JSON 到 Objective-C 字典

Posted 2023-03-08

【中文标题】JSON 到 Objective-C 字典

【英文标题】：JSON to Objective-C Dictionary

【发布时间】：2012-09-16 23:05:42

【问题描述】：

我正在向 API 发出 URL 请求，但我不知道如何呈现 JSON，它会生成一个由多个用户组成的数组，例如 ["user": "value", "user":"value"]，我正在尝试使用 TableView，所以我需要一个 NSDictionary 但我认为最好渲染像users: ["user": "value", "user":"value"] 这样的JSON。我有这个代码来提出请求

#import "JSONKit.h"
NSError *error = nil;
NSURLResponse *response = nil;
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL: [NSURL URLWithString: @"http://localhost:3000/getusers"]];
[request setHTTPMethod:@"GET"];
NSData *jsonData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
users = [[jsonData objectFromJSONData] objectForKey:@"users"];
usersKeys = [users allKeys];

但我收到此错误

2012-09-16 18:51:11.360 tableview[2979:c07] -[JKArray allKeys]：无法识别的选择器发送到实例 0x6d30180 2012-09-16 18:51:11.362 tableview[2979:c07] * 由于未捕获的异常“NSInvalidArgumentException”而终止应用程序，原因：“-[JKArray allKeys]：无法识别的选择器发送到实例 0x6d30180”

我真的不知道如何做到这一点，所以任何帮助都是有用的，谢谢

【参考方案1】：

您收到该错误是因为从“jsonData”中解析出来的内容不一定是您所期望的（即字典）。

也许您需要在您的代码中进行一些错误检查。

例如：

NSData *jsonData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
if(jsonData)

    id objectReturnedFromJSON = [jsonData objectFromJSONData];
    if(objectReturnedFromJSON)
    
        if([objectReturnedFromJSON isKindOfClass:[NSDictonary class]])
        
            NSDictionary * dictionaryFromJSON = (NSDictionary *)objectReturnedFromJSON;
            // assuming you declared "users" & "usersKeys" in your interface,
            // or somewhere else in this method
            users = [dictionaryFromJSON objectForKey:@"users"];
            if(users)
            
                usersKeys = [users allKeys];
             else 
                NSLog( @"no users in the json data");
            
         else 
            NSLog( @"no dictionary from the data returned by the server... check the data to see if it's valid JSON");
        
     else 
        NSLog( @"nothing valid returned from the server...");
    
 else 
    NSLog( @"no data back from the server");

【参考方案2】：

我在想这样的事情

NSError *error = nil;
NSURLResponse *response = nil;
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL: [NSURL URLWithString: @"http://localhost:3000/getusers"]];
[request setHTTPMethod:@"GET"];
NSData *jsonData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

JSONDecoder *decoder = [[JSONDecoder alloc]
                        initWithParseOptions:JKParseOptionNone];
NSArray *json = [decoder objectWithData:jsonData];

NSMutableArray *objects = [[NSMutableArray alloc] init];
NSMutableArray *keys = [[NSMutableArray alloc] init];
for (NSDictionary *user in json) 
    [objects addObject:[user objectForKey:@"user" ]];
    [keys addObject:[user objectForKey:@"value" ]];

users = [[NSDictionary alloc] initWithObjects:objects forKeys:keys];
NSLog(@"users: %@", users);
usersKeys = [users allKeys];

但是对于许多项目来说它看起来效率不高还是我错了？