CODEWITHSUNDEEP
phparrays
ariel
ariel

Reputation: 389

Variables not going to array

I got problem with array, this is my code:

$random_string = array();
for($i=0;$i<10;$i++)
{
$zmienna = generatePassword();
//echo $zmienna;
if (in_array($zmienna, $random_string))
$random_string[$i] = $zmienna;
}

var_dump($random_string);

on screen i see olny

array(0) { }

What's wrong? Becouse I'm sure that generatePassword(); working well

Upvotes: 0

Views: 53

Answers (3)

seeker
seeker

Reputation: 3333

In_array function always returns false In this code statement - its does not add to array, just checks. That's why he is mistake.

Upvotes: 1

andrewsi
andrewsi

Reputation: 10732

This code:

if (in_array($zmienna, $random_string)) 
    $random_string[$i] = $zmienna;

Will add $zmienna to $random_string if it's already in there; since it's a blank array at the start of the code, it's never going to be added, so the array will stay empty.

Do you mean:

if (! in_array($zmienna, $random_string)) {

Which will add it if it's not already there?

Upvotes: 4

sachleen
sachleen

Reputation: 31131

This line fails.

if (in_array($zmienna, $random_string))

$random_string is empty so $zmienna will never be "in" it.

I think you might have meant to put a NOT operator in there? if it is not in the array, add it? If so:

if (!in_array($zmienna, $random_string))

Also, your array will skip some keys. You'll end up with something like 

array(3) {
  [0]=>
  string(1) "a"
  [2]=>
  string(1) "b"
  [4]=>
  string(1) "c"
}

If you don't specify an index with $i (just do $random_string[] = $zmienna;), it'll do the keys automatically, so you get

array(3) {
  [0]=>
  string(1) "a"
  [1]=>
  string(1) "b"
  [2]=>
  string(1) "c"
}

Upvotes: 5

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