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Thread: Slash in variable

  1. #1
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    Slash in variable

    Hi,
    I am writing following script:
    Code:
    huangyingw@laptop:~/bashrc$ cat fr.sh
    #! /bin/sh
    file=/home/huangyingw/myproject/git/test_fr/makefile
    sed -i "s/"$1"/"$2"/gi" $file
    huangyingw@laptop:~/bashrc$
    And passing following parameter to it, then I got the error.
    Code:
    huangyingw@laptop:~/myproject/git/test_fr$ fr ~/myproject/git/makefile/GNU_makefile_template ~/myproject/git/GNU_makefile_template
    sed: -e expression #1, char 9: unknown option to `s'
    huangyingw@laptop:~/myproject/git/test_fr$
    I think it is caused by the special character "Slash" in the parameter.
    How could I work around this?

  2. #2
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    Re: Slash in variable

    Quotes don't nest and it looks like you had them nested. Try different quotes instead.

    Code:
    sed -i 's/"$1"/"$2"/gi' $file
    Last edited by Lars Noodén; September 29th, 2012 at 01:27 PM.

  3. #3
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    Re: Slash in variable

    I'm not sure what those inner quotes are meant to be doing. Are they intentionally part of the pattern? If so, for this particular case, this should get around your slash problem.
    Code:
    sed -i "s#\"$1\"#\"$2\"#gi" $file
    However, if $1 or $2 contains a # then you will get the same problem.

    You would have to update your variables to escape whatever character you use as the separator.

    @ Lars Noodén If a single quote is used for the outer quote, then $1 and $2 are not subsituted by the shell.

  4. #4
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    Re: Slash in variable

    Quote Originally Posted by spjackson View Post
    I'm not sure what those inner quotes are meant to be doing. Are they intentionally part of the pattern? If so, for this particular case, this should get around your slash problem.
    Code:
    sed -i "s#\"$1\"#\"$2\"#gi" $file
    However, if $1 or $2 contains a # then you will get the same problem.

    You would have to update your variables to escape whatever character you use as the separator.

    @ Lars Noodén If a single quote is used for the outer quote, then $1 and $2 are not subsituted by the shell.
    I have tried both of above solution, neither of them works
    Code:
    huangyingw@laptop:~/bashrc$ cat fr.sh
    #! /bin/sh
    file=/home/huangyingw/myproject/git/test_fr/makefile
    #sed -i "s/"$1"/"$2"/gi" $file
    #sed -i 's/"$1"/"$2"/gi' $file
    sed -i "s#\"$1\"#\"$2\"#gi" $file
    ....

  5. #5
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    Re: Slash in variable

    spjackson asked if the quotes are a part of the string (s/"something"/"something else"/). that would require escaping literal quotes and he he showed how to obtain it (\")

    in case there are no quotes to worry about, the whole thing is straightforward

    Code:
    sed -i "s#$1#$2#gi" "$file"
    # will solve slash issue, but you can use any character as a delimiter. It's best to use one that can't possibly exist in patterns and replacements.
    if your question is answered, mark the thread as [SOLVED]. Thx.
    To post code or command output, use [code] tags.
    Check your bash script here // BashFAQ // BashPitfalls

  6. #6
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    Re: Slash in variable

    Quote Originally Posted by Vaphell View Post
    spjackson asked if the quotes are a part of the string (s/"something"/"something else"/). that would require escaping literal quotes and he he showed how to obtain it (\")

    in case there are no quotes to worry about, the whole thing is straightforward

    Code:
    sed -i "s#$1#$2#gi" "$file"
    # will solve slash issue, but you can use any character as a delimiter. It's best to use one that can't possibly exist in patterns and replacements.
    I have tried this:
    Code:
    huangyingw@laptop:~/bashrc$ cat fr.sh
    #! /bin/sh
    file=/home/huangyingw/myproject/git/test_fr/makefile
    #sed -i "s/"$1"/"$2"/gi" $file
    #sed -i 's/"$1"/"$2"/gi' $file
    #sed -i "s#\"$1\"#\"$2\"#gi" $file
    sed -i "s#$1#$2#gi" "$file"
    But nothing change after I use in this way:
    Code:
    huangyingw@laptop:~/myproject/git/test_fr$ fr ~/myproject/git/makefile/GNU_makefile_template temp
    huangyingw@laptop:~/myproject/git/test_fr$ gs
    # On branch master
    nothing to commit (working directory clean)
    the git status command tell me that, nothing changed. And I have actually open the target file to check.

