cannot pass php data to mysql

Question

I try to pass some data to mysql but i didnt work.did i miss something. i already check my database name and table.

<?php
$dbServername = "localhost";
$dbUsername   = "root";
$dbPassword   = "";
$dbName       = "pelanggan";

$conn = mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);
if (!$conn) {
    die('error connecting to database');
}

It keeps giving me the "connection error" message:

<?php
include_once 'includes/dbh.inc.php';
?>
    <html>
    <body>
    <form action='form.php' method='POST'>
        NAMA: <br> <input type='text' name='nama'><br>
        NO TELP : <br> <input type='text' name='telp'><br>
        PAKET DATA: <br> <input type='text' name='paket'><br>
        <button value='submit' name='submit'>submit</button>
    </form>
    </body>
    </html>
<?php
if (isset($_POST['nama']) && isset ($_POST['telp']) && isset($_POST['paket'])) {
    $nama  = $_POST['nama'];
    $telp  = $_POST['telp'];
    $paket = $_POST['paket'];
    if (empty ($nama) || empty ($telp) || empty ($paket)) {
        echo '*FIELDS MUST BE FILLED';
    } else {
        $sqlinsert = "INSERT INTO daftarpelanggan (nama,telp,paket) VALUES ('$nama','$telp','$paket')";
        if (!mysqli_query($conn, $sqlinsert)) {
            die ('connection error');
        } else {
            echo '1 record has been added successfully';
        }
    }
}
?>

"Fatal error: Uncaught exception 'mysqli_sql_exception' with message 'Table 'daftarpelanggan' is read only

Try using die ('connection error' . mysqli_error($conn)); and tell what you get. — Praveen Kumar Purushothaman
– Praveen Kumar Purushothaman, Commented May 10, 2018 at 14:50
Your code is vulnerable to SQL Injection by the way. Make sure to sanitise your user input before executing a SQL query with it embedded within. Anyway, do post the results from the query @Praveen Kumar posted as it will give us some more detailed information. — Cillian Collins
– Cillian Collins, Commented May 10, 2018 at 14:53
Add ini_set('display_errors', 1); ini_set('log_errors',1); error_reporting(E_ALL); mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT); to the top of your script. This will force any mysqli_ errors to generate an Exception that you can see on the browser and other errors will also be visible on your browser. — RiggsFolly
– RiggsFolly, Commented May 10, 2018 at 15:01
@Praveen Kumar it says "connection errorTable 'daftarpelanggan' is read only" — hendry
– hendry, Commented May 10, 2018 at 15:03
@RiggsFolly i did tried you suggestion and it says '"Fatal error: Uncaught exception 'mysqli_sql_exception' with message 'Table 'daftarpelanggan' is read only' in C:\xampp\htdocs\coba\form.php:22 Stack trace: #0 C:\xampp\htdocs\coba\form.php(22): mysqli_query(Object(mysqli), 'INSERT INTO daf...') #1 {main} thrown in C:\xampp\htdocs\coba\form.php on line 22"' — hendry
– hendry, Commented May 10, 2018 at 15:10

Shahid Kamal · Accepted Answer · 2018-05-10 15:01:24Z

1

If connection to db gives error your queries won't evaluate. try to get what error it gives using mysqli_connect_error() function.

$conn = mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);
//if (!$conn) {
//    die('error connecting to database');
//}
if (mysqli_connect_errno()){ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

Solve the error based on the error message.

edited May 10, 2018 at 15:01

answered May 10, 2018 at 14:55

Shahid Kamal

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1 Comment

RiggsFolly Over a year ago

Actually there is nothing wrong with the concatenation. Remember $variables are automatically expanded in a double quoted string literal in PHP

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