Question

我有以下有效的JSON到Parse：

[{"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"}]

解析之后我想在employee对象上设置它的属性。我试过以下代码无效：

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

我不明白有什么问题请帮助我。

Answer 1

如果firstParseData是您提供的json，那么您将在NSArray变量中获得dict，而不是NSDictionary。

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

如果只需要一个名称 - 在返回的对象上使用objectAtIndex:和valueForKey:来获取顶级数组的给定索引的名称值。

解析JSON无法正常工作

1 个答案: