Question

我从我的网络服务中获取此JSON对象

0:  {
Username: "John"
Password: "12345"
Position: "Admin"
Job: "Developer"
ResourceJobId: 1
}

当我尝试将值分配给Username JSON字符串时，我收到语义错误：

id obj = [NSJSONSerialization JSONObjectWithData:data
                                                 options:NSJSONReadingAllowFragments
                                                   error:&error];
            for (NSDictionary *dictionary in array) {
                NSString *usernameString = @"";
//this line of code gives me an error: Expression is not assignable               
[dictionary objectForKey:@"Username"] = usernameString   ;

我知道如何使用NSJSONSerialization类获取JSON对象，但我的目标是如何将值@“”分配给JSON对象。那么如何将值@“”分配给Username JSON字符串？

Answer 1

当您尝试从dictionary抓取array时。它不是可变字典，因此您需要在从NSMutableDictionary获取数据时创建Array。以下代码可帮助您更改username。

for (NSMutableDictionary *dictionary in array)
{
   NSMutableDictionary* dictTemp = [NSMutableDictionary dictionaryWithDictionary:dictionary];

   [dictTemp setObject:usernameString forKey@"Username"];

   [_arrayList replaceObjectAtIndex:index withObject:dictTemp];
}

Answer 2

你的作业操作写错了，应该是这样的：

如果你需要从字典中获取价值

usernameString = [dictionary objectForKey:@"Username"];

如果您想为字典设置一个值，首先您的字典应为NSMutableDictionary
```
[dictionary setObject:usernameString forKey:@"Username"];
```

Answer 3

 NSData *data = [NSData dataWithContentsOfURL:url];
        NSError *error;
        id obj = [NSJSONSerialization JSONObjectWithData:data
                                                 options:NSJSONReadingAllowFragments
                                                   error:&error];

        if(!error && [obj isKindOfClass:[NSArray class]]){
            NSArray *array = (NSArray *)obj;
            Booking *booking = nil;
            for (NSMutableDictionary *dictionary in array) {
               NSString *usernameString = @"";
//this line of code gives me an error: Expression is not assignable               
[dictionary objectForKey:@"Username"] = usernameString;

是的，是编译错误

Answer 4

您需要有一个可变字典才能更改该值。试试这个

for (NSDictionary *dictionary in array) {
   NSMutableDictionary *mutableDictionary = [NSMutableDictionary dictionaryWithDictionary:dictionary];
   NSString *usernameString = @"";     
   [mutableDictionary setValue:@"" forKey:usernameString];
}

Answer 5

即使答案是正确的，我想补充一点，你不必自己动手。 NSJSONSerialization已经有了解决方法。只需传递以下选项之一：

NSJSONReadingMutableContainers = (1UL << 0),
NSJSONReadingMutableLeaves = (1UL << 1),

阅读JSON时。

如何为JSON字符串赋值？

5 个答案: