Error in running shell script while loop

Question

I wrote this shellscript file. But I get error near unexpected tokendone'`

#!/bin/bash
i=1
while [ $i -lt 12 ]; do
    echo Hi
    i=$[$i+1]
done

Previously, there was no ; before do. I read stackexchange answers and wrote ; after while [], still I get error. I could not found the resolution online. Any ideas?

Are you sure your script is run by bash? If you are executing it with sh myscript, and sh isn't a link to bash, your script is being run by some other shell (probably dash) which doesn't support bash's ancient $[...] syntax. — chepner
– chepner, Commented Nov 26, 2014 at 18:29

Jakub Kotowski · Accepted Answer · 2014-11-25 23:56:45Z

1

i=$[$i+1] this is wrong. You probably meant i=$((i+1))

answered Nov 25, 2014 at 23:56

Jakub Kotowski

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2 Comments

chepner Over a year ago

It's not so much wrong as antiquated and undocumented. $[...] was the bash notation for arithmetic expressions before POSIX $((...)) was adopted.

Jakub Kotowski Over a year ago

ah, nice, I didn't know

neveragny · Accepted Answer · 2014-11-25 23:56:50Z

0

path to bash is incorrect, you have missed a root in path change first line to

#!/bin/bash

answered Nov 25, 2014 at 23:56

neveragny

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1 Comment

b2850624 Over a year ago

fixed that. But still I have problem

user2243670 · Accepted Answer · 2014-11-26 07:46:05Z

0

Correct placement of ; in a bash script can be tricky indeed. Here's a one-liner version of your script with semicolons in the right place:

i=1; while (( i < 12 )); do echo "Hi"; (( i++ )); done

Note that the $ in variables becomes redundant for integer comparisons if placed between double parentheses (( )).

answered Nov 26, 2014 at 7:46

user2243670

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Comments

tvCa · Accepted Answer · 2014-11-26 17:56:23Z

0

Works perfectly

$ ./ttt
Hi
Hi
Hi
Hi 
Hi
Hi
Hi
Hi
Hi
Hi 
Hi
$ cat ttt
#!/bin/bash
i=1
while [ $i -lt 12 ]; do
    echo Hi
    i=$[$i+1]
done  
$

answered Nov 26, 2014 at 17:56

tvCa

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