在数据输入表单中，我希望当单击“提交”按钮时，它不会离开表单所在的页面。提交论坛后，会向用户显示一个弹出窗口，其中包含表单已发送的信息！在搜索中，我知道AJAX存在，但我不知道如何完全实现它。我是 PHP 新手，我希望有人能帮助我！对不起我的英语！HTML：<form action="php/newsletter.php" method="post" class="formulario"> <input type="text" name="email" placeholder="Email" required> </div> <div class="ss-item-required" style="text-align:center;"> <button type="submit" class="site-btn">Send</button> </div> </form>PHP： <?php$servername = "localhost";$username = "root";$password = "";$dbname = "DB";$conexao = new mysqli($servername, $username, $password, $dbname);if ($conexao->connect_error) { die("Erro na conexão: " . $conexao->connect_error);}if (!$conexao) { die("Erro de ligação: " . mysqli_connect_error());}$sql = "INSERT INTO newsletter (email) VALUES ('email')";if (mysqli_query($conexao, $sql)) { echo '<div id="form-submit-alert">Submitted!</div>'; } else { echo "Erro: " . $sql . "<br>" . mysqli_error($conexao);}?>
4 回答
皈依舞
TA贡献1851条经验 获得超3个赞
使用 Ajax，您可以调用 html 请求并使用返回代码执行某些操作。
例如，您可以拦截提交表单并就地创建 ajax 请求。这个ajax请求调用你的php函数，并获取结果来显示/更新页面中的一些数据，而无需重新加载所有html。
https://api.jquery.com/jquery.ajax/
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.4.1/jquery.min.js"></script>
</head>
<form action="php/newsletter.php" method="post" class="formulario">
<input type="text" name="email" placeholder="Email" required>
<div class="ss-item-required" style="text-align:center;">
<button type="submit" class="site-btn">Send</button>
</div>
</form>
<script>
jQuery(document).ready(function() {
let form = $('body').find('form');
$(form).submit(function(e){
e.preventDefault();
e.stopPropagation();
var formData = new FormData($(form)[0]); // serialize form data
$.ajax({
type:"POST",
url:$(form).attr('action'), // use your form action as ajax url
data:formData,
contentType: false,
processData: false,
success: function(response){
// test if response is json array or html content
var is_JSON = true;
try{ var json = $.parseJSON(response);
}catch(err){ is_JSON = false; }
if(is_JSON){ // json resonse : if your php return json (for handle error )
console.log('response is JSON')
}else{
// response is your html of php return
console.log('response is not JSON')
console.log(response)
}
}
});
});
});
</script>
未经测试，但这应该有效。
汪汪一只猫
TA贡献1898条经验 获得超8个赞
我会使用 JavaScript 来实现这种方法。
假设您的表单具有 id my-form；您的 JavaScript 需要阻止表单正常提交并接管该过程本身。
let form = document.getElementById("my-form"); // find the form
form.addEventListener("submit", function (event){ // react to the submit try
event.preventDefault(); // prevent the sending at this point
let data = new FormData(form); // collect data, as we want to send it later
// We'll use an XMLHttpRequest, which basically means we send a normal web
// request from JavaScript and can interpret the answer afterwards
let xhr = new XMLHttpRequest();
xhr.onreadystatechange = function() {
// here we can define code to be executed when the request is running
if(this.readyState == 4 && this.status == 200){
// we know the request has been successfull
window.alert("Data sent!"); // Popup
}
};
// finally we need to execute the xhr
let method = form.getAttribute("method").toUpperCase();
let url = form.getAttribute("action"); // reuse from form
xhr.open(method, url, true);
xhr.send(data);
});
这种方法的好处是它完全在后台运行，即用户可以在请求运行时继续使用该站点，并且不涉及另一个库。
翻过高山走不出你
TA贡献1875条经验 获得超3个赞
那么您所说的是在提交按钮之后您实际上会发送到后端页面吗？我要做的是（在后端页面）当代码执行时，简单地放置一个 php 语句：
location:('index.html');这样您将立即返回到您提交表单的页面
烙印99
TA贡献1829条经验 获得超13个赞
当您提交表单时，我们将设置一个 $result var 来保存结果。
然后我们再次使用查询url参数重定向到index.php ?result=success
然后我们确保 if 已设置 (isset()) 并比较该值以显示正确的消息。
所以假设表单位于 /index.php 中
<?php
if(isset($_GET['result') && $_GET['result') == "success") {
echo '<div id="form-submit-alert">Submitted success!</div>';
} else {
echo '<div id="form-submit-alert">Submitted error!</div>';
}
?>
<form action="php/newsletter.php" method="post" class="formulario">
<div>
<input type="text" name="email" placeholder="Email" required>
</div>
<div class="ss-item-required" style="text-align:center;">
<button type="submit" class="site-btn">Send</button>
</div>
</form>
// php/newsletter.php
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "DB";
$conexao = new mysqli($servername, $username, $password, $dbname);
// Init result
$result = "error";
if ($conexao->connect_error) {
die("Erro na conexão: " . $conexao->connect_error);
}
if (!$conexao) {
die("Erro de ligação: " . mysqli_connect_error());
}
$sql = "INSERT INTO newsletter (email) VALUES ('email')";
if (mysqli_query($conexao, $sql)) {
// if the result is OK
$result = "success";
} else {
echo "Erro: " . $sql . "<br>" . mysqli_error($conexao);
}
// Redirect to initial page with the query url param
header("Location: /index.php?result={$result}");
?>
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