A variable is not being set in my PHP code

Question

I am having issues submitting content from my form to my database. When using isset function, it showing that the variables are not being set.

Here is my PHP code and html code.

PHP

<?PHP
$title = $_POST['title']['name']; 
if(isset($title)) {
    echo 'all is well';
} else {
    echo 'all is not well';
}

?>

HTML

<div class="container">
    <form action="newEmptyPHP.php" method="post" enctype="multipart/form-data">
        <input type="text" name="title" placeholder="title">
        <input type="text" name="artist" placeholder="artist">
        <input type="file" name="cover" placeholder="Upload Picture">
        <input type="file" name="song" placeholder="Upload Song">
    </form> 
</div>

When I refresh the browser I am receiving "all is not well". What am I missing?

Also, you don't need to use isset, just $title is fine. Like this: if ($title) — frosty
– frosty, Commented Mar 12, 2016 at 0:14
You can cook it down with usage of a ternary operator, which is a one-liner for that, like this: echo (isset($_POST['title']) ? "All is well!" : "All is not well.."); -- your title in the form isn't an array, so you must use $_POST['title'] and nothing more to get that value. — Qirel
– Qirel, Commented Mar 12, 2016 at 0:17

BLiu1 · Accepted Answer · 2016-03-12 00:31:27Z

1

In the PHP, $_POST['title'] should be used to get the value of the title input.

Also, you should always check if the variable in $_POST is set before assigning it to a variable.

<?php

if(isset($_POST['title'])) {
    $title = $_POST['title'];
}

if($title) {
    echo 'all is well';
}else{
    echo 'all is not well';
}

?>

answered Mar 12, 2016 at 0:31

BLiu1

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Comments

Panda · Accepted Answer · 2016-03-12 00:42:43Z

1

This line caused an error: $title = $_POST['title']['name'];.

$_POST should be followed by the HTML name attribute, in your case, name="title".

Also, you should check if the form is posted before assigning it to a variable.

Thus, it should be:

if(isset($_POST['title'])) {
    $title = $_POST['title'];

answered Mar 12, 2016 at 0:42

Panda

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