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Old 05-22-2005, 10:28 AM   #1
chii-chan
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sed doesn't accept $variable in bash script


I made this script to replace text using 'sed':

Code:
#!/bin/bash

txt_ori="original text"
txt_rplc="replacement text"

for text in "$@"
do
    sed 's/$txt_ori/$txt_rplc/g' $text > text.out
done
But 'sed' doesn't seem to accept the variable. Anyway to make 'sed' to read the variables?
 
Old 05-22-2005, 10:40 AM   #2
rjlee
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Re: sed doesn't accept $variable in bash script

Quote:
Originally posted by chii-chan
I made this script to replace text using 'sed':

Code:
#!/bin/bash

txt_ori="original text"
txt_rplc="replacement text"

for text in "$@"
do
    sed 's/$txt_ori/$txt_rplc/g' $text > text.out
done
But 'sed' doesn't seem to accept the variable. Anyway to make 'sed' to read the variables?
You are passing two arguments to sed. The first is the literal string s/$txt_ori/$txt_rplc/g, the second is the value of $text

I assume that the value it's not accepting is $txt_ori or $txt_rplc (you can find out exactly what you are passing by using echo instead of sed).

BASH treats anything in single-quotes as a literal string, and will not try to interpolate any variables. You probably want to be using double-quotes instead.
 
Old 05-22-2005, 11:21 AM   #3
chii-chan
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Thanks for noticing me the single quotes. It seems that 'sed' needs this kind of line:

Code:
sed -e 's/SEARCH_STRING/REPLACE_STRING/g' file > file.out
Using double-quotes i.e.:

Code:
sed -e "s/"$txt_ori"/"$txt_rplc"/g" $text > text.out
doesnt't work. But instead I did this on try and error:

Code:
sed -e s/"$txt_ori"/"$txt_rplc"/g $text > text.out
then it works.
 
Old 05-22-2005, 12:26 PM   #4
keefaz
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Or use eval bash function like :
Code:
eval "sed -e s/$txt_ori/$txt_rplc/g $text > text.out"
This way you just quote at the beginning and at the end of
the sed line, not inside
 
Old 05-22-2005, 01:38 PM   #5
eddiebaby1023
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Registered: May 2005
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Re: sed doesn't accept $variable in bash script

Quote:
Originally posted by chii-chan
I made this script to replace text using 'sed':

Code:
#!/bin/bash

txt_ori="original text"
txt_rplc="replacement text"

for text in "$@"
do
    sed 's/$txt_ori/$txt_rplc/g' $text > text.out
done
But 'sed' doesn't seem to accept the variable. Anyway to make 'sed' to read the variables?
All you need to do (apart from learning how to treat metacharacters in the shell ) is use double quotes instead of the single ones, so the $ variables are expanded:
Code:
    sed "s/$txt_ori/$txt_rplc/g" $text > text.out
 
Old 05-25-2005, 07:33 AM   #6
chii-chan
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Registered: Sep 2003
Location: chikyuu (E103N6)
Distribution: Redhat 8.0 (2.4.25-custom), Fedora Core 1 (2.4.30-custom)
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Original Poster
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Thanks for the suggestions. So far the only working one is:

Code:
sed -e s/"$txt_ori"/"$txt_rplc"/g $text > text.out
Code:
sed -e "s/$txt_ori/$txt_rplc/g" $text > text.out
I can't get these to work though:

Code:
eval "sed -e s/$txt_ori/$txt_rplc/g $text > text.out"
But, really, thanks for all of the suggestions since I can apply that to other scripts later.

Last edited by chii-chan; 05-25-2005 at 07:49 AM.
 
Old 05-28-2005, 07:07 AM   #7
rjlee
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Posts: 1,990

Rep: Reputation: 67
Quote:
Originally posted by chii-chan
I can't get these to work though:

Code:
eval "sed -e s/$txt_ori/$txt_rplc/g $text > text.out"
This may or may not work, depending on the contents of the values of the variables you are quoting.

If any of them contains a space, this will create a new parameter, which probably isn't what you want. If they contain <chevrons>, then you will start a file redirect. If they contain a pipe (|) then you will start a file redirect.

If you run
Code:
man bash
then you can read all about how and when BASH copes with special characters. For example:
Quote:
eval [arg ...]
The args are read and concatenated together into a single command. This command is then read and exe_cuted by the shell, and its exit status is returned as the value of eval. If there are no args, or only null arguments, eval returns 0.
 
  


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