PHP fails to post to MySQL Database

Question

I have a formText.php file that contains a form with the following code form code:

<form action="insert.php" method="post">
    <p>
        <label for="theNames">Name:</label>
        <input type="text" name="theName" id="theName">
    </p>
    <p>
        <label for="theCitys">City:</label>
        <input type="text" name="theCity" id="theCity">
    </p>
    <p>
        <label for="theAges">Are you over eighteen?(Y/N)</label>
        <input type="text" name="theAge" id="theAge">
    </p>
    <p>
    <label for="theDates">Date:</label>
    <input type="text" name="theDate" id="theDate">
    </p>
    <input type="submit" value="Submit">
</form>

Then I have an insert.php file with the following script:

<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "root","phpteste");

// Check connection
if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

// Escape user inputs for security (EDITED)
$theName = mysqli_real_escape_string($link, $_POST['theName']);
$theCity = mysqli_real_escape_string($link, $_POST['theCity']);
$theAge = mysqli_real_escape_string($link, $_POST['theAge']);
$theDate = mysqli_real_escape_string($link, date("Y-m-d h:i:s",$_POST['theDate']));


// attempt insert query execution
$sql = "INSERT INTO tabelateste (id, name, city, overeighteen, date) VALUES (NULL, '$theName', '$theCity', '$theAge', '$theDate')";
if(mysqli_query($link, $sql)){
    echo "Records added successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// close connection
mysqli_close($link);
?>

My database is called phpteste and my table name is tabelateste. What am I doing wrong here? Whenever I click Submit nothing comes up and nothing gets added to the database.

First verify that the form generated by formText.php is sending the proper data to insert.php by replacing insert.php with a test script that simply dumps/renders the data POSTed. Then verify that insert.php can actually modify the database by hardcoding it to insert a dummy record in your database. Then get them to work together. — Peter
– Peter, Commented Jan 2, 2016 at 5:56
your column name is signup_date but in your insert it is date — myselfmiqdad
– myselfmiqdad, Commented Jan 2, 2016 at 7:03

Hardik Solanki · Accepted Answer · 2016-01-02 06:13:04Z

2

Your post data name fields are wrong. SO you need to change below line:

// Escape user inputs for security
$theName = mysqli_real_escape_string($link, $_POST['theName']);
$theCity = mysqli_real_escape_string($link, $_POST['theCity']);
$theAge = mysqli_real_escape_string($link, $_POST['theAge']);
$theDate = mysqli_real_escape_string($link, date("Y-m-d h:i:s",$_POST['theDate']));

You need to change date to signup_date as per your database table structure.

$sql = "INSERT INTO tabelateste (name, city, overeighteen, signup_date) VALUES ('$theName', '$theCity', '$theAge', '$theDate')";

edited Jan 2, 2016 at 6:13

answered Jan 2, 2016 at 5:53

Hardik Solanki

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9 Comments

Konstantin Kulakov Over a year ago

Chanage sql to: $sql = "INSERT INTO tabelateste (name, city, overeighteen, date) VALUES ('$theName', '$theCity', '$theAge', '$theDate')";

Hardik Solanki Over a year ago

Did you get any error while data inserted into database? Or echo your query and try to run it directly into database.

Johnny Over a year ago

Still did not work. I had the NULL in there because the id was a NULL value. But yet, no luck. Have no idea what I might be doing wrong.

Hardik Solanki Over a year ago

What is your database structure? Is id auto increment and primary key?

Johnny Over a year ago

Yes it is. tabelateste (id INT NOT NULL PRIMARY KEY AUTO_INCREMENT, name VARCHAR(20), city VARCHAR (20), overeighteen CHAR(1), signup_date DATE);

|

Vinothkumar Nataraj · Accepted Answer · 2016-01-02 06:29:24Z

0

$sql = "INSERT INTO tabelateste (`name`, `city`, `overeighteen`, `date`) VALUES ('$theName', '$theCity', '$theAge', '$theDate')";

Use this code

answered Jan 2, 2016 at 6:29

Vinothkumar Nataraj

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2 Comments

Johnny Over a year ago

Tried it and didn't fix it. Damn it!

Vinothkumar Nataraj Over a year ago

print_r($_POST); use to debug.values are correctly post or not

Yeikel · Accepted Answer · 2016-01-02 06:45:59Z

0

I just tested your code (copied and pasted) and it works perfectly under my server configuration (Windows 10 - PHP 5.6) . My best guess is that you have a typo in either the table name or the MySQL configuration.

If you copied this code from another site. Please check that you created the database and the table , and that the MySQL configuration is correct.

A good to check for this kind of mistakes so is to read the PHP error logs

edited Jan 2, 2016 at 6:45

answered Jan 2, 2016 at 6:43

Yeikel

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2 Comments

Johnny Over a year ago

I'm running Xubuntu 14.04 with PHP 5.5.9 and I'm not getting this to work. What could be wrong?

Yeikel Over a year ago

Can you check the PHP error logs? They tend to be useful

Renegade Rob · Accepted Answer · 2016-01-02 07:54:25Z

0

Try it like this maybe

if(isset($_POST['submit']) && !empty($_POST) ){

$theName = $_POST['theName'];
$theCity  = $_POST['theCity'];
$theAge = $_POST['theAge'];
$theDate = $_POST['theDate'];

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

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

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



        $sql = "INSERT INTO tabelateste (name, city, overeighteen, date)
        VALUES ('$theName ', '$theCity ', '$theAge ', '$theDate ')";

            if ($conn->query($sql) === TRUE) {
            $last_id = $conn->insert_id;

            } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
        }
    }

edited Jan 2, 2016 at 7:54

answered Jan 2, 2016 at 6:19

Renegade Rob

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4 Comments

Johnny Over a year ago

Still returns an empty page. No error but no output nor entry on db.

Renegade Rob Over a year ago

Do you have an ID on your DB with auto increment?

Johnny Over a year ago

Yes I do, first column.

Renegade Rob Over a year ago

I have just taken out id from the sql line. Try that.

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