Bash script 'if' statement that calls a function with parameters

Question

I want an if statement in my bash script that does:

if [[ "$v" == "A" || my_func $x $y ]] ; then

but I get the error "conditional binary operator expected". I've tried putting quotes round the parameters in the call to my_func, but still no good. Tried playing around with eval, which didn't help either.

Thanks for any help.

[[ ]] is not part of if syntax; if takes a command, and [[ happens to be a builtin which can act as a comand, but not all things passed to if need to be within [[ ]] -- only tests and non-numeric comparisons (numeric comparisons should use (( ))). — Charles Duffy
– Charles Duffy, Commented Mar 13, 2012 at 14:30
I don't understand why you are using [[ instead of [ here. You can simply do: if [ "$v" = A ] || my_func $x $y; then ... or even 'test $v = A || my_func $x $y && ...' — William Pursell
– William Pursell, Commented Mar 13, 2012 at 14:57

kev · Accepted Answer · 2012-03-13 14:21:29Z

7

You can try:

if [[ "$v" == "A" ]] || my_func $x $y ; then

answered Mar 13, 2012 at 14:21

kev

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