Comparing two variables. Second contains first [closed]

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var_1="hi"
var_2="bye,hi,hello"
if [[ "$var_1" == *"$var_2"* ]];
then
   echo "value of var_1 in var_2";
else
   echo "sad =(";
fi

I would like check second variable contains first. That prints "Sad =(", why?

It's a typo: you should revert var1 and var2 as it's var_1 that should be into var_2 — Elikill58
– Elikill58, Commented Feb 28, 2023 at 13:49

choroba · Accepted Answer · 2023-02-28 13:50:36Z

0

The condition checks that $var_1 consists of $var_2, possibly preceded and followed by something. But it's the other way round.

if [[ $var_2 = *"$var_1"* ]]
#          ~          ~

answered Feb 28, 2023 at 13:50

choroba

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5 Comments

Elikill58 Over a year ago

it should be == not = but yes (it's a typo question)

choroba Over a year ago

@Elikill58: Why? They both work (and it's documented).

Elikill58 Over a year ago

it will change the variable, no ?

tripleee Over a year ago

@Elikill58 No. Read choroba's comment again.

markp-fuso Over a year ago

@Elikill58 perhaps you're thinking of awk where x=3 is an assignment and x==3 is a test/conditional ... ?

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