PHP "if" , "or" variable equals to

Question

Looking for a bit of PHP help here please..

I am looking to check to see if the contents of a few variables match defined criteria, and if so, to set the value of an additional variable accordingly.

So, I was thinking the below should work... but it has syntax errors which confuse me!

if ($var1 == "chicken" || $var == "banana") { $outputVar = "fail" } else { $outputVar = "success" }

What is wrong with the above?

Thanks

Error I get when testing on phpcodechecker.com is "PHP Syntax Check: Parse error: syntax error, unexpected '}' in your code on line 1" — dubbs, Aug 19 '16 at 11:50
`if (...) { $outputVar = "fail"; } else { $outputVar = "success"; }` — Steve, Aug 19 '16 at 11:51
[*This is your friend during testing/debugging...*](http://php.net/manual/en/function.error-reporting.php) and [*another good read...*](http://php.net/manual/en/language.basic-syntax.instruction-separation.php) — Funk Forty Niner, Aug 19 '16 at 11:54

score 1 · Accepted Answer · answered Aug 19 '16 at 11:52

1

You haven't put semicolons in any of your statements. It will solve your problem.

if ($var1 == "chicken" || $var == "banana") { $outputVar = "fail"; } else { $outputVar = "success"; }

answered Aug 19 '16 at 11:52

Imesha Sudasingha

3,462
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thanks yep - didnt realise they were needed after declaring the outputvar – dubbs Aug 19 '16 at 11:52

score 0 · Answer 2 · answered Aug 19 '16 at 11:47

0

You are evalutaing ($var and $var1 ) this is true? ... and you don't have semicolon at the end

if ($var1 == "chicken" || $var == "banana") {
    $outputVar = "fail" ;
 } else {
   $outputVar = "success" ; 
}

answered Aug 19 '16 at 11:47

ScaisEdge

131,976
10
91
107

PHP "if" , "or" variable equals to

2 Answers2