[5.3] php warning in BackupcodesModel (#45170)

_this is an AI generated explanation_

The warning Only the first byte will be assigned to the string offset occurs because $temp1 is being treated as a string, and assigning a value to $temp1[$i] attempts to modify a specific character (byte) of the string. However, the value being assigned (random_int(0, 99999999)) is an integer, which is incompatible with this operation.

Root Cause:
In PHP, when you use array-like syntax (e.g., $temp1[$i]) on a variable that has not been explicitly declared as an array, PHP may treat it as a string. In this case, $temp1 is being treated as a string, and $temp1[$i] is interpreted as modifying a specific character of the string. Since random_int(0, 99999999) generates an integer, PHP raises a warning because you cannot assign an integer to a specific byte of a string.

Fix:
To resolve this issue, you need to explicitly declare $temp1 as an array before using it. This ensures that $temp1[$i] is treated as an array element rather than a string offset.

Explanation of the Fix:
Explicit Declaration:

$temp1 = []; ensures that $temp1 is treated as an array from the start.
This avoids PHP's automatic type juggling, which might treat $temp1 as a string if it is not explicitly declared.
Assigning Values:

$temp1[$i] = random_int(0, 99999999); now correctly assigns the random integer to the $ith index of the $temp1 array.

Summary:
This fix ensures that $temp1 is treated as an array, eliminating the warning and allowing the code to function as intended.

Signed-off-by: BrianTeeman <[email protected]>

Author: brianteeman

administrator/components/com_users/src/Model/BackupcodesModel.php

@@ -252,7 +252,7 @@ public function isBackupCode($code, ?User $user = null): bool
          */
         $otherResult = false;
 
-        $temp1 = '';
+        $temp1 = [];
 
         for ($i = 0; $i < 10; $i++) {
             $temp1[$i] = random_int(0, 99999999);