php上传文件的要求
PHP大文件分块上传功能实例详解本文实例讲述了PHP大文件分块上传功能。分享给大家供大家参考，具体如下：
前端代码
使用file.slice将文件进行分割，然后分别进行异步上传。
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<!DOCTYPE html><html lang="zh-cn"> <head> <meta charset="utf-8"> <meta http-equiv="X-UA-Compatible" content="IE=edge"> <meta name="viewport" content="width=device-width, initial-scale=1"> <title><%= title %></title> <!-- zui --> <link rel="stylesheet" href="http://zui.sexy/dist/css/zui.min.css" rel="external nofollow" > </head> <body> <li class="container" style="margin-top: 30px;"> <form class="form-inline" method="post" enctype="multipart/form-data"> <li class="form-group"> <input type="file" id="fileBig" class="form-control"> </li> <button type="submit" class="btn btn-primary">提交</button> </form> </li> <!-- ZUI Javascript 依赖 jQuery --> <script src="http://zui.sexy/assets/jquery.js"></script> <!-- ZUI 标准版压缩后的 JavaScript 文件 --> <script src="http://zui.sexy/dist/js/zui.min.js"></script> <script type="text/javascript"> $('form').submit(function() { let file = $(":file")[0].files[0]; let fileName = file.name; let fileSize = file.size; console.log('fileSize',fileSize); let blockSize = 0.9*1024*1024; let num = Math.ceil(fileSize/blockSize); let start = 0; let end = 0; for(let i=1;i<=num;i++){ end = blockSize*i; if(end > fileSize){ end = fileSize; } let block = file.slice(start,end); start = end; let fd = new FormData(); fd.append('block',block); fd.append('name',fileName); fd.append('total',num); fd.append('index',i); $.ajax({ url:"upload.php", type:"POST", data:fd, async:true, processData:false, contentType:false, success:(res)=>{ console.log('res_'+i+":"); console.log(res); } }) } return false; }); </script> </body></html> |
后端代码
在所有的文件上传成功之后，合并生成原来的大文件
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<?php$name = $_POST['name'];$index = $_POST['index'];$total = $_POST['total'];echo "name:".$name.PHP_EOL;echo "index:".$index.PHP_EOL;echo "total:".$total.PHP_EOL;move_uploaded_file($_FILES['block']['tmp_name'],'upload/'.$name."_".$index);$list = scandir('upload');$num = count($list)-2;echo "cur_num:".$num.PHP_EOL;if($num == $total){ echo "upload done".PHP_EOL; echo $cmd = "cat upload/'{$name}_'* > upload/'{$name}'"; shell_exec($cmd);} |
希望本文所述对大家PHP程序设计有所帮助。
原文链接：https://blog.csdn.net/koastal/article/details/78984569