JSON 解析错误

发布于 2024-11-28 03:12:15 字数 615 浏览 2 评论 0原文

我在 iOS 上使用 SBJson 框架（也称为 json-framework）。

解析某个 JSON 文件时，出现以下错误： -JSONValue 失败。错误是：未转义的控制字符 [0x09]'

我已经多次使用这个框架，并且还在同一个应用程序中解析一个非常相似的 JSON 文件（甚至更长），并且它工作正常。

我尝试扔掉一堆 NSLogs，一切似乎都很好。有人可以告诉我这个错误意味着什么，或者至少如何继续调试这样的错误吗？

以下是显示错误的代码：

- (void)downloadSchedule:(NSString *)jsonString {

    // Get JSON feed URL and instantiate a dictionary object with its content
    NSDictionary *topDic = [jsonString JSONValue];

    NSLog(@"topDic count %d", [topDic count]);

topDic 显示计数为 0。错误位于 [jsonString JSONValue] 行。

谢谢

原文

I am using SBJson framework (also known as json-framework) for the iOS.

When parsing a certain JSON file, I am getting the following error:
-JSONValue failed. Error is: Unescaped control character [0x09]'

I have used this framework many times and I am also parsing a very similar JSON file (that is even much longer) in that same app and it's working fine.

I tried throwing around a bunch of NSLogs and everything seems to be fine. Can someone please point me to what this error means, or at least how to go ahead in debugging such an error?

Here is the code that displays the error:

- (void)downloadSchedule:(NSString *)jsonString {

    // Get JSON feed URL and instantiate a dictionary object with its content
    NSDictionary *topDic = [jsonString JSONValue];

    NSLog(@"topDic count %d", [topDic count]);

topDic is showing a count of 0. The error is at the [jsonString JSONValue] line.

Thank you

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神经大条 2024-12-05 03:12:15

我有一个很好的解决方案。应用此方法删除转义字符。

-(NSString *)removeUnescapedCharacter:(NSString *)inputStr
{

NSCharacterSet *controlChars = [NSCharacterSet controlCharacterSet];

NSRange range = [inputStr rangeOfCharacterFromSet:controlChars];

  if (range.location != NSNotFound) 
  {

      NSMutableString *mutable = [NSMutableString stringWithString:inputStr];

      while (range.location != NSNotFound) 
      {

          [mutable deleteCharactersInRange:range];

          range = [mutable rangeOfCharacterFromSet:controlChars];

      }

      return mutable;

   }

  return inputStr;
}

通过像这样传递输出字符串来调用此方法

NSString *output = [NSString stringWithContentsOfURL:[NSURL URLWithString:@"yourUrlString"] encoding:NSUTF8StringEncoding error:nil];

output = [self removeUnescapedCharacter:output];

I have a great solution for it. Apply this method for remove escaped characters.

-(NSString *)removeUnescapedCharacter:(NSString *)inputStr
{

NSCharacterSet *controlChars = [NSCharacterSet controlCharacterSet];

NSRange range = [inputStr rangeOfCharacterFromSet:controlChars];

  if (range.location != NSNotFound) 
  {

      NSMutableString *mutable = [NSMutableString stringWithString:inputStr];

      while (range.location != NSNotFound) 
      {

          [mutable deleteCharactersInRange:range];

          range = [mutable rangeOfCharacterFromSet:controlChars];

      }

      return mutable;

   }

  return inputStr;
}

Call this method with passing your output string like this

NSString *output = [NSString stringWithContentsOfURL:[NSURL URLWithString:@"yourUrlString"] encoding:NSUTF8StringEncoding error:nil];

output = [self removeUnescapedCharacter:output];

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