Error "[: too many arguments" from bash script - bash

I have a bash script:
#!/bin/bash
pathToTestCasesFile=$1
while IFS=';' read -r col1 col2
do
if [[ $col1 == V[0-9]:* ]]; then
echo "decrypt"
if [[ "$(decrypt "$col1")" == "$col2" ]] ; then
echo "$col1 is equal to $col2"
else
echo "$col1 is not equal to $col2"
fi
else
echo "encrypt"
if [[ "$(encrypt $col1)" == "$col2" ]] ; then
echo "$col1 is equal to $col2"
else
echo "$col1 is not equal to $col2"
fi
fi
done < $pathToTestCasesFile
exit 0
Here is the test file:
alex;V1:IVjd9qcAbUrR954gyPDbKw==
V1:IVjd9qcAbUrR954gyPDbKw==;alex
The output looks like this:
encrypt
alex is not equal to V1:IVjd9qcAbUrR954gyPDbKw==
decrypt
V1:IVjd9qcAbUrR954gyPDbKw== is not equal to alex
decrypt
But the output should say that everything is equal.
I am sure that after the commands de -encrypt the value is equal to the other one. I tested it separately.
Maybe there is an issue with the comparisons..
Thanks a lot for helping.

This should work:
#!/bin/bash
while IFS=, read -r col1 col2
do
echo "I got|$col1|$col2|"
if [[ "$col1" =~ V.:.* ]]; then
echo "decrypt"
if [[ "$(decrypt "$col1")" == "$col2" ]] ; then
echo "[$col1] is equal to [$col2]"
else
echo "[$col1] is not equal to [$col2]"
fi
else
echo "encrypt"
if [[ "$(encrypt "$col1")" == "$col2" ]] ; then
echo "[$col1] is equal to [$col2]"
else
echo "[$col1] is not equal to [$col2]"
fi
fi
done < "$pathToTestCasesFile"
for the field separator is used the , comma - as in your code. In the example data you have ;. If you need ; change the IFS=, to IFS=';'.

Related

Continue while loop in Bash Script even if a condition fails

#!/bin/bash
var="true"
i=1
while $var
do
read -p "Enter value (true/false): " var
if [[ $var == "true" ]]
then
echo "Iteration : $i"
((i++))
elif [[ $var == "false" ]]
then
echo "Exiting the process"
elif [[ $? -eq 1 ]]
then
echo "Invalid Choice."
echo "Avaialable Choices are true or false"
exit
fi
done
Script is Working Fine. I Enter true the loop will iterate for false the script stops.
I want the script will continue asking "Enter Value" if any other value instead of true or false will be entered.
This would do the same with a more academic syntax:
i=0
while :; do
printf 'Enter value (true/false): '
read -r var
case $var in
true)
i=$((i + 1))
printf 'Iteration : %d\n' $i
;;
false)
printf 'Exiting the process\n'
break
;;
*)
printf 'Invalid Choice.\nAvaialable Choices are true or false\n'
;;
esac
done
You might find this to be a cleaner solution:
i=0
while true; do
read -p "enter value: " myinput
if [[ $myinput = true ]]; then
echo "iteration $i"
i=$((i+1))
elif [[ $myinput = false ]]; then
echo "exiting"
exit
else
echo "invalid input"
fi;
done;
The issue I see with your current code is that it is unclear which command's exit status $? refers to. Does it refer to the echo in the previous elif block? Or the last condition check? Or something else entirely?
I'm new in bash. I tried that:
#!/bin/bash
i=1
while [[ $var != "false" ]]
do
read -p "Enter value (true/false): " var
if [[ $var == "true" ]]
then
echo "Iteration : $i"
((i++))
elif [[ $var == "false" ]]
then
echo "Exiting the process"
elif [[ $? -eq 1 ]]
then
echo "Invalid Choice."
echo "Avaialable Choices are true or false"
fi
done
I changed while $var with while [[ $var ]] because while works like if. It runs the given command. In there it is $var's value.
And I moved exit to first elif expression's end. So if user type false program will exit.

Get user input and check it, linux

I am a beginner at unix and I am trying to use a while loop to get a user integer input for 2 numbers, but I need to check if it is an integer and reask if it's not, so I attempted to utilize if statements inside the while loop. I cannot get this to work, what am I doing wrong with the while and if loop?
#! /bin/bash
echo “Enter first number:“
while read number1
do
if[[ $number1 ]] && [ $input -eq $input 2>/dev/null ]
then
echo “$number1”
else
echo "$number1 is not an integer or not defined.Try again”
fi
done
echo “Enter second number:“
while read number2
do
if[[ $number2 ]] && [ $input -eq $input 2>/dev/null ]
then
echo “$number2”
else
echo "$number2 is not an integer or not defined.Try again”
fi
done
If you are trying to get the number and checking it until it becomes integer, Please try the below code.
#!/bin/bash
echo "enter number"
while read number1
do
if ! [[ $number1 =~ ^[0-9]+$ ]]
then
echo "number is not an integer"
else
echo "number is an integer"
exit;
fi
done

Bash Script : Check if number starts with

$var=1545334345
The $var is dynamic. How do I check if it starts with 15? I tried:
if [[ $var == 15* ]]; then
echo "Yes"
fi
and
if (( $var -eq 15* )); then
echo "Yes"
fi
Neither works.
Remove the $ from the assignment.
var=1545334345
The first if statement you wrote is correct.
if [[ $var == 15* ]]; then
echo "Yes"
fi

bash check is string contains substring not working

When I execute below script it works fine:
if [[ "[1,2,3]" =~ .*1.* ]]; then
techStatus=1
else
techStatus=0;
fi
echo $techStatus
Output is 1
But when we changes it to variable it does not work.
var1=[1,2,3]
var2=1
if [[ "$var1" =~ .*"$var2".* ]]; then
techStatus=1
else
techStatus=0;
fi
echo $techStatus
Output is 0.
Please help me figure out what is wrong here.
A better & readable approach would be to convert var1 to array and loop through var1.
var1=(1 2 3)
var2=1
for elem in "${var1[#]}"; do
if [[ "$elem" -eq "$var2" ]]; then
techStatus=1
break
else
techStatus=0
fi
done
echo "$techStatus"

IF condition in bash

I have just started learning to write bash scripts. This a simplified form of what I want to write.
The problem is despite of the input, it prints only "YES".
#! /usr/bin/bash
read input
if (("$input"== "y" || "$input" == "Y"))
then
echo "YES";
elif (("$input" == "n" || "$input" == "N"))
then
echo "NO";
else
echo "Not a valid input!!"
fi
Use [[ instead of (( like,
if [[ "$input" == "y" || "$input" == "Y" ]]
and also there must be a space exists before == operator.
ie,
input="n"
if [[ "$input" == "y" || "$input" == "Y" ]]
then
echo "YES";
elif [[ "$input" == "n" || "$input" == "N" ]]
then
echo "NO";
else
echo "Not a valid input!!"
fi
You could use regular expression also for condition checking purpose.
if [[ "$input" =~ ^[yY]$ ]]
then
echo "YES";
elif [[ "$input" =~ ^[nN]$ ]]
then
echo "NO";
else
echo "Not a valid input!!"
fi
When you automaticly convert the input to lowercase (using typeset), you do not have to bother with the the uppercases.
When you use an elif, always think 10 seconds for another solution. In this case you might want to use a "switch", in shell written as a case-statement:
#!/usr/bin/bash
typeset -l input
read input
case ${input} in
"y") echo "Yes";;
"n") echo "NO";;
*) echo "Not a valid input!!";;
esac

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