bash while [: too many arguments

Question

#!/bin/bash

number=0
while [ $number -lt 10 ];do
echo "$number"
number='expr $number + 1'
done
echo 'script complete!'

=> this results in

0
./while.sh: line 4: [: too many arguments
script complete!

I've tried ...

while [ $number -lt 10 ]

to

while [ '$number' -lt 10 ]

and

while [[ $number -lt 10 ]]

which doesn't work at all..

vidit · Accepted Answer · 2015-08-08 06:56:37Z

2

You are using wrong quotes for expression evaluation; single quote(') instead of backquote(`). Try this..

number=`expr $number + 1`

answered Aug 8, 2015 at 6:56

vidit

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2 Comments

chepner Over a year ago

Two points about the original: in general you should use number=$(expr $number + 1) in place of backquotes, and you should use number=$(( number + 1)) in place of expr.

user2350426 Over a year ago

Or, even better, ((number++)). Option added to answer. :)

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