Bash script: test command -o option

Question

I have a piece of code like this:

count=0
until [ -f $filename -o $count -ge 60 ]
do
  count=$((count+1))
  sleep 1
done

I know [ -f $filename -o $count -ge 60 ] is a test command, "-f $filename" tests if the file exists or not, "$count -ge 60" tests if count is greater than or equal to 60, what does "-o" mean?

I googled, can't find the answer, anyone can help?

Try the help command for builtins: help test (help [ will tell you it's actually test). — muru
– muru, Commented Sep 22, 2014 at 18:53

cuonglm · Accepted Answer · 2014-09-22 18:09:51Z

2

It's logical OR operator.

From bash documentation:

expr1 -o expr2
    True if either expr1 or expr2 is true.

edited Sep 22, 2014 at 18:09

answered Sep 22, 2014 at 18:04

cuonglm

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Michael Mrozek · Accepted Answer · 2014-09-22 22:34:37Z

0

It is a logical or operation. The code checks if $filename exists once/second for 60 seconds.

edited Sep 22, 2014 at 22:34

Michael Mrozek

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answered Sep 22, 2014 at 21:00

hardeep

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