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我有以下有效的 JSON 来解析:

[{"name":"kkkk","empid":"55628","address":"mumbai","mobile":"9525878558"},{"name":"xtreme","empid":"20","address":"stripes","mobile":"9999597"},{"name":"xtreme","empid":"20","address":"stripes","mobile":"9699597"},{"name":"xtreme","empid":"20","address":"stripes","mobile":"9699597"},{"name":"xtreme","empid":"20","address":"stripes","mobile":"9699597"},{"name":"xtreme","empid":"20","address":"stripes","mobile":"9689597"},{"name":"xtreme","empid":"20","address":"stripes","mobile":"9699597"},{"name":"xtreme","empid":"20","address":"stripes","mobile":"9689597"},{"name":"xtreme","empid":"20","address":"stripes","mobile":"9699597"},{"name":"vx","empid":"96","address":"addre","mobile":"9999596"},{"name":"vxx","empid":"96","address":"addre","mobile":"96899"},{"name":"vx","empid":"96","address":"addre","mobile":"9689596"}]

解析后,我想在员工对象上设置它的属性。我尝试了以下不起作用的代码:

NSDictionary *dict=[parser objectWithString:firstParseData];
NSString *secondParseData=[dict objectForKey:@"name"];
NSLog(@"name=%@",secondParseData);

我不明白是什么问题,请帮助我。

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1 回答 1

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如果firstParseData是您提供的 json,那么您将获得NSArrayindict变量,而不是NSDictionary.

如果您需要所有名称,请使用valueForKey:该数组的方法。它将返回dict变量中所有名称的数组。

如果您只需要一个名称 - 在返回的对象上使用objectAtIndex:andvalueForKey:来获取顶级数组的给定索引上的名称值。

于 2012-06-01T12:04:29.187 回答