我寫了一個AJAX代碼並鏈接了php文件,但代碼在最後不接受return false;
函數。它將我帶到行動檔案。我的Ajax代碼ajax代碼,jquery返回false不工作。嘗試其他答案
$("#submit").click(function() {
var data = $("#form :input").serializeArray();
$.post($("#form").attr("action"), data, function(info) { $("#result").html(info);});
});
$("#form").submit(function() {
return false;
});
我的index.php代碼
<?php
include('new.php');
?>
<!DOCTYPE html>
<html>
<head>
<title>Ajax practice</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script type="text/javascript" src="ajax.js"></script>
</head>
<body>
<form action="new.php" method="post" id="form">
<label>Name</label><br>
<input type="text" name="name"/><br>
<label>Email</label><br>
<input type="Email" name="email"/><br>
<button type="submit" name="submit" id="submit">submit to database</button>
</form>
<span id="result"></span>
</body>
</html>
我的PHP文件
<?php
$con = mysqli_connect("localhost","root","","ajax") or die ("couldnt connect to database.");
if (isset($_POST['submit'])) {
global $con;
$title = $_POST['name'];
$email = $_POST['email'];
$insert_user = "INSERT INTO ajax (title, email) VALUES ('$title', '$email')";
$run_user = mysqli_query($con,$insert_user);
if ($run_user) {
echo "Success";
} else {
echo "Failed";
}
}
?>
裏面是什麼new.php? – mplungjan