321X pax gt     Just be careful to ensure that   replace  doesn t have any characters of significance to sed  like   for instance  since it will cause confusion unless escaped  But if  as you say  you re replacing one number with another  that shouldn t be a problem

User · Answer

Not related to question but in case if someone is passing the string as an argument to bash script this might help  Use    bash script sh  quot path to file quot   Instead of    bash script sh path to file  For your reference

User · Answer

Single quotes are very strong  Once inside  there s nothing you can do to invoke variable substitution  until you leave  Use double quotes instead   echo  LINE   sed -e  s 12345678  replace g

User · Answer

Found a graceful solution   echo   LINE  12345678  replace

User · Answer

Not specific to the question  but for folks who need the same kind of functionality expanded for clarity from previous answers     create some variables str  someFileName foo  find   foo  replace   bar    notice the the str isn t prefixed with        this is just how this feature works    result   str   find  replace  echo  result       result is  someFileName bar  str  someFileName sally  find   foo  replace   bar  result   str   find  replace  echo  result       result is  someFileName sally because   foo  was not found

User · Answer

I prefer to use double quotes   as single quptes are very powerful as we used them if dont able to change anything inside it or can invoke the variable substituion    so use double quotes instaed   echo  LINE   sed -e  s 12345678  replace g

User · Answer

To let your shell expand the variable  you need to use double-quotes like  sed -i  s 12345678  replace g  file txt   This will break if  replace contain special sed characters         But you can preprocess  replace to quote them   replace quoted   printf   s    replace    sed  s         0 g   sed -i  s 12345678  replace quoted g  file txt

User · Answer

you can use the shell  bash ksh       var  12345678abc    replace  test    echo   var  12345678  replace  testabc

User · Answer

Use this instead  echo  LINE   sed -e  s 12345678  replace g    this works for me just simply remove the quotes

[unix] Replace a string in shell script using a variable

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