Can you append strings to variables in PHP?

PhpStringOperators

Php Problem Overview

Why does the following code output 0?

It works with numbers instead of strings just fine. I have similar code in JavaScript that also works. Does PHP not like += with strings?

<?php
    $selectBox = '<select name="number">';
    for ($i=1; $i<=100; $i++)
    {
        $selectBox += '<option value="' . $i . '">' . $i . '</option>';
    }
    $selectBox += '</select>';

    echo $selectBox;
?>

Php Solutions

Solution 1 - Php

This is because PHP uses the period character . for string concatenation, not the plus character +. Therefore to append to a string you want to use the .= operator:

for ($i=1;$i<=100;$i++)
{
    $selectBox .= '<option value="' . $i . '">' . $i . '</option>';
}
$selectBox .= '</select>';

Solution 2 - Php

In PHP use .= to append strings, and not +=.

> Why does this output 0? [...] Does PHP not like += with strings?

+= is an arithmetic operator to add a number to another number. Using that operator with strings leads to an automatic type conversion. In the OP's case the strings have been converted to integers of the value 0.

More about operators in PHP:

Solution 3 - Php

PHP syntax is little different in case of concatenation from JavaScript. Instead of (+) plus a (.) period is used for string concatenation.

<?php

$selectBox = '<select name="number">';
for ($i=1;$i<=100;$i++)
{
    $selectBox += '<option value="' . $i . '">' . $i . '</option>'; // <-- (Wrong) Replace + with .
    $selectBox .= '<option value="' . $i . '">' . $i . '</option>'; // <-- (Correct) Here + is replaced .
}
$selectBox += '</select>'; // <-- (Wrong) Replace + with .
$selectBox .= '</select>'; // <-- (Correct) Here + is replaced .
echo $selectBox;

?>
Categories
Recommended Php Solutions
Recommended String Solutions
Recommended Operators Solutions

Previous Solution:

Android Share Intent for a Bitmap - is it possible not to save it prior sharing?

AndroidAndroid Intent

Next Solution:

How to use andWhere and orWhere in Doctrine?

PhpSqlDoctrine

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJamesView Question on Stackoverflow
Solution 1 - PhpJeremyView Answer on Stackoverflow
Solution 2 - PhpHenrikView Answer on Stackoverflow
Solution 3 - PhpAli HaiderView Answer on Stackoverflow