CODEWITHSUNDEEP
php
Norman
Norman

Reputation: 6365

How can I solve Fatal error: Cannot use [] for reading in this case

I get the error Fatal error: Cannot use [] for reading on line at return $array[]; inside the function.

I'm using PHP Version 7.0.1 on Windows 10.

I'm trying to get the output as follows:

<i>Error Line One...</i><br>
<i>Error Line Two...</i><br>

, in json_encode

{"s":"Error Line One...<\/i>Error Line Two...<\/i>","success":false}

I previously used a different approach, but switched to this yesterday.

How can I get this work?

$errorCount = 1;

$errors[] = '*Error Line One...';
$errors[] = '*Error Line Two...';

$errors['success'] = ($errorCount == 0 ? True : False);

$errors[] = ajax($errors);

function ajax($array) {
    $array = preg_replace('#\*(.+?)(?![^*])#','<i>$1</i><br>',$array);
    return $array[];
}

json_encode($errors);

Below is how it was done previously, then changed to the above to avoid the .= parts. This approach works. See Fiddle.

$errorCount = 1;

$errors['s'] = '*Error Line One...';
$errors['s'] .= '*Error Line Two...';


$errors['s'] = ajax($errors);

function ajax($array) {
    $array = preg_replace('#\*(.+?)(?![^*])#','<i>$1</i>',$array);
    return $array['s'];
}
$errors['success'] = ($errorCount == 0 ? True : False);

echo json_encode($errors);

Upvotes: 0

Views: 750

Answers (1)

Mark van Herpen
Mark van Herpen

Reputation: 347

You are returning $array[] instead of $array,

Change this:

function ajax($array) {
    $array = preg_replace('#\*(.+?)(?![^*])#','<i>$1</i><br>',$array);
    return $array[];
}

into this:

function ajax($array) {
    return preg_replace('#\*(.+?)(?![^*])#','<i>$1</i><br>',$array);
}

edit:

This fixes it I guess:

<?php

$errorCount = 1;

$errors[] = '*Error Line One...';
$errors[] = '*Error Line Two...';
$errors = ajax($errors);
$errors['success'] = ($errorCount == 0 ? True : False);

function ajax($array) {
    foreach ($array as $key => $value)
        $array[$key] = preg_replace('#\*(.+?)(?![^*])#','<i>$1</i><br>', $value);
    return $array;
}

echo json_encode($errors);

?>

Upvotes: 1

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