我是一名計算機科學專業的學生，大二。為了獨立學習，我決定創建一個使用 SQL、PHP、JS、AJAX、BOOTSTRAP 等技術的網站。我正在嘗試向數據庫添加內容，我使用 AJAX - 我不想刷新頁面，所以我使用 AJAX。我設法將內容添加到數據庫 - 但頁面刷新。我嘗試使用 jquery - 當我添加內容時 - 來防止刷新。該代碼有效 - 但仍然需要刷新。訪問數據庫的代碼：<?php$DBConInfo = [ 'server' => '127.0.0.1', 'username' => 'root', 'password' => '', 'name' => 'test',];mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);$conn = new mysqli($DBConInfo['server'],$DBConInfo['username'], $DBConInfo['password'],$DBConInfo['name']);if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error);}echo "Connected successfully";// php code to Insert data into mysql database from input textif(isset($_POST['insert'])){ $hostname = "127.0.0.1"; $username = "root"; $password = ""; $databaseName = "test"; // get values form input text and number $name = $_POST['name']; $description = $_POST['description']; $price = $_POST['price']; $picture = $_POST['picture'];// mysql query to insert data $query = "INSERT INTO `product`(`name`,`description`, `price`, `picture`) VALUES ('$name','$description','$price','$picture')"; $result = mysqli_query($conn,$query);// check if mysql query successful if($result) { echo 'Data Inserted'; } else{ echo 'Data Not Inserted'; var_dump($conn->error); } //mysqli_free_result($result); mysqli_close($conn);}?><!DOCTYPE html><html><head> <title> PHP INSERT DATA </title> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"></head>
1 回答
慕的地10843
TA貢獻1785條經驗 獲得超8個贊
提交表格后，您必須使用event.preventDefault()
$("#submit-insert").click( function(event) {
event.preventDefault();
});
$('#form-insert').on('submit', function (event) {
event.preventDefault();
$.ajax({
type : 'post',
url : 'NameOfPHPFile.php',
data : $('#form-insert').serialize(),
success : function () {
alert('form was submitted');
}
});
});
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