您好，我是php的初学者，谁能告诉我这里是什么问题？我似乎无法解决。这是我的PHP编码：
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "tempahperalatan";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
if(isset($_POST['submit']))
{
$pemohon = $_POST['namaPemohon'];
$trkhMula = $_POST['tmula'];
$trkhAkhir = $_POST['takhir'];
$n_program = $_POST['namaProgram'];
$lokasi = $_POST['lokasi'];
$n_anjuran = $_POST['namaAnjuran'];
$catatan = $_POST['catatan'];
$sql = "INSERT INTO daftartempah (pemohon, trkhMula, trkhAkhir, n_program, lokasi, n_anjuran, catatan) VALUES ('$namaPemohon', '$tmula', '$takhir', '$namaprogram', '$lokasi', '$namaAnjuran', '$catatan')";
}
if (mysqli_query($conn, $sql)) { //this is line 30
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
它和我的Xampp有关系吗？先感谢您 ：）
最佳答案
把你的代码块
if (mysqli_query($conn, $sql)) { //this is line 30
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
在 - 的里面
if(isset($_POST['submit']))
{
就在$ sql = [...]之下
否则，如果未提交表单，脚本将尝试执行不存在的查询