我有一个用于输入信息和注册帐户的表格。在网站上输入信息，当按下“注册”按钮时，它会通过外部 JavaScript 方法进行验证，然后使用 ajax 调用 PHP 方法，该方法应从文本框中获取信息并将其输入到数据库中。我似乎无法让 PHP 获取信息。<?php$mysqli = new mysqli('localhost:8080', 'root', null, 'salmonhouse');/* check connection */if (mysqli_connect_errno()) { printf("Connect failed: %s\n", mysqli_connect_error()); exit();}$sql = "INSERT INTO clients (name, surname, email, address, password) VALUES (?,?,?,?,?)";$name = $_POST['name'];$surname = $_POST['surname'];$email= $_POST['email'];$address= $_POST['Address'];$pass= $_POST['Password'];$stmt = $mysqli->prepare($sql);$stmt->bind_param("sssss", $name, $surname, $email, $address, $pass);$stmt->execute();?>HTML 文本框<form class="quote"> <div class = inliner> <label>Name</label> <input id="name" type="text" placeholder="Name"> </div> <div class = inliner> <label>Surname</label> <input id="surname" type="text" placeholder="Surname"> </div> <div> <label>Email</label><br> <input id="email" type="email" placeholder="Email Address"><br> </div> <div> <label>Address</label><br> <input id="Address" type="email" placeholder="Home Address"><br> </div> <div class = inliner> <label>Password</label> <input id="Password" type="text" placeholder="Password"> </div> <div class = inliner> <label>Verify Password</label> <input id="vPassword" type="text" placeholder="Password"> </div> <br><button class="button_1" type="button" onclick="Validate()">Register</button></form>这个 PHP 文件是一个单独的文件，只有这个代码。我已经测试过，如果我手动设置变量而不是尝试检索它们，则数据会成功插入到数据库中。所以从我的测试来看，它只是检索不起作用。我也试过$_REQUEST['name']这是 ajax/xmlhttprequest 代码。xmlhttp = new XMLHttpRequest();var url = "asset/php/inserting.php";xmlhttp.open("POST",url,true);xmlhttp.send();
2 回答
慕虎7371278
TA贡献1802条经验 获得超4个赞
您可以在ajax调用中附加变量，您需要在 php 页面中使用它&来分隔变量和=分配值 .i,e ：
//attaching values to pass
var data = "name=" + name + "&email=" + email + "&surname=" + surname + "&Address=" + Address + "&Password=" + Password;
if (window.XMLHttpRequest) {
request = new XMLHttpRequest();
} else if (window.ActiveXObject) {
request = new ActiveXObject("Microsoft.XMLHTTP");
}
request.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
//if success do something here
alert("save");
}
};
var url = "asset/php/inserting.php";
request.open("POST", url, true);
//send data to php page
request.send(data);
慕容3067478
TA贡献1773条经验 获得超3个赞
我的建议是使用 jQuery 库而不是 XMLHttpRequest。您需要在<head>HTML 部分包含一个<script>标签，以便从某些 CDN（内容交付网络）加载 jQuery 库。也添加id="f"到您的<form>标签。那么你的 Ajax 调用就可以很简单：
$.ajax({
type: "POST",
url: 'asset/php/inserting.php',
data: $('#f').serialize(), // serializes the form's elements.
success: function(msg)
{
alert(msg); // show response from the php script.
}
});
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