我尝试将一些数据传递给mysql,但我没有工作。我想念一些东西。我已经检查了我的数据库名称和表格。
<?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:未捕获的异常'mysqli_sql_exception',消息'表'daftarpelanggan'是只读的
答案 0 :(得分:1)
如果与db的连接出错,则您的查询将无法评估。
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();
}
根据错误消息解决错误。