CODEWITHSUNDEEP
bash
johnbakers
johnbakers

Reputation: 24771

Simple bash script; what's wrong with this syntax?

I'm reading bash script tutorials and this seems like it should work, but I clearly am missing something:

isframecount=0
framecount=0
while read p; do
  if ["$isframecount" -eq 0]
    then 
    echo $p
    $isframecount = 1
    else
    echo "frame count"
    echo $p
    $isframecount = 0
  fi
done < $filenam

I get "command not found errors" when I try to run the above; any pointers?

Upvotes: 1

Views: 80

Answers (3)

James
James

Reputation: 4716

Watch out for spaces where they matters and where there should be not. In this case, if ["$isframecount" -eq 0] should be if [ "$isframecount" -eq 0 ] (see the spaces after [ and before ]).

The reason for this is that [ is actually the name of a program... See by yourself... Type ls /bin/[... Now, if there is no space, then bash will look for a program named ["0" or something similar, which most certainly does not exist in your path.

Then, the opposite situation... There must be no space around the = in variable assignment. So $isframecount = 1 should be isframecount=1. Note that I also removed the dollar sign, that's the way to go.

Upvotes: 2

jaypal singh
jaypal singh

Reputation: 77175

Two issues:

Issue 1:

You need to have spaces in test operator. Change the following line: 

if ["$isframecount" -eq 0]

to 

if [ "$isframecount" -eq 0 ]

Issue 2:

$isframecount = 1

There should be no $ sign before variable and no spaces in the assignment operator

Change it to isframecount=1

Upvotes: 2

John Kugelman
John Kugelman

Reputation: 362037

if [ "$isframecount" -eq 0 ]

Spaces are required on both sides of the square brackets.

isframecount=1

No dollar sign, no spaces around = in an assignment statement.

Upvotes: 2

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