Compare 1 with another string in bash

Question

I've spent 2 hours with an if statement, that never works like I want:

#should return true
if [ "$1" == "355258054414904" ]; then 

Here is the whole script:

#!/bin/bash

param=$1
INPUT=simu_900_imei_user_pass.csv
OLDIFS=$IFS
IFS=,
[ ! -f $INPUT ] && { echo "$INPUT ime not found"; exit 99; }
while read imei email pass 
do
    echo "First Parameter-IMEI: $1"
    if [ "$1" == "355258054414904" ]; then
        echo "GOOD"
    fi
done < $INPUT
IFS=$OLDIFS

This is the output of the script:

First Parameter-IMEI: 355258054414904
First Parameter-IMEI: 355258054414904
First Parameter-IMEI: 355258054414904

I have seen a lot of pages about the subject, but I can't make it work :(

EDIT: I Join the content of csv for better understanding ! Tx for your help !

4790057be1803096,user1,pass1
355258054414904,juju,capp
4790057be1803096,user2,pass2
358854053154579,user3,pass3

Answer 1

The explanation $1 doesn't match is on the grounds that $1 implies the main boundary given to the content on the order line, while you believe it should match the principal field read from the record. That worth is in $imei.

You most likely implied:

on the off chance that [ "$imei" == "355258054414904" ];
    reverberation "Great"
fi
Since it is inside the circle where you read input record line by line.

To actually look at content of $1 use:

feline - vet <<< "$1"
UPDATE: To strip \r from $1 have this at top:

param=$(tr - d '\r' <<< "$1")
And afterward use "$param" in rest of your content.