PHP - 提交表单而不退出页面
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);
}
?>
料青山看我应如是浏览 73回答 4
4回答
-
皈依舞
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
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 formform.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
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
当您提交表单时，我们将设置一个 $result var 来保存结果。然后我们再次使用查询url参数重定向到index.php ?result=success然后我们确保 if 已设置 (isset()) 并比较该值以显示正确的消息。所以假设表单位于 /index.php 中<?phpif(isset($_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 paramheader("Location: /index.php?result={$result}");?>
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