CODEWITHSUNDEEP
bashshell
user2886545
user2886545

Reputation: 713

the code ignore the if/else statement in bash

I don't know if you can help me... I have a problem with the following code:

#!/bin/bash

if [ $# = 0 ]; then
    awk '{for (x=5; x<=NF; x++) {printf "%s ", $x } printf "\n" }' affy.txt
else
    columns=""
    for i in $@; do
        if [ $i = "-g" ]; then
            word=$2
            is_there_g="True"
            break
        else
            is_there_g="False"
        fi
        columns+="$i,"
        shift
    done
    columns=${columns%,}
    if [ $is_there_g="False" ]; then
        cat affy.txt | cut -d" " -f$columns 
    else
        grep $word affy.txt | cut -d" " -f$columns
    fi
fi

echo $is_there_g

So, there's a problem obtaining the results I want. It's about the last If/else written.

When $is_there_g="False", I wanna do this: "cat affy.txt | cut -d" " -f$columns".

And if it's True, then: "grep $word affy.txt | cut -d" " -f$columns".

I assess it using "echo $is_there_g" at the end of my code .

So the problem is that the results I obtain are always for a False case even though "echo $is_there_g" gives me a True value. Why???

Upvotes: 0

Views: 1582

Answers (1)

Carl Norum
Carl Norum

Reputation: 225032

You need spaces around the operator:

if [ $is_there_g = "False" ]; then

man bash or man test for more information.

Upvotes: 3

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