How to insert strings containing slashes with sed?

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别那么骄傲
别那么骄傲 2020-11-22 04:44

I have a Visual Studio project, which is developed locally. Code files have to be deployed to a remote server. The only problem are the URLs they contain, which are hard-cod

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  •  孤城傲影
    2020-11-22 05:17

    In a system I am developing, the string to be replaced by sed is input text from a user which is stored in a variable and passed to sed.

    As noted earlier on this post, if the string contained within the sed command block contains the actual delimiter used by sed - then sed terminates on syntax error. Consider the following example:

    This works:

    $ VALUE=12345
    $ echo "MyVar=%DEF_VALUE%" | sed -e s/%DEF_VALUE%/${VALUE}/g
    MyVar=12345
    

    This breaks:

    $ VALUE=12345/6
    $ echo "MyVar=%DEF_VALUE%" | sed -e s/%DEF_VALUE%/${VALUE}/g
    sed: -e expression #1, char 21: unknown option to `s'
    

    Replacing the default delimiter is not a robust solution in my case as I did not want to limit the user from entering specific characters used by sed as the delimiter (e.g. "/").

    However, escaping any occurrences of the delimiter in the input string would solve the problem. Consider the below solution of systematically escaping the delimiter character in the input string before having it parsed by sed. Such escaping can be implemented as a replacement using sed itself, this replacement is safe even if the input string contains the delimiter - this is since the input string is not part of the sed command block:

    $ VALUE=$(echo ${VALUE} | sed -e "s#/#\\\/#g")
    $ echo "MyVar=%DEF_VALUE%" | sed -e s/%DEF_VALUE%/${VALUE}/g
    MyVar=12345/6
    

    I have converted this to a function to be used by various scripts:

    escapeForwardSlashes() {
    
         # Validate parameters
         if [ -z "$1" ]
         then
                 echo -e "Error - no parameter specified!"
                 return 1
         fi
    
         # Perform replacement
         echo ${1} | sed -e "s#/#\\\/#g"
         return 0
    }
    

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