Shell script - remove first and last quote from a variable

Question

Below is the snippet of a shell script from a larger script  It removes the quotes from the string that is held by a variable  I am doing it using sed  but is it efficient  If not  then what is the efficient way      bin sh  opt    html  test      temp  echo  opt   sed  s          1     sed  s          1    echo  temp

User · Answer

The easiest solution in Bash     s   abc     echo  s  abc    echo    s 1 -1   abc   This is called substring expansion  see Gnu Bash Manual and search for   parameter offset length    In this example it takes the substring from s starting at position 1 and ending at the second last position  This is due to the fact that if length is a negative value it is interpreted as a backwards running offset from the end of parameter

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You can do it with only one call to sed     echo    html  test        sed  s             1   html test

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The shortest way around - try   echo  opt   sed  s     g    It actually removes all  s  double quotes  from opt  are there really going to be any more double quotes other than in the beginning and the end though  So it s actually the same thing  and much more brief  -

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This is the most discrete way without using sed   x   fish   printf     quotes   s nno quotes    s n    x     x          Or  echo  x echo   x         Output      quotes   fish  no quotes   fish   I got this from a source

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I know this is a very old question  but here is another sed variation  which may be useful to someone  Unlike some of the others  it only replaces double quotes at the start or end     echo   opt    sed -r  s        g

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If you try to remove quotes because the Makefile keeps them  try this    subst    quot     YOUR VARIABLE    Based on another answer  https   stackoverflow com a 10430975 10452175

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If you re using jq and trying to remove the quotes from the result  the other answers will work  but there s a better way  By using the -r option  you can output the result with no quotes     echo    foo    bar      jq   foo   bar     echo    foo    bar      jq -r   foo  bar

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There is another way to do it  Like   echo   opt 1 -1

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There s a simpler and more efficient way  using the native shell prefix suffix removal feature   temp    opt      temp    temp      echo   temp      opt     will remove the suffix    escaped with a backslash to prevent shell interpretation      temp     will remove the prefix    escaped with a backslash to prevent shell interpretation    Another advantage is that it will remove surrounding quotes only if there are surrounding quotes   BTW  your solution always removes the first and last character  whatever they may be  of course  I m sure you know your data  but it s always better to be sure of what you re removing    Using sed   echo   opt    sed -e  s       -e  s          Improved version  as indicated by jfgagne  getting rid of echo   sed -e  s       -e  s        lt  lt  lt   opt    So it replaces a leading   with nothing  and a trailing   with nothing too  In the same invocation  there isn t any need to pipe and start another sed  Using -e you can have multiple text processing

User · Answer

Use tr to delete      echo   opt    tr -d       Note  This removes all double quotes  not just leading and trailing

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Update  A simple and elegant answer from Stripping single and double quotes in a string using bash   standard Linux commands only   BAR   eval echo  BAR  strips quotes from BAR                                                                  Based on hueybois s answer  I came up with this function after much trial and error   function stripStartAndEndQuotes       cmd  temp     1            eval echo  cmd     temp    temp          eval echo   1  temp      If you don t want anything printed out  you can pipe the evals to  dev null 2 gt  amp 1   Usage     BAR  FOO BAR    echo BAR  FOO BAR    stripStartAndEndQuotes  BAR    echo BAR FOO BAR

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This will remove all double quotes   echo    opt

User · Answer

There is a straightforward way using xargs    gt  echo   quoted     xargs quoted   xargs uses echo as the default command if no command is provided and strips quotes from the input  See e g  here

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My version  strip quotes         while       -gt 0     do         local value    1          local len    value               value 0 1         amp  amp    value  len-1 1            amp  amp  declare -g  1    value 1  len-2           shift     done     The function accepts variable name s  and strips quotes in place  It only strips a matching pair of leading and trailing quotes  It doesn t check if the trailing quote is escaped  preceded by   which is not itself escaped     In my experience  general-purpose string utility functions like this  I have a library of them  are most efficient when manipulating the strings directly  not using any pattern matching and especially not creating any sub-shells  or calling any external tools such as sed  awk or grep   var1    test       end     var2 test var3   test var4 test   echo before  for i in var 1 2 3 4   do     echo  i     i   done strip quotes var 1 2 3 4  echo echo after  for i in var 1 2 3 4   do     echo  i     i   done

User · Answer

In Bash  I would use the following one-liner      quot   str  quot       quot    quot      quot   str  quot               amp  amp  str  quot   str 1 -1  quot   This will remove surrounding quotes  both single and double  while keeping quoting characters inside the string intact  Also  it won t do anything if there s only a single leading quote or only a single trailing quote  which is usually what you want in my experience  Wrapped in a function     usr bin env bash    Strip surrounding quotes from string   1  variable name  function strip quotes         local -n var  quot  1 quot          quot   var  quot       quot    quot      quot   var  quot               amp  amp  var  quot   var 1 -1  quot     str  quot    quot  echo  quot Before    str  quot  strip quotes str echo  quot After     str  quot

[string] Shell script - remove first and last quote (") from a variable

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