Java 将变量从android类移动到php端时出错

我目前正在开发一个应用程序，在将变量从android传输到php脚本时遇到了一个问题。然而，连接已建立。任何帮助都将不胜感激

这是我的Java代码

// open a URL connection to the Servlet
FileInputStream fileInputStream = new FileInputStream(sourceFile);
URL url = new URL(upLoadServerUri);

// Open a HTTP  connection to  the URL
conn = (HttpURLConnection) url.openConnection(); 
conn.setDoInput(true); // Allow Inputs
conn.setDoOutput(true); // Allow Outputs
conn.setUseCaches(false); // Don't use a Cached Copy
conn.setRequestMethod("POST");
conn.setRequestProperty("Connection", "Keep-Alive");
conn.setRequestProperty("ENCTYPE", "multipart/form-data");
conn.setRequestProperty("Content-Type", "multipart/form-data;boundary=" + boundary);
conn.setRequestProperty("uploaded_file", fileName); 

dos = new DataOutputStream(conn.getOutputStream());

dos.writeBytes(twoHyphens + boundary + lineEnd); 
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";filename=\""
                    + fileName + "\"" + lineEnd);

dos.writeBytes(lineEnd);

dos.writeBytes(twoHyphens + boundary + lineEnd); 
dos.writeBytes("Content-Disposition: form-data; name=\"uploaded_file\";name=\""
                    + username + "\"" + lineEnd);

dos.writeBytes(lineEnd);

// create a buffer of  maximum size
bytesAvailable = fileInputStream.available(); 

bufferSize = Math.min(bytesAvailable, maxBufferSize);
buffer = new byte[bufferSize];

// read file and write it into form...
bytesRead = fileInputStream.read(buffer, 0, bufferSize);  

while (bytesRead > 0) {
  dos.write(buffer, 0, bufferSize);
  bytesAvailable = fileInputStream.available();
  bufferSize = Math.min(bytesAvailable, maxBufferSize);
  bytesRead = fileInputStream.read(buffer, 0, bufferSize);   
}

// send multipart form data necesssary after file data...
dos.writeBytes(lineEnd);
dos.writeBytes(twoHyphens + boundary + twoHyphens + lineEnd);

这是我的php脚本

<?php
  require("config.inc.php");
    $file_path = "uploads/";

    $file_path = $file_path . basename( $_FILES['uploaded_file']['filename']);
    $user = $_FILES['username'];

    if(move_uploaded_file($_FILES['uploaded_file']['tmp_name'], $file_path)) {
        $query = "INSERT INTO users ( username, password, picture ) VALUES ( '1234567', $user, $file_path ) ";
        $stmt   = $db->prepare($query);
        $stmt->execute();
        echo "success";

    } else{
        echo "fail";
    }
 ?>

---

失败的原因是您没有正确准备查询：

$query = "INSERT INTO users ( username, password, picture ) VALUES ( ?, ?, ? ) ";
$stmt = $db->prepare($query);
$stmt->execute(array('1234567', $user, $file_path));

一点格式将大大有助于你的问题被看到，而不是隐藏在你的代码中。谢谢你的这句话，我编辑了它，减少了代码，这样人们可以很容易地通过它。事实上这不是问题，因为在修复了小语法问题后，它确实正确插入了。主要的问题是图像确实移动到了php脚本和我可以提取，而字符串参数我不能提取：/@AbdelrahmanMaged你在说哪个字符串你应该回显，看看哪个是空的字符串$user但是图像工作正常，我可以得到它的路径@meda@AbdelrahmanMaged执行
var\u转储（$\u文件）
在第二个
内容配置中，您还编写了两次
name
属性。。你正在上传多个文件吗？