PHP isset() function returns false

Question

Please look at my code...

$sample = 'done';

$sample1 = 'welcome';

echo isset($sample) ? $sample : isset($sample1) ? $sample1 : '';

this will returns as welcome. But how variable $sample is present in this sense result would be done .. kindly help me?

Sorry. typing mistake.. please give the answer for this – Jagadeesh Feb 28 '15 at 20:46 — Jagadeesh, Feb 28 '15 at 20:46
duplicate: http://stackoverflow.com/q/28716043/3933332 – Rizier123 Feb 28 '15 at 20:48 — Rizier123, Feb 28 '15 at 20:48
exactly David, but now i used $sample1 – Jagadeesh Feb 28 '15 at 20:49 — Jagadeesh, Feb 28 '15 at 20:49

score 3 · Answer 1 · answered Feb 28 '15 at 20:49

I suspect this is a problem with the order of operations in your last line. This is, in a word, confusing:

isset($sample) ? $sample : isset($sample1) ? $sample1 : '';

If you want separate conditional clauses, explicitly separate them with parentheses:

isset($sample) ? $sample : (isset($sample1) ? $sample1 : '');

Otherwise I suspect everything before the second conditional statement is collectively evaluating to true and thus outputting $sample1.

score 0 · Answer 2 · answered Feb 28 '15 at 20:44

0

$sample = 'done';

$sample1 = 'welcome';

echo isset($sample) ? $sample : isset($sample) ? $sample : '';

You're missing a $ in front of sample1 =

answered Feb 28 '15 at 20:44

Jesper

2 Answers2