任何人都可以在这个 POST NSMutableURLRequest 到 PHP 脚本中找到错误吗？

Question

我尝试将数据发送到这个 php 脚本：

 mb_internal_encoding("UTF-8");

    if(isset($_POST['format']) && isset($_POST['category']) && isset($_POST['title']) && isset($_POST['description']) && isset($_FILES['photo']) {
     save($_POST['category'], $_POST['title'], $_POST['description'], $_FILES['photo']); 
    } else {
     echo "There was an error, a field does not exist, please try again!<br />";
     echo "format = " . $_POST['format'] . "<br />";
     echo "category = " . $_POST['category'] . "<br />";
     echo "title = " . $_POST['title'] . "<br />";
     echo "description = " . $_POST['description'] . "<br />";
     echo "photo =" . $_FILES['photo'] . ;
    }
...
...

我正在尝试从 iPhone 发送带有此 Objc 代码的同步请求：

//creating the url request:
 NSURL *cgiUrl = [NSURL URLWithString:@"http://www.hhh.com/uploading.php"];
 NSMutableURLRequest *postRequest = [NSMutableURLRequest requestWithURL:cgiUrl];

 //adding header information:
 [postRequest setHTTPMethod:@"POST"];

 NSString *stringBoundary = [NSString stringWithString:@"0xKhTmLbOuNdArY"];
 NSString *contentType = [NSString stringWithFormat:@"multipart/form-data; boundary=%@",stringBoundary];
 [postRequest addValue:contentType forHTTPHeaderField: @"Content-Type"];


 //setting up the body:
 NSMutableData *postBody = [NSMutableData data];
 [postBody appendData:[[NSString stringWithFormat:@"--%@\r\n",stringBoundary] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithString:@"Content-Disposition: form-data; name=\"category\"\r\n\r\n"] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithString:@"iPhone"] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithFormat:@"\r\n--%@\r\n",stringBoundary] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithString:@"Content-Disposition: form-data; name=\"format\"\r\n\r\n"] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithString:@"jpg"] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithFormat:@"\r\n--%@--\r\n",stringBoundary] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithString:@"Content-Disposition: form-data; name=\"title\"\r\n\r\n"] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithString:self.theTitle] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithFormat:@"\r\n--%@--\r\n",stringBoundary] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithString:@"Content-Disposition: form-data; name=\"description\"\r\n\r\n"] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithString:self.theCaption] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithFormat:@"\r\n--%@\r\n",stringBoundary] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithFormat:@"Content-Disposition: form-data; name=\"photo\"; filename=\"%@.jpg\"\r\n", self.theTitle] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[[NSString stringWithString:@"Content-Type: application/octet-stream\r\n\r\n"] dataUsingEncoding:NSUTF8StringEncoding]];
 [postBody appendData:[NSData dataWithData:UIImageJPEGRepresentation(resizedImage(self.myPhoto, CGRectMake(0, 0, 600, 800)), 0.5)]];
 [postBody appendData:[[NSString stringWithFormat:@"\r\n--%@--\r\n",stringBoundary] dataUsingEncoding:NSUTF8StringEncoding]];

 [postRequest setHTTPBody:postBody];

 NSError *theError;
 NSData *returnData = [ NSURLConnection sendSynchronousRequest: postRequest returningResponse: nil error:&theError ];
 NSString *returnDataString = [[NSString alloc] initWithData:returnData encoding:NSUTF8StringEncoding];

 NSLog(@"[DEBUG]... response from request: %@", returnDataString);

但答案是：

[DEBUG]... 来自请求的响应： 出现错误，所有字段都没有 存在，请重试！
格式化 =
类别 =
标题 =
描述 =
照片 =纬度 =
经度 =

我做错了什么？因为我看到的是没有收到任何字段。

Answer 1

如果您的 PHP 脚本记录了它实际接收到的内容，而不是它如何解释它，那么您可能会获得更多信息。

Answer 2

嗯...解决了一个问题。参数之间的格式必须是这样的：

@"\r\n--%@\r\n"

我正在发送：

@"\r\n--%@--\r\n" // 多两个分数

现在我还不能发送数据。我在同一台服务器上制作了一个只有此内容的新脚本。

<?php

        echo "format = " . $_POST['format'];
      echo "[NEXT]... category = " . $_POST['category'];
      echo "[NEXT]... title = " . $_POST['title'];
      echo "[NEXT]... description = " . $_POST['description'];
      echo "[NEXT]... latitude = " . $_POST['latitude'];
      echo "[NEXT]... longitude =" . $_POST['longitude'];

?>

并将代码复制到原始脚本（由其他人编写）。

但现在我看到一个脚本接收数据而原来的不能。

会不会是文件编码问题？