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php - 无法将php数据传递给mysql

问题描述

我尝试将一些数据传递给 mysql，但我没有工作。我错过了什么。我已经检查了我的数据库名称和表。

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

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

它不断给我“连接错误”消息：

<?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';
        }
    }
}
?>

“致命错误：带有消息‘表’daftarpelanggan’的未捕获异常‘mysqli_sql_exception’是只读的

标签： phpmysql

解决方案

如果与 db 的连接出现错误，则您的查询将不会评估。尝试使用mysqli_connect_error()函数获取它给出的错误。

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

根据错误信息解决错误。

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