我有标题中提到的错误。当我单击表单上的提交按钮时会发生这种情况。这是我的表单句柄文件(我认为没有必要复制表单代码):
<?php
$servername = "localhost";
$username = "sabashel_sabaadm";
$password = "saba1365%karaj@*";
$dbname = "sabashel_saba";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$fname = $lname = $gender = $birthdate = $organization = $degree = $field = $address = $post_code = $mobile = $email = $check_1 = $check_2 = $check_3 = $check_4 = $check_5 = $check_6 = $check_7 = $check_8 "";
$check_9 = $check_10 = $check_11 = $check_12 = $check_13 = $description = $person_image = "";
if(isset($_POST['fname']) && isset($_POST['lname']) && isset($_POST['gender']) && isset($_POST['birthdate']) && isset($_POST['degree']) && isset($_POST['filed-of-study']) && isset($_POST['address']) && isset($_POST['post-code']) && isset($_POST['mobile']) && isset($_POST['email']) && isset($_POST['check-1']) && isset($_POST['check-2']) && isset($_POST['check-3']) && isset($_POST['check-4']) && isset($_POST['check-5']) && isset($_POST['check-6']) && isset($_POST['check-7']) && isset($_POST['check-8']) && isset($_POST['check-9']) && isset($_POST['check-10']) && isset($_POST['check-11']) && isset($_POST['check-12']) && isset($_POST['check-13']) && isset($_POST['description']) && isset($_POST['person-iamge'])){
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$gender = $_POST['gender'];
$birthdate = $_POST['birdthdate'];
$organization = $_POST['organization'];
$degree = $_POST['degree'];
$field = $_POST['field-of-study'];
$address = $_POST['address'];
$post_code = $_POST['post-code'];
$mobile = $_POST['mobile'];
$email = $_POST['email'];
$check_1 = $_POST['check-1'];
$check_2 = $_POST['check-2'];
$check_3 = $_POST['check-3'];
$check_4 = $_POST['check-4'];
$check_5 = $_POST['check-5'];
$check_6 = $_POST['check-6'];
$check_7 = $_POST['check-7'];
$check_8 = $_POST['check-8'];
$check_9 = $_POST['check-9'];
$check_10 = $_POST['check-10'];
$check_11 = $_POST['check-11'];
$check_12 = $_POST['check-12'];
$check_13 = $_POST['check-13'];
$description = $_POST['description'];
$person_image = $_POST['person-image'];
$iftest = true;
}
if ($iftest == true) {
$query = "INSERT INTO volunteer (fname, lname, gender, organization, degree, field, address, post_code, mobile, email, check_1, check_2, check_3, check_4, check_5, check_6, check_7, check_8, check_9, check_10, check_11, check_12, check_13, description, person_image, birthdate) VALUES ('$fname', '$lname', '$gender', '$organization', '$degree', '$field', '$address', '$post_code', '$mobile', '$email', '$check_1', '$check_2', '$check_3', '$check_4', '$check_5', '$check_6', '$check_7', '$check_8', '$check_9', '$check_10', '$check_11', '$check_12', '$check_13', '$description', '$person_image', '$birthdate')";
}
$result = mysqli_query($conn, $query);
if ($result) {
header('Location: http://sabashelter.com/success');
}
else {
header('Location: http://sabashelter.com/fail');
}
}
$conn->close();
?>
并且提到:我与另一个页面具有完全相同的问题,该页面执行相同的操作并尝试使用相同的代码将大量值添加到数据库中。我想知道这个页面中的问题是否解决了,同样的方法也可以用于其他页面。
答案 0 :(得分:1)
正如@CBroe所说,检查您的日志文件首先。您似乎错过了第14行的=
。
$fname = $lname = $gender = $birthdate = $organization = $degree = $field = $address = $post_code = $mobile = $email = $check_1 = $check_2 = $check_3 = $check_4 = $check_5 = $check_6 = $check_7 = $check_8 = "";
此外,你在第60行有一个迷路}
。
您的错误日志文件将帮助您解决这些问题。