Question

在android客户端中，我已经为http post请求设置了多个参数。（我做了确定在android客户端和我的webservice之间有连接。）

Android代码：

Content-Disposition: form-data; name="password" Content-Type: text/plain 123

Content-Disposition: form-data; name="username" Content-Type: text/plain tuyenpx

Content-Disposition: form-data; name="file"; filename="studio.png" Content-Type: image/png Content-Transfer-Encoding: binary

[＆＃34;此处的数据，已写入连接的输出流＆＃34;]

PHP代码：

 if (isset($_GET['username'])) {
        $userName = $_GET['username'];
    }
    if (isset($_GET['password'])) {
        $passWord = $_GET['password'];
    }

    move_uploaded_file($_FILES['file']['filename'], "uploads/" . $_FILES['file']['filename']);


  //  if (!($userName == '' && $passWord == '')) {
        $json = array("status" => TRUE, "username" => $userName,"password"=>$passWord);
        header('HTTP/1.1 200 OK', true, 200);
        response($json);
//    } else {
//        header('HTTP/1.1 400 Bad Request', true, 400);
//        header('Content-type: text/plain; charset = UTF8');
//        echo'invalid request , username or pass wo`enter code here`rd must not null';

问题：我在PHP代码中出错了，为什么$_GET['username']总是返回null，我也尝试使用$_POST['username']，它也会返回null。

感谢您的帮助！

Answer 1

感谢您的时间，我在我的PHP服务器中解决了我的问题，并且在我的库中也缺少一个边界，以完成写入输出流。

$ userName = $ _POST [＆＃39; username＆＃39;]; $ passWord = $ _POST [＆＃39;密码＆＃39;];

if (is_uploaded_file($_FILES['file']['tmp_name'])) {
    $uploads_dir = './';
    $tmp_name = $_FILES['file']['tmp_name'];
    $pic_name = $_FILES['file']['name'];
    move_uploaded_file($tmp_name, $uploads_dir . $pic_name);
    $json = array("status" => TRUE, "msg" => "upload success","username"=>$userName,"password"=>$passWord);
    response($json);
} else {
    header('HTTP/1.1 400 Bad Request', true, 400);
}

在PHP帖子中处理Mutil Part Param

1 个答案: