首页 > 开发 > Php > 正文

PHP响应post请求上传文件的方法

2020-02-18 22:57:13
字体:
来源:转载
供稿:网友

本文实例讲述了PHP响应post请求上传文件的方法。分享给大家供大家参考,具体如下:

function send_file($url, $post = '', $file = '') {  $eol = "/r/n";  $mime_boundary = md5 ( time () );  $data = '';  $confirmation = '';  date_default_timezone_set ( "Asia/Shanghai" );  $time = date ( "Y-m-d H:i:s " );  $post ["filename"] = $file [filename];  foreach ( $post as $key => $value ) {    $data .= '--' . $mime_boundary . $eol;    $data .= 'Content-Disposition: form-data; ';    $data .= "name=" . $key . $eol . $eol;    $data .= $value . $eol;  }  $data .= '--' . $mime_boundary . $eol;  $data .= 'Content-Disposition: form-data; name=' . $file [name] . '; filename=' . $file [filename] . $eol;  $data .= 'Content-Type: text/plain' . $eol;  $data .= 'Content-Transfer-Encoding: binary' . $eol . $eol;  $data .= $file [filedata] . $eol;  $data .= "--" . $mime_boundary . "--" . $eol . $eol;  $params = array ('http' => array ('method' => 'POST', 'header' => 'Content-Type: multipart/form-data;boundary=' . $mime_boundary . $eol, 'content' => $data ) );  $ctx = stream_context_create ( $params );  $response = file_get_contents ( $url, FILE_TEXT, $ctx );  return $response;}

希望本文所述对大家PHP程序设计有所帮助。

发表评论 共有条评论
用户名: 密码:
验证码: 匿名发表