如何将 JSON 解析为 Objective C - SBJSON

问题描述

能否请您告诉我如何传递如下所示的 JSON 字符串:

Could you please tell me how to pass a JSON String which looks like this:

{"lessons":[{"id":"38","fach":"D","stunde":"t1s1","user_id":"1965","timestamp":"0000-00-00 00:00:00"},{"id":"39","fach":"M","stunde":"t1s2","user_id":"1965","timestamp":"0000-00-00 00:00:00"}]}

我试过这样:

SBJSON *parser =[[SBJSON alloc] init];
    NSArray *list = [[parser objectWithString:JsonData error:nil] copy];
    [parser release];
    for (NSDictionary *stunden in list)
    {
        NSString *content = [[stunden objectForKey:@"lessons"] objectForKey:@"stunde"];

    }

提前致谢

最好的问候

推荐答案

请注意，您的 JSON 数据具有以下结构:

Note that your JSON data has the following structure:

1. 顶级值是一个对象(字典)，它有一个名为课程"的属性
2. 课程"属性是一个数组
3. lessons"数组中的每个元素都是一个对象(包含课程的字典)，具有多个属性，包括stunde"

对应的代码是:

SBJSON *parser = [[[SBJSON alloc] init] autorelease];
// 1. get the top level value as a dictionary
NSDictionary *jsonObject = [parser objectWithString:JsonData error:NULL];
// 2. get the lessons object as an array
NSArray *list = [jsonObject objectForKey:@"lessons"];
// 3. iterate the array; each element is a dictionary...
for (NSDictionary *lesson in list)
{
    // 3 ...that contains a string for the key "stunde"
    NSString *content = [lesson objectForKey:@"stunde"];

}

几个观察:

• 在-objectWithString:error:中，error参数是一个指向指针的指针.在这种情况下，使用 NULL 而不是 nil 更为常见.not 传递 NULL 并使用 NSError 对象检查错误也是一个好主意，以防方法返回 nil