Failing empty check

Question

I have simple PHP code like this

$revenue = $this->dailyCounts['revenue'][$date]['net_revenue'] + (!empty($activity[$date]) ? $activity[$date] : 0) * 1000;

But here it is always showing warnings when the $activity variable is NULL even though I am checking for empty. I am running this inside a loop and it is always showing undefined index when $activity has null value.

score 1 · Answer 1 · answered Oct 27 '15 at 07:10

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Using isset instead of empty won't solve any problem in that case.

No warning is generated if the variable does not exist. That means empty() is essentially the concise equivalent to !isset($var) || $var == false.

http://php.net/manual/en/function.empty.php

I guess the notice comes from $this->dailyCounts['revenue'][$date]['net_revenue']

answered Oct 27 '15 at 07:10

Theo SCHAEFFER

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No it is not. When I added the isset() check , it has gone. I am using this in Laravel 5 – Happy Coder Oct 27 '15 at 08:26
@HappyCoder: can you paste your fixed code? This shows that both do not produce warnings and give the same results: http://codepad.org/HjgrZ3RZ – HectorJ Oct 27 '15 at 08:39

score 0 · Answer 2 · edited May 23 '17 at 10:28

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use isset() instead of empty(),

(!isset($activity[$date])

empty() will directly check the value, but isset() will see if the index is there or not.

See this and this for more information, its explained nicely.

edited May 23 '17 at 10:28

Community

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answered Oct 27 '15 at 05:14

Niranjan N Raju

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with a `null` array, `isset` and `!empty` give the same result – HectorJ Oct 27 '15 at 07:45

score 0 · Answer 3 · edited Oct 27 '15 at 07:11

0

Try this :

$revenue = $this->dailyCounts['revenue'][$date]['net_revenue'] + ((isset($activity[$date]) && !empty($activity[$date])) ? $activity[$date] : 0) * 1000;

edited Oct 27 '15 at 07:11

Piotr Olaszewski

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answered Oct 27 '15 at 05:17

Bipin Kareparambil

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Failing empty check

3 Answers3