您好我将记录添加到数据库中的问题。当我将记录插入数据库时。消息显示记录插入失败。我仔细检查了编码和编程语法都很好。但是在运行问题时我没有得到任何错误。有人能解决问题吗会得到错误。异常处理我也放了但不幸的是它没有帮助我显示错误。
<?php
if (isset($_POST['submit'])) {
$bassname = $_POST['bassname'];
$cat = $_POST['cat'];
$p_img1 = $_FILES['p_img1']['name'];
$p_img2 = $_FILES['p_img2']['name'];
$p_img3 = $_FILES['p_img3']['name'];
$temp_name1 = $_FILES['p_img1']['tmp_name'];
$temp_name2 = $_FILES['p_img2']['tmp_name'];
$temp_name3 = $_FILES['p_img3']['tmp_name'];
move_uploaded_file($temp_name1, "product_images/$p_img1");
move_uploaded_file($temp_name2, "product_images/$p_img2");
move_uploaded_file($temp_name3, "product_images/$p_img3");
$mobile = $_POST['mobile'];
$address = $_POST['address'];
$job_desc = $_POST['job_desc'];
$work_video = $_POST['work_video'];
$insert_product = "insert into
bass_registation(
bassname,
category,
job_image1,
job_image2,
job_image3,
mobile,
address,
job_des,
video
) values (
'$bassname',
'$cat',
'$p_img1',
'$p_img2',
'$p_img3',
'$mobile',
'$address',
'$job_desc',
'$work_video'
)";
$run_product = mysqli_query($con, $insert_product);
try {
if ($run_product) {
echo "<script>alert('Data has been inserted successfully')</script>";
} else {
echo "<script>alert('Data has been Failed')</script>";
}
}
catch (Exception $e) {
echo 'Message: ' . $e->getMessage();
}
}
?>