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我已经在我的应用程序中使用了该项目，并在我想在视图控制器的其他部分中使用的部分中使用了AFJSONRequest它返回的数据。success我这辈子都想不通。我创建了一个属性来保存数据，但我无法在其他方法中访问它。这有什么原因吗？这可能是我做错了什么。如何在其他方法中使用 nameArray、emailArray 和 passArray？它们已经被声明为属性。这是我的代码：

- (void)viewDidLoad
{
[super viewDidLoad];
NSURL *url = [NSURL URLWithString:@"http://10.247.245.87/it/emailmanager.php"];
NSURLRequest *request = [NSURLRequest requestWithURL:url];
[AFJSONRequestOperation addAcceptableContentTypes:[NSSet setWithObject:@"text/html"]];
AFJSONRequestOperation *operation = [AFJSONRequestOperation JSONRequestOperationWithRequest:request success:^(NSURLRequest *request, NSHTTPURLResponse *response, id JSON)
                                     {
                                         NSLog(@"email.last_name is of type: %@", [[JSON valueForKeyPath:@"email.lastname" ] class]);
                                         nameArray = [JSON valueForKeyPath:@"email.last_name"];
                                         emailArray = [JSON valueForKeyPath:@"email.email_address"];
                                         passArray = [JSON valueForKeyPath:@"email.password"];
                                         NSLog(@"%@", nameArray);
                                     }
                                                                                    failure:^(NSURLRequest *request, NSHTTPURLResponse *response, NSError *error, id JSON)
                                     {
                                         NSLog(@"error %@ %@", [error description], JSON);
                                     }];

[operation start];
}

这是其余的。

- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath
{
EMailCell* cell = [tableView dequeueReusableCellWithIdentifier:@"EmailCell"];
cell.nameLabel.text = @"%@", nameArray;
cell.emailLabel.text = @"%@", emailArray;
cell.passLabel.text = @"%@", passArray;
return cell;

}

以下是 JSON 结果的结构：

2012-11-13 07:43:46.026 IT Tools[13987:c07] Names: (
        {
    code = 0496;
    department = Management;
    "email_address" = "????@???.com";
    "first_name" = John;
    id = 227;
    "last_name" = Doe;
    password = "T0496)$(^";
    store = Toyota;
},

调试器控制台的结果： 2012-11-13 09:04:37.416 IT Tools[15386:c07] email.last_name is of type: __NSArrayI

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1 回答 1

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You are assigning arrays to string values. You should loop through the JSON results and create arrays of strings... It would help to know the structure of your JSON results. If you are using core data, there is a handy method for parsing JSON into core data objects.

于 2012-11-13T13:33:16.917 回答