Error in array/IndexOf

[Jakub Grufik]

Tried also copy your code and it works fine for me

[Jakub Grufik]

Hey for me it works just fine...

[yuri]

The version is required in bug reports. https://forum.espocrm.com/forum/bug-...o-report-a-bug I move the post out of bug reports.

[yuri]

$index = array\indexOf(list(0, 1), 2);

if ($index == null) {
    output\print('is null');
}​

prints "is null"

[esendino]

Hi.

When it doesn't find the value in the array it is not returning null. Is returning something else.

To test this I wrote this script

$testList = list(1,2,3,4,5,6);

$verify = 9;

$position = array\indexOf($testList, $verify);


$answer = 'unanswered';
ifThen($position > 0,
$answer = '>0');

ifThen($position == 0,
$answer = '0');

ifThen($position == null,
$answer = 'null');

ifThen($position == '',
$answer = 'nothing');


//********** INICIO LOG DE PRUEBAS
$tstDescription = null;
$tstName = "Test 1 ";
$tstDescription = string\concatenate(
' $position ', $position ,
' $verify ', $verify ,
' $answer ', $answer ,
' $testList ', array\join($testList, '-') ,
" ");
$tstLogId = record\create('LogPruebas',
'name', $tstName,
'description', $tstDescription);
//*********** FIN LOG DE PRUEBAS


I ran this script changing the line with the verify line to

$verify = 1;
$verify = 3;
$verify = 9;

And he output was:



So when $verify = 9; ir shows $position apparentely as a null, but when I ask if $position == null, it doesn't make $answer = 'null'