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NeverStopLearning
Asked: 2018-03-22 02:04:06 +0800 CST

SED on Bash Script : how to replace text with special character that come from variable on Bash script

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first, there is a variable that contain special character on bash script. here is the example. list.txt :

user = root
password = 1234

Bash Script :

#!/bin/sh
var1 = var&2018+&
find . -name "list.txt" -exec sed -i "s/1234/"$var1"/g" {} +

when i executed the script, the result is : var12342018+1234.

Update 1 (answer by Oliv) :

var1='var&2018+&'
find . -name "*list.txt" -exec bash -c 'sed -i "s/1234/$2/g" $1' _ {} "$var1" \;

and the result still : password = var12342018+1234

maybe anyone can help me. thank you

bash

2 Answers

  1. Best Answer

    First you need to quote your var1 variable.

    You can use bash -c in your find command in order to give the password as a variable:

     var1='var\&2018+\&'
 find . -name "list.txt" -exec bash -c 'sed -i "s/1234/$2/g" $1' _ {} "$var1" \;
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  2. it might help you understand the problem you are having to know that & is a special character that represent the whole match of your regexp.

    since 1234 is found, you are replacing it with var12342018+1234, the two & is becoming the matched string.

    the other point I'd like to make is really a personal preference, but using find ... -exec with {}, i don't like this syntax, I usually end it with \;

    I prefer using xargs

    find . -name list.txt | xargs sed -i 's/1234/var&2018+&/g'

    note the ending g would replace it as many times as it is found...

    also this might be obvious.. I don't know how many list.txt file you have in your subtree, but if you only have one, you don't need find.

    sed -i 's/pattern/replacement/options' filename

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