ios - JSON 从 PHP 到 iOS 并从 UILabel 更新

Question

在另一位 SO 专家用户的帮助下，我正在实现一种方法，将 viewController 连接到远程 PHP 文件以更新 MySQL 字段值并让它也更新 UILabel .

我从 PHP 文件中回显的 JSON 值是这种格式:["34"]。

这是方法:

- (IBAction)votarAction:(id)sender {
    //URL definition where php file is hosted
    dispatch_queue_t backgroundQueue = dispatch_queue_create("com.mycompany.myqueue", 0);

    dispatch_async(backgroundQueue, ^{

        int categoriaID = [[detalleDescription objectForKey:@"idEmpresa"] intValue];

        NSString *string = [NSString stringWithFormat:@"%d", categoriaID];
        NSLog(@"ID EMPRESA %@",string);
        NSMutableString *ms = [[NSMutableString alloc] initWithString:@"http://mujercanariasigloxxi.appgestion.eu/app_php_files/cambiarvaloracionempresa.php?id="];
        [ms appendString:string];
        // URL request
        NSLog(@"URL = %@",ms);
        NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:ms]];
        //URL connection to the internet
        NSData *responseData = [[NSURLConnection alloc]initWithRequest:request delegate:self];
        NSLog(@"responseData= %@",responseData);

        //Parse JSON
        NSError *error = nil;
        NSArray  * json = [NSJSONSerialization
                                       JSONObjectWithData:responseData
                                       options: NSJSONReadingMutableContainers
                           error: &error];           

        int valoracionNumber = [[json objectAtIndex:0] integerValue];      



        dispatch_async(dispatch_get_main_queue(), ^{
            //update your label

            NSString *labelValue = [NSString stringWithFormat:@"%d", valoracionNumber];
            self.valoracionLabel.text = labelValue;
        });    
    });
}

这是抛出的异常:

2014-02-28 22:45:16.921 Vive Gran Canaria[928:582b] responseData= <NSURLConnection: 0xf050bc0> { request: <NSURLRequest: 0x9de3d40> { URL: http://mujercanariasigloxxi.appgestion.eu/app_php_files/cambiarvaloracionempresa.php?id=25 } }
2014-02-28 22:45:16.922 Vive Gran Canaria[928:582b] -[NSURLConnection bytes]: unrecognized selector sent to instance 0xf050bc0
2014-02-28 22:45:16.947 Vive Gran Canaria[928:582b] *** Terminating app due to uncaught exception 'NSInvalidArgumentException', reason: '-[NSURLConnection bytes]: unrecognized selector sent to instance 0xf050bc0'
*** First throw call stack:
(
    0   CoreFoundation                      0x00fca5e4 __exceptionPreprocess + 180
    1   libobjc.A.dylib                     0x02c5c8b6 objc_exception_throw + 44
    2   CoreFoundation                      0x01067903 -[NSObject(NSObject) doesNotRecognizeSelector:] + 275
    3   CoreFoundation                      0x00fba90b ___forwarding___ + 1019
    4   CoreFoundation                      0x00fba4ee _CF_forwarding_prep_0 + 14
    5   Foundation                          0x029a308c -[_NSJSONReader findEncodingFromData:withBOMSkipLength:] + 36
    6   Foundation                          0x029a323b -[_NSJSONReader parseData:options:] + 63
    7   Foundation                          0x029a3800 +[NSJSONSerialization JSONObjectWithData:options:error:] + 161
    8   Vive Gran Canaria                   0x00009bca __44-[DetalleEmpresaViewController votarAction:]_block_invoke + 602
    9   libdispatch.dylib                   0x02f5b7f8 _dispatch_call_block_and_release + 15
    10  libdispatch.dylib                   0x02f704b0 _dispatch_client_callout + 14
    11  libdispatch.dylib                   0x02f5e07f _dispatch_queue_drain + 452
    12  libdispatch.dylib                   0x02f5de7a _dispatch_queue_invoke + 128
    13  libdispatch.dylib                   0x02f5ee1f _dispatch_root_queue_drain + 83
    14  libdispatch.dylib                   0x02f5f137 _dispatch_worker_thread2 + 39
    15  libsystem_c.dylib                   0x03288e72 _pthread_wqthread + 441
    16  libsystem_c.dylib                   0x03270daa start_wqthread + 30
)
libc++abi.dylib: terminating with uncaught exception of type NSException (lldb) 

这是 PHP 文件中与 JSON 数组相呼应的部分:

$arr = array();

$rs = mysql_query("SELECT * FROM tbempresas where idEmpresa='$id'");
while ($