  7. #7
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    Re: Slash in variable

    test it without -i first

    or in terminal with
    echo "line that should be changed" | sed 's#a#b#g'
    i always do that while finetuning sed expressions

    edit: wait a minute, won't ~ get expanded to /home/username/ ?
    in that case sed won't be able to find literal "~/file/something"
    what do you have in file and what do you get if you echo "$1"?

    do yourself a favor and normalize it, expand everything in file to full paths (sed "s#~/#$HOME/#g")
    or do the reverse - leave ~, but explicitly replace /home/whatever/ with ~ in $1 ( sed "s#${1/$HOME/\~}#$2#g" )
    Last edited by Vaphell; September 29th, 2012 at 06:25 PM.
    if your question is answered, mark the thread as [SOLVED]. Thx.
    To post code or command output, use [code] tags.
    Check your bash script here // BashFAQ // BashPitfalls

  8. #8
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    Re: Slash in variable

    Quote Originally Posted by Vaphell View Post
    test it without -i first

    or in terminal with
    echo "line that should be changed" | sed 's#a#b#g'
    i always do that while finetuning sed expressions

    edit: wait a minute, won't ~ get expanded to /home/username/ ?
    in that case sed won't be able to find literal "~/file/something"
    what do you have in file and what do you get if you echo "$1"?

    do yourself a favor and normalize it, expand everything in file to full paths (sed "s#~/#$HOME/#g")
    or do the reverse - leave ~, but explicitly replace /home/whatever/ with ~ in $1 ( sed "s#${1/$HOME/\~}#$2#g" )
    Currently it works when the file content is
    Code:
    huangyingw@laptop:~/myproject/git/test_fr$ cat makefile 
    OBJECTS = bitsort.exe
    include /home/huangyingw/myproject/git/makefile/GNU_makefile_template
    LOCFLAGS = -I../bitsort
    And the fr.sh is
    Code:
    huangyingw@laptop:~/bashrc$ cat fr.sh
    #! /bin/sh
    file=/home/huangyingw/myproject/git/test_fr/makefile
    #sed -i "s/"$1"/"$2"/gi" $file
    #sed -i 's/"$1"/"$2"/gi' $file
    #sed -i "s#\"$1\"#\"$2\"#gi" $file
    sed "s#$1#$2#g" "$file"
    So, if wrapped in fr.sh, the special character like "~", and "..", could not be recognized by sed?

  9. #9
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    Re: Slash in variable

    there is no problem with .. but when you type fr.sh ~/some/stuff bash automatically replaces ~ with /home/me

    Code:
    $ cat params.sh
    #!/bin/bash
    param=( "$@" )
    for(( i=0; i<${#param[@]}; i++ )); do echo $((i+1)) ${param[$i]}; done
    $ ./params.sh ~/test/ .. "~"
    1 /home/me/test/
    2 ..
    3 ~
    as you can see the script never gets to see ~, shell replaces it with /home/me before passing parameters to the script. You have to quote the string so bash ignores the expansion or change $1 back to ~ inside the script with "${1/$HOME/~}" (it will replace $HOME=/home/me with literal ~)
    if your question is answered, mark the thread as [SOLVED]. Thx.
    To post code or command output, use [code] tags.
    Check your bash script here // BashFAQ // BashPitfalls

  10. #10
    Join Date
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    Re: Slash in variable

    Quote Originally Posted by Vaphell View Post
    there is no problem with .. but when you type fr.sh ~/some/stuff bash automatically replaces ~ with /home/me

    Code:
    $ cat params.sh
    #!/bin/bash
    param=( "$@" )
    for(( i=0; i<${#param[@]}; i++ )); do echo $((i+1)) ${param[$i]}; done
    $ ./params.sh ~/test/ .. "~"
    1 /home/me/test/
    2 ..
    3 ~
    as you can see the script never gets to see ~, shell replaces it with /home/me before passing parameters to the script. You have to quote the string so bash ignores the expansion or change $1 back to ~ inside the script with "${1/$HOME/~}" (it will replace $HOME=/home/me with literal ~)
    Great thanks for your explanation.
    Finally, my working fr.sh is as bellow:
    Code:
    huangyingw@laptop:~/bashrc$ cat fr.sh
    #! /bin/sh
    FILE_POSTFIX=$HOME/bashrc/postfix
    PRUNE_POSTFIX=$HOME/bashrc/prunefix
    PRUNE_FILE=$HOME/bashrc/prunefile
    find_params=();
    prune_params=();
    prune_files=();
    or="";
    grep_params="";
    if [ -n "$3" ]
    then grep_params=" -A"$3" -B"$3;
    fi
    while read suf
    do
      find_params+=( $or "-iname" "*.$suf" )
      or="-o"
    done < "$FILE_POSTFIX"
    or="";
    while read suf
    do
      prune_params+=( $or "-iname" "*.$suf" )
      or="-o"
    done < "$PRUNE_POSTFIX"
    while read suf
    do
      prune_files+=( $or "-iname" "$suf" )
      or="-o"
    done < "$PRUNE_FILE"
    find "$1" "(" "${prune_params[@]}" "${prune_files[@]}" ")" -prune -o "(" "${find_params[@]}" "-o" "-iname" "makefile" ")" -type f | while read -r file
    do
      sed -i "s#$2#$3#g" "$file"
    done

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