PHP Checkbox checked is not working correctly when comparing 2 Arrays

Question

I trying to compare 2 Arrays and then let the Checkbox to be checked if there are common values between them.

coding

$array_1[]="value1";
$array_1[]="value2";
$array_1[]="value3";
$array_1[]="value4";
$array_1[]="value5";
$array_1[]="value6";
$array_1[]="value7";
$array_1[]="value8";

$array_2[]="value1";
$array_2[]="value3";
$array_2[]="value4";


for($i=0;$i<count($array_1);$i++){
    $checked = isset($array_2[$i])? 'checked' : '';
    echo '<input type="checkbox" ' . $checked .' name="zzz[] "value="'.$array_1[$i].'"> '.$array_1[$i].'<br>';
}

as you can see from the screenshot, the correct result should be right side with value1, value3 and value4. However, my php output is leftside of the screenshot

Anyone knows what's wrong ?

Maxim Krizhanovsky · Accepted Answer · 2016-11-08 07:45:34Z

1

Since you are working on values, and not on keys, you should use in_array check

for($i=0;$i<count($array_1);$i++){
    $checked = in_array($array_1[$i], $array_2) ? 'checked' : '';
    echo '<input type="checkbox" ' . $checked .' name="zzz[] "value="'.$array_1[$i].'"> '.$array_1[$i].'<br>';
}

Alternatively, you can set the keys to be the same, as Nadir Latif suggests

answered Nov 8, 2016 at 7:45

Maxim Krizhanovsky

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Nadir Latif · Accepted Answer · 2016-11-08 07:43:28Z

1

The value2 array initialization is wrong. It should be:

$array_2[1]="value1";
$array_2[3]="value3";
$array_2[4]="value4";

answered Nov 8, 2016 at 7:43

Nadir Latif

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