php Data not stored in mySQL database using XAMPP

Question

I have this code and I am using the latest version of XAMPP:

filename: store.html

<!DOCTYPE html>

<html>
<body>

<form action="store.php" method="post">
User input: <input type="text" name="userinput"><br>
<input type="submit">
</form>

</body>
</html>

filename: store.php

<?php

$servername = "localhost";
$username = "root";
$password = "";
$dbname = "database";


// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);

// Check connection
if ($conn->connect_error) {
     die("Connection failed: " . $conn->connect_error);
 } 
 echo "Connected successfully";

$input = $_POST["userinput"];


$sql = "INSERT INTO table_1 (s_num)
VALUES ('$input')";


 ?>

Whatever I do, no data is added to the database. Please help. Thank you.

score 2 · Accepted Answer · edited May 23 '17 at 12:15

2

The problem here is that you never executed the query.

$sql = mysqli_query($conn,"INSERT INTO table_1 (s_num)
VALUES ('$input')");

if(!sql){

   echo "Error: " . mysqli_error($conn);

}

else{

   echo "Success";

}

Reference:

http://php.net/manual/en/mysqli.query.php

Your present code is open to SQL injection if user-input (other than yourself) ever gets involved.

Use prepared statements, or PDO with prepared statements, they're much safer.

Plus, seeing you are running this from your own machine, make sure you are accessing as http://localhost/file.php as opposed to file:///file.php.

edited May 23 '17 at 12:15

Community

1
1

answered Mar 03 '16 at 22:27

Funk Forty Niner

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thanks! it worked! i was wondering about this symbol "->" what does that mean? im quite new to php.. – Juan dela Cruz Mar 03 '16 at 22:46
@JuandelaCruz You're welcome. Do you mean the `->` in `$conn->connect_error`? If so, that is the syntax for an object oriented method. – Funk Forty Niner Mar 03 '16 at 22:58

score 1 · Answer 2 · answered Mar 03 '16 at 22:27

You are not executing your sql insertcode. After this line:

$sql = "INSERT INTO table_1 (s_num)
VALUES ('$input')";

Add these line:

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();

php Data not stored in mySQL database using XAMPP

2 Answers2