在單個PHP中使用AFNetworking 2.0上傳圖像和信息（兩者）

Question

我正在嘗試使用AFNetworking 2.0上傳多個圖像（產品圖片）和信息（品牌名稱，價格等）。

我確實使用兩個單獨的php文件上傳了圖像和信息
1. upload_product_info.php
2. upload_product_images.php，

首先在成功調用upload_product_images.php上調用upload_product_info.php。 一切工作正常，但我希望使用單個php文件。

我嘗試使用單個PHP，但是它只能部分工作並且僅上傳圖像時沒有出現任何錯誤，數據庫中的產品信息字段（產品品牌，名稱等）始終為空白。我不明白我在做什么錯。 我是php的新手。

這是我的iOS代碼，用於上傳圖像和信息。

- (IBAction)uploadProduct:(id)sender {

// for Testing purpose just taken 2 fields.
  NSString *productbrand  = @"xyz";
  NSString *productname   = @"pqr";

  NSDictionary *infoDictionary = @{@"pbrand": productbrand, @"pname": productname};

  __block NSUInteger success = 0;
  __block NSString *message;
  static int count = 1;

// returns array of product images url from temp Directory.
  productImages = [self returnImagesFromTemporaryDirectory];

  AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];

  manager.requestSerializer = [AFJSONRequestSerializer serializerWithWritingOptions:NSJSONWritingPrettyPrinted];
  manager.responseSerializer = [AFJSONResponseSerializer serializerWithReadingOptions:NSJSONReadingAllowFragments];

  [manager POST:@"http://localhost/~abc/Website/uploadProduct.php" parameters:infoDictionary constructingBodyWithBlock:^(id<AFMultipartFormData> formData)
  {
    for (NSURL *filePath in productImages)
    {

      CFStringRef pathExtension = (__bridge_retained CFStringRef)[filePath pathExtension];
      CFStringRef type = UTTypeCreatePreferredIdentifierForTag(kUTTagClassFilenameExtension, pathExtension, NULL);
      CFRelease(pathExtension);

      NSString *mimeType = (__bridge_transfer NSString *)UTTypeCopyPreferredTagWithClass(type, kUTTagClassMIMEType);
      NSLog(@" Mime Type : %@", mimeType);

      NSString *imageName = [NSString stringWithFormat:@"IMG00%i.png",count];
      count++;

      [formData appendPartWithFileURL:filePath name:@"uploaded_file[]" fileName:imageName mimeType:mimeType error:nil];
    }
  }
    success:^(AFHTTPRequestOperation *operation, id responseObject)
   {
     NSDictionary *responseDic = (NSDictionary *)responseObject;

     success = [responseDic[@"success"] integerValue];
     NSLog(@"Success: %ld",(long)success);
     message = responseDic[@"message"];

     if (success == 1)
     {
       UIAlertView *successAlert = [[UIAlertView alloc] initWithTitle:@" Success " message:message delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];

       [successAlert show];
     }

   }
    failure:^(AFHTTPRequestOperation *operation, NSError *error)
   {
     NSLog(@"Error: %@ ***** %@", operation.responseString, error);

     UIAlertView *failedAlert = [[UIAlertView alloc] initWithTitle:@" Failed " message:message delegate:nil cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];

     [failedAlert show];
  }];

這是我的PHP for Images + Information上傳。

   <?php

    header('Content-type: application/json');

    $json = file_get_contents('php://input'); // Catching input
    $value= json_decode($json, true);   // Decode JSON into Dictionary.

    $response = array();

// retrieve values from Dictionary using key.
    $productBrand = $value['pbrand'];
    $productname  = $value['pname'];


    // Database Connection.
    $mysqlserver="localhost";
    $mysqlusername="abc123";
    $mysqlpassword="pqr123";
    $link=mysql_connect($mysqlserver, $mysqlusername, $mysqlpassword) or die ("Error connecting to mysql server: ".mysql_error());

    $dbname = 'myDatabase';    // change this to the name of your database
    mysql_select_db($dbname, $link) or die ("Error selecting specified database on mysql server: ".mysql_error()); 

    // Insert Data into Table.
     $insertQuery = "INSERT INTO user_test (productBrand, productName) VALUES ('$productBrand', '$productname')";

    $result = mysql_query($insertQuery);

    if($result) 
    {
        $count=0;

             foreach ($_FILES['uploaded_file']['name'] as $filename) 
            {
                $file_path="uploads/";             

                $tmp=$_FILES['uploaded_file']['tmp_name'][$count];

                move_uploaded_file($tmp,$file_path.$filename);         

             $count=$count + 1;
            }

            $response["success"] = 1;
            $response["message"] = "Images uploaded Successfully.";
    } 
    else
    {
            $response["success"] = 0;
            $response["message"] = "Failed to upload Images";
    }

    echo json_encode($response);

     ?>

Answer 1

問題似乎在PHP方面

$json = file_get_contents('php://input'); // Catching input
$value= json_decode($json, true);   // Decode JSON into Dictionary.

不工作。 用此行替換上面的塊

$value= $_REQUEST;

應該解決您的問題。