Updating a variable inside a loop in bash shell

Question

Everytime I run this script, it prints 0 and then aborts with the error

./hw4_1: line 6: syntax error near unexpected token `let'
./hw4_1: line 6: `  let COUNTER=0'

Source:

#!/bin/bash
COUNTER=0
echo $COUNTER
for i in {$@:2}do
  let COUNTER=0
  while [COUNTER -ne $1]; do
      echo "$i"
      let COUNTER+=1
    done;
done
exit

I've tried getting rid of let and adding a dollar sign before COUNTER, but no combination of those things work... This bash syntax is killing me. Changing 'let COUNTER=0' to COUNTER=0 just returns the error

./hw4_1: line 6: syntax error near unexpected token `let'  
./hw4_1: line 6: `  let COUNTER=0'

rici · Accepted Answer · 2013-10-05 21:22:54Z

2

You need a semicolon (or newline) after the value list in the for statement:

for i in "${@:2}"; do

I also added the quotes because you probably want them (but maybe not).

edited Oct 5, 2013 at 21:22

answered Oct 5, 2013 at 20:42

rici

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3 Comments

Zach Over a year ago

thanks for the help. For anyone who might read this, I also had to add spaces: [COUNTER -ne $1] to [ COUNTER -ne $1 ]

rici Over a year ago

@Zach: Oh, yeah. I didn't notice that. In bash, a better version is while ((COUNTER != $1)); do (Unlike [ and {, ( is a self-delimiting character.

Zach Over a year ago

...and I also messed up my attempt at array slicing ($ sign should be outside {})

Carl Norum · Accepted Answer · 2013-10-05 20:30:07Z

0

There's no need to use let. Just use:

COUNTER=0

You will need $COUNTER elsewhere, for example:

while [ $COUNTER -ne $1 ]

and to do the increment:

COUNTER=$(($COUNTER + 1))

answered Oct 5, 2013 at 20:30

Carl Norum

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1 Comment

Zach Over a year ago

It gives the error './hw4_1: line 6: syntax error near unexpected token let' ./hw4_1: line 6: let COUNTER=0''

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