【发布时间】：2015-12-05 04:24:34
【问题描述】：
我正在尝试将用户详细信息输入数据库并存储数据..我希望成功消息淡入我已经尝试了一些代码，但遗憾的是它不起作用...请帮助我摆脱这个..乞求如果我错了，请原谅.. 这是我的 register.php 代码
<?php
require_once 'DB_Functions.php';
$db = new DB_Functions();
// json response array
$response = array("error" => false);
if (!empty($_POST['fname']) && !empty($_POST['lname']) && !empty($_POST['email']) && !empty($_POST['password']) && !empty($_POST['mobile'])){
// receiving the post params
$fname = trim($_POST['fname']);
$lname = trim($_POST['lname']);
$email = trim($_POST['email']);
$password = $_POST['password'];
$mobile = trim($_POST['mobile']);
// validate your email address
if(filter_var($email, FILTER_VALIDATE_EMAIL)) {
// valid email address
if ($db->isUserExisted($email)) {
// user already existed
$response["error"] = true;
$response["error_msg"] = "User already existed with " . $email;
echo json_encode($response);
} else {
// create a new user
$user = $db->storeUser($fname, $lname, $email, $password, $mobile);
if ($user) {
// user stored successfully
$response["error"] = false;
$response["uid"] = $user["id"];
$response["user"]["fname"] = $user["fname"];
$response["user"]["lname"] = $user["lname"];
$response["user"]["email"] = $user["email"];
$response["user"]["created_at"] = $user["created_at"];
$response["user"]["updated_at"] = $user["updated_at"];
echo json_encode($response);
} else {
// user failed to store
$response["error"] = true;
$response["error_msg"] = "Unknown error occurred in registration!";
echo json_encode($response);
}
}
} else {
// invalid email address
$response["error"] = true;
$response["error_msg"] = "invalid email address";
echo json_encode($response);
}
} else {
$response["error"] = true;
$response["error_msg"] = "Required parameters are missing!";
echo json_encode($response);
}
?>
这里是带有 jquery 的 .html 文件..
<html>
<head>
<title>jQuery Test</title>
<script type="text/javascript" src="http://code.jquery.com/jquery-1.9.1.js"></script>
<script type="text/javascript" src = "register.js"></script>
</head>
<body>
<!--html body-->
<form name = "register" id = "register" method = "POST">
<label>First name:</label>
<input type = text name = "fname" id = "fname" required>
<label>Last name:</label>
<input type = "text" name = "lname" id = "lname" required>
<label>E-mail:</label>
<input type = "email" name = "email" id = "email" required>
<label>Password</label>
<input type = "password" name = "password" id = "password" required>
<label>Mobile no:</label>
<input type = "number" name = "mobile" id = "mobile" required>
<input type="submit" value="Insert" name="submit" id = "submit">
</form>
<div id = "result" align = "right"></div>
</body>
</html>
这是我的 /.js/ 文件
$(document).ready(function(){
$("#submit").click(function(e){
e.preventDefault();
$.ajax({
url: "register.php",
type: "POST",
data: {
fname: $("#fname").val(),
lname: $("#lname").val(),
email: $("#email").val(),
password: $("#password").val(),
mobile: $("#mobile").val()
},
dataType: "JSON",
success: function (json) {
$("#result").html(json.user.email); // like that you can display anything inside #result div
$("#result").fadeOut(1500);
},
error: function(jqXHR, textStatus, errorThrown){
alert(errorThrown);
}
});
});
});
【问题讨论】：
-
为什么你的输入中有
href = "#"？ -
先生..！！我已经删除了它，但它对我的输出没有影响..!!:(
-
检查 PHP/SQL 上的错误并检查您的控制台。
-
@krishna 你能解释一下在
success中使用jsonStr究竟想达到什么目的吗？ -
奇怪的是你说它保存了用户但你没有收到成功回调。如果您在
$response = array("error" => false);下方添加echo "Register check";，您的success函数中是否有任何响应