简单实现ajax拖拽上传文件

AJAX拖拽上传功能实现，供大家参考，具体内容如下

<!DOCTYPE html>
<html lang="en">

<head>
 <meta charset="UTF-8">
 <meta name="viewport" content="width=device-width, initial-scale=1.0">
 <meta http-equiv="X-UA-Compatible" content="ie=edge">
 <title>Document</title>
 <style>
 .box {
  width: 300px;
  height: 300px;
  border: 1px solid #000;
  text-align: center;
  line-height: 300px;
  font-size: 40px;
 }
 </style>
</head>

<body>
 <div class="box">+</div>

 <script>
 var box = document.querySelector('.box');

 box.ondragover = function (e) {
  e.preventDefault();
 }
 box.ondrop = function (e) {
  console.log(e.dataTransfer)
  e.preventDefault();
  var xhr = new XMLHttpRequest();
  xhr.onreadystatechange = function () {
  if (xhr.readyState == 4 && xhr.status == 200) {
   console.log(xhr.responseText)
  }
  }
  xhr.open('POST', './server.php', true);

  var formdata = new FormData();
  formdata.append('pic', e.dataTransfer.files[0]);
  formdata.append('name', 'luyao');
  xhr.send(formdata);

 }
 </script>
</body>

</html>

//server.php

<?php
 $rand = rand(1,1000).'.jpg';
 move_uploaded_file($_FILES['pic']['tmp_name'], './uploads/'.$rand);
 echo '/uploads/'.$rand;

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。

简单实现ajax拖拽上传文件

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