用AJAX写一个表单提交
HTML代码:
<form id="myForm">
<label for="name">Name:</label>
<input type="text" id="name" name="name"><br><br>
<label for="email">Email:</label>
<input type="email" id="email" name="email"><br><br>
<label for="message">Message:</label>
<textarea id="message" name="message"></textarea><br><br>
<button type="submit" id="submitBtn">Submit</button>
</form>
JavaScript代码:
const form = document.querySelector('#myForm');
const submitBtn = document.querySelector('#submitBtn');
submitBtn.addEventListener('click', (e) => {
e.preventDefault(); // 阻止表单默认提交行为
const formData = new FormData(form);
const xhr = new XMLHttpRequest();
xhr.open('POST', '/submit-form');
xhr.onreadystatechange = function() {
if (xhr.readyState === 4) {
if (xhr.status === 200) {
console.log(xhr.responseText);
alert('Form submitted successfully!');
} else {
alert('Form submission failed.');
}
}
};
xhr.send(formData);
});
这段代码监听了表单提交按钮的点击事件,阻止了表单默认提交行为,使用FormData对象获取表单数据,使用XMLHttpRequest对象发送POST请求。在请求完成后,根据请求状态和响应状态码判断表单提交是否成功。如果成功,弹出提示框;如果失败,也弹出提示框
原文地址: http://www.cveoy.top/t/topic/d5rh 著作权归作者所有。请勿转载和采集!