if with mulitiple empty checks and OR

Question

i'm wondering why this condition dosn't work

if(!empty($b['boxText1']) || !empty($b['boxText2']) || !empty($b['boxText3']))) {
 // do something
}

i get this error:

Parse error: syntax error, unexpected '!' in

am i missing something?

best

You are probably missing to take a look here first: [Reference - What does this error mean in PHP?](http://stackoverflow.com/q/12769982/367456) ;) — hakre, Feb 20 '13 at 10:05

score 2 · Answer 1 · answered Feb 20 '13 at 09:45

2

Try this..........you typed extra ')'.....

if(!empty($b['boxText1']) || !empty($b['boxText2']) || !empty($b['boxText3'])) {
 // do something
}

answered Feb 20 '13 at 09:45

Venkata Krishna

4,287
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score 1 · Answer 2 · answered Feb 20 '13 at 09:45

1

if(!empty($b['boxText1']) || !empty($b['boxText2']) || !empty($b['boxText3']) ) ) {
                                                                                ^

Remove the extra bracket you have used.

answered Feb 20 '13 at 09:45

Rikesh

26,156
14
79
87

score 0 · Answer 3 · answered Feb 20 '13 at 09:47

0

if(!empty($b['boxText1']) || !empty($b['boxText2']) || !empty($b['boxText3'])) {
 // do something
}

answered Feb 20 '13 at 09:47

Prasanth Bendra

31,145
9
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73

if with mulitiple empty checks and OR

3 Answers3