Question

我正在做一个小学校项目我需要在数据库上上传图片我只是不知道如何在上传图片时添加用户ID条件因为我想以特定的用户ID发送图像，所以该ID已连接到应用程序。

       public static final String UPLOAD_URL = "urlto/upload.php";

       public static final String UPLOAD_KEY = "image";

        private void uploadImage() {

        class UploadImage extends AsyncTask<Bitmap,Void,String> {

        ProgressDialog loading;
        RequestHandler rh = new RequestHandler();

        @Override
        protected void onPreExecute() {
            super.onPreExecute();
            loading = ProgressDialog.show(UserProfilActivity.this, "Uploading Image", "Please wait...",true,true);
        }

        @Override
        protected void onPostExecute(String s) {
            super.onPostExecute(s);
            loading.dismiss();
            Toast.makeText(getApplicationContext(),s,Toast.LENGTH_LONG).show();
        }

        @Override
        protected String doInBackground(Bitmap... params) {
            Bitmap bitmap = params[0];
            String uploadImage = getStringImage(bitmap);

            HashMap<String,String> data = new HashMap<>();
            data.put(UPLOAD_KEY, uploadImage);

            String result = rh.sendPostRequest(UPLOAD_URL,data);

            return result;
        }
    }

    UploadImage ui = new UploadImage();
    ui.execute(bitmap);
}

PHP代码

if($_SERVER['REQUEST_METHOD']=='POST'){

    $image = $_POST['image'];

    require_once('dbConnect.php');

    $sql = "INSERT INTO users (image) VALUES (?)";                 

    $stmt = mysqli_prepare($con,$sql);

    mysqli_stmt_bind_param($stmt,"s",$image);
    mysqli_stmt_execute($stmt);

    $check = mysqli_stmt_affected_rows($stmt);

    if($check == 1){
        echo "Image Uploaded Successfully";
    }else{
        echo "Error Uploading Image";
    }
    mysqli_close($con);
}else{
    echo "Error";
}

应该类似于“更新用户图像=：new_image其中（：user_id = user_id）” 我只是不明白如何将user_id参数发送到php代码，以及在php中进行哪些修改。

谢谢

编辑：

好的，这个php可以正常工作

if($_SERVER['REQUEST_METHOD']=='POST'){

    $image = $_POST['image'];
    $user_id = $_POST['user_id'];

    require_once('dbConnect.php');

    $sql = "Update users SET image=? WHERE user_id=?";                 

    $stmt = mysqli_prepare($con,$sql);

    mysqli_stmt_bind_param($stmt,"si",$image,$user_id);
    mysqli_stmt_execute($stmt);

    $check = mysqli_stmt_affected_rows($stmt);

    if($check == 1){
        echo "Image Uploaded Successfully";
    }else{
        echo "Error Uploading Image";
    }
    mysqli_close($con);
}else{
    echo "Error";
}

现在我需要发送用户ID参数的Java代码，我已经使用在线POST工具进行了测试

如何使用用户ID将图像上传到数据库？

0 个答案: