Question

代码：

NSURL * url=[NSURL URLWithString:@"alamghareeb.com/mobileData.ashx?catid=0&No=10"]; 
NSData * data=[NSData dataWithContentsOfURL:url]; 
NSError * error; 
NSMutableDictionary * json = [NSJSONSerialization JSONObjectWithData:data options: NSJSONReadingMutableContainers error: &error];
NSMutableArray * referanceArray=[[NSMutableArray alloc]init];
NSMutableArray * periodArray=[[NSMutableArray alloc]init]; 
NSArray * responseArr = [NSArray arrayWithArray:json[@"item"]];

for (NSDictionary * dict in responseArr) { 
    [referanceArray addObject:[dict objectForKey:@"created"]]; 
}

JSON看起来像：

{ 
  "item" = [ 


  {
  "id" : "2292",

  "created" : "10/01/2015 10:21:18 ص",

  "title" : "الإمارات: ملابس رياضية فاخرة لتحسين أداء الإبل في السباقات ", 

  "image_url" : "http://alamghareeb.com/Photos/L63556482078696289044.jpg",

  "image_caption" : ""

   },

  {

  "id" : "2291",

  "created" : "10/01/2015 09:28:11 ص",

  "title" : "طبيبة تجميل 'مزورة' تحقن مرضاها بالغراء والاسمنت",

  "image_url" : "http://alamghareeb.com/Photos/L63556478891290039015.jpg",

  "image_caption" : ""
  } 
 ]
}

Answer 1

此JSON无效。你手动构建它了吗？ "item" = ...应为"item" : ...。

将来，您可以通过http://jsonlint.com运行JSON来验证任何问题。同样，您应该在Objective-C代码中添加错误检查，例如

NSMutableDictionary *json = [NSJSONSerialization JSONObjectWithData:data options: NSJSONReadingMutableContainers error: &error];
if (!json) {
    NSLog(@"JSON parsing error: %@", error);
}

我还建议将服务器代码更改为不手动构建JSON，但要利用JSON函数（例如，如果使用PHP，构建关联数组，然后使用json_encode函数生成JSON输出）。

因此，一旦修复了JSON错误，解析结果可能如下所示：

NSError *error;
NSDictionary *dictionary = [NSJSONSerialization JSONObjectWithData:data options:0 error:&error];
if (!dictionary) {
    NSLog(@"NSJSONSerialization error: %@", error);
    return;
}

NSArray *items = dictionary[@"item"];

for (NSDictionary *item in items) {
    NSString *identifier = item[@"id"];
    NSString *created    = item[@"created"];
    NSString *title      = item[@"title"];
    NSString *urlString  = item[@"image_url"];
    NSString *caption    = item[@"image_caption"];

    NSLog(@"identifier = %@", identifier);
    NSLog(@"created = %@", created);
    NSLog(@"title = %@", title);
    NSLog(@"urlString = %@", urlString);
    NSLog(@"caption = %@", caption);
}

Answer 2

您可以使用NSJSONSerialization。无需导入任何类。

NSDictionary *result = [NSJSONSerialization JSONObjectWithData:[strResult dataUsingEncoding:NSUTF8StringEncoding] options:0 error:nil];

或直接使用NSData

NSDictionary *result = [NSJSONSerialization JSONObjectWithData:theJSONDataFromTheRequest options:0 error:nil];

来源：Stackoveflow question

如何在ios7中解析这个json数据

2 个答案: