PHP errors. Trying to connect a form in PHP to my MySQL Database

Question

I'm trying to connect my website to a database, by adding whatever someone writes from the form to a table in my database, but so far I've only gotten errors and nothing appears on my MySQL database.

The error that I keep getting are these two:

Notice: Undefined index: fill

Warning: mysql_close() expects parameter 1 to be resource, object given in (folder)

How do I proceed to solve this problem?

<form action="insert.php" method="post">
        <label for="fill">Add:</label>
        <input type="text" name="fill">
    
    <input type="submit" value="Insert!">
</form>



<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "database";

$conn = mysqli_connect($servername,$username,$password,$dbname);


if ($conn->connect_error) {
    die ("connection failed: " . $conn->connect_error);
}

else
{
    echo 'Connected to database';
}

mysql_select_db($dbname);


$sql = "INSERT INTO card_refill ('refill') VALUES ('$_POST[fill]')";


mysql_close($conn)

?>

You're mixing mysql_* and mysqli_* functions. That won't work. — Jay Blanchard
– Jay Blanchard, Commented Oct 12, 2015 at 16:32
$conn is mysqli object. You are trying to close as mysql object — Subin Thomas
– Subin Thomas, Commented Oct 12, 2015 at 16:35
How do I change it so that It'll work? I tried changing the mysql functions over to mysqli functions and it didnt work. — PHP Scrub
– PHP Scrub, Commented Oct 12, 2015 at 16:42
You need to wrap the word 'fill' in your $_POST with single quotes. — Filthy_Rich
– Filthy_Rich, Commented Oct 12, 2015 at 17:24
@Filthy_Rich Ive already done that but i keep getting the error: Notice: Undefined index: fill — PHP Scrub
– PHP Scrub, Commented Oct 12, 2015 at 18:15

Mm-Art-In · Accepted Answer · 2015-10-12 19:44:02Z

1

Since the dbname is already provided in mysqli_connect function, mysql_select_db($dbname) is not necessary. As for the execution of the query, SQL string should be run via mysqli_query function. However before running SQL queries, the variables need to be sanitised to prevent SQL injections and XSS attacks.

Try something like this:

$conn = mysqli_connect($servername,$username,$password,$dbname);


if ($conn->connect_error) {
    die ("connection failed: " . $conn->connect_error);
}else{
    echo 'Connected to database';
}

$sql = "INSERT INTO card_refill ('refill') VALUES ({$_POST['fill']})";

$result = mysqli_query($conn,$sql);
//do whatever you need to do with the result

mysqli_close($conn);

edited Oct 12, 2015 at 19:44

Mm-Art-In

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answered Oct 12, 2015 at 16:53

kunoichi

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8 Comments

kunoichi Over a year ago

by the way you should never use post data directly in your queries.

Jay Blanchard Over a year ago

Why should the OP "try this"? A good answer will always have an explanation of what was done and why it was done that way, not only for the OP but for future visitors to SO.

kunoichi Over a year ago

Ok, I'll clarify. Since the dbname is already provided in mysqli_connect function, mysql_select_db($dbname) is not necessary. As for the execution of the query, sql string should be run via mysqli_query function. However before running sql queries, the variables need to be sanitized to prevent sql injections.

kunoichi Over a year ago

One more thing, you should use quotation marks around the key string to access array members. ex: $_POST['fill'] rather than $_POST[fill]

kunoichi Over a year ago

I'll edit my post but please try not to "dictate" next time.

|

Subin Thomas · Accepted Answer · 2015-10-12 19:00:31Z

1

This database insert part should be done only if you press the submit of form. So use isset to detect submit.

Put all the php code inside a if loop.

if(isset($_POST['fill']))   //execute only if fill is set
{
$conn = mysqli_connect......;
// all your codes here
mysqli_close($conn);
}

answered Oct 12, 2015 at 19:00

Subin Thomas

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