Integer expression expected BASH scripting IF statement

Question

I am trying to insert names and numbers in a text file. I have wrote a short script for the learning purpose.

v= expr $# % 2
echo $v
if [ "$v" -eq 0 ]; then
    i=1
    while [ "$i" -lt $# ]
    do
        echo "$i    $i+1" >> database
        i=$((i+1))
    done
echo "User(s) successfully added \n\v"
else
    echo "Arguments are not complete";
fi

When i enter two arguments, the shell output is as follows

0                     # (The value of variable v)
./myscript: line 3: [: : integer expression expected
Arguments are not complete  # (else statement is executed)

When i replace -eq to == in line 3 (if statement), error msg is gone but still the IF statement doesn't execute as i expect.

0              # (output of variable v)
Arguments are not complete # (else statement is executed)

chepner · Accepted Answer · 2016-10-16 13:23:46Z

3

You need to enclose the variable assignment in $(...) ("command substitution"):

v=$(expr $# % 2)

In the if statement, -eq should be correct. Also, to make sure it works, I would use double square brackets (this might depend on the shell you use):

if [[ ${v} -eq 0 ]]; then

edited Oct 16, 2016 at 13:23

chepner

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answered Oct 16, 2016 at 10:33

deefff

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2 Comments

Nouman Tajik Over a year ago

Thank you, I have added double brackets in the IF statement and it worked. Kindly explain whats the difference in the single and double brackets in if statement. I also want to read the arguments passed to the script. I want to read it in such manner e.g for i = 1, $($i) should read $1 (argument first).

deefff Over a year ago

The difference between single and double brackets is nicely explained here: (serverfault.com/questions/52034/…) Your idea for the for loop is almost complete, you just need a sequence to iterate over, you can also do this with command substitution: for i in $(seq 1 $#); do <code> ; done

chepner · Accepted Answer · 2016-10-16 13:25:18Z

0

The immediate problem is the failure to use command substitution to capture the result of the expr command: v=$( expr $# % 2 ). However, expr is no longer needed for arithmetic; use an arithmetic expression just as you did to increment i.

v=$(( $# % 2 ))

answered Oct 16, 2016 at 13:25

chepner

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