Bash problems with string comparison

Question

I have a problem with writing bash script. The problem is in comparison of strings. When I launch it, there's no errors. However in result, it is always changing the variable client. So if for an example we have two lines in file

apple A
orange D

and if I give the who=A I expect to see in result apple, or if at D - orange

But no matter of what I choose A or D it is always giving me the result - orange

No matter of the strings, it always change the variable client, like ignoring the comparison. Please help.

while read line
do
    IFS=" "
    set -- $line
    echo $2" "$who":"$1
    if [[ "$2"="$who" ]]
    then
        echo "change"
        client=$1
    fi
done < $file

echo $client

So now I changed the code as in one of the comment below, but now the caparison always false therefore the variable client is always empty

    while read -r line
do
    #IFS=" "
    #set -- $line
    #echo $2" "$who":"$1
    #if [[ "$2" = "$who" ]]
    a="${line% *}"
    l="${line#* }"
    if [[ "$l" == "$who" ]]
    then
        echo "hi"
        client="$a"
    fi
done < $file

using unquoted variables inside `[ ]` isn't a good idea...you can use unquoted variables inside `[[ ]]` though.. It deals with white spaces nicely — Jahid, May 18 '15 at 05:48
I know that it is not failing, my question how to make it failed and work properly — ZeroVash, May 18 '15 at 13:07

Jahid · Answer 1 · 2015-05-18T06:00:24.747

1

Change if [[ "$2"="$who" ]] to

if [[ "$2" = "$who" ]]

spaces around =

Example (for clarification):

who=A
while read line
do

    IFS=" "
    set -- $line
    echo $2" "$who":"$1
    if [[ "$2" = "$who" ]]
    then
        echo "change"
        client=$1
    fi
done < file #this is the file I used for testing

echo $client

Output:

A A:apple
change
D A:orange
apple

For who=D:

A D:apple
D D:orange
change
orange

edited May 18 '15 at 06:00

answered May 18 '15 at 05:41

Jahid

21,542
10
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108

This should work, are you sure there's no other problem in your code. It gives correct result in my case.. – Jahid May 18 '15 at 05:53
I changed my code and it is still doesn't work, just look my edit code in the original post – ZeroVash May 18 '15 at 13:07
your new code works on me, output is apple for A and Orange for D. You sure you haven't forgotten to put `who=A` or `who=D` in the code? I don't see that assignment in your code... – Jahid May 18 '15 at 14:49
1

Yes I find a mistake it is was in argument, thanks for your help – ZeroVash May 18 '15 at 16:24

David C. Rankin · Accepted Answer · 2015-05-18T06:20:20.520

If you have data in a file with each line like apple D and you want to read the file and separate then items, the parameter expansion/substring extraction is the correct way to process the line. For example (note $who is taken from your problem statement):

while read -r line
do
    fruit="${line% *}"       # remove from end to space
    letter="${line#* }"      # remove from start to space
    if [[ "$letter" == "$who" ]]
    then
        echo "change"
        client="$fruit"
    fi
done < $file

Short Example

Here is a quick example of splitting the words with parameter expansion/substring extraction:

#!/bin/bash

while read -r line
do
    fruit="${line% *}"
    letter="${line#* }"
    echo "fruit: $fruit  letter: $letter"
done

exit 0

input

$ cat dat/apple.txt
Apple A
Orange D

output

$ bash apple.sh <dat/apple.txt
fruit: Apple  letter: A
fruit: Orange  letter: D

yes it is work, as same as my method, but the problem in comparison not in reading strings... — ZeroVash, May 18 '15 at 13:01

score 0 · Answer 3 · edited May 23 '17 at 11:58

0

You do need spaces around that = operator.

However, I think you're facing yet another issue as you're trying to change the value of the client variable from inside the while loop (which executes in a subshell). I don't think that will work; see this quesion for details.

edited May 23 '17 at 11:58

Community

1
1

answered May 18 '15 at 06:05

Costi Ciudatu

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Bash problems with string comparison

3 Answers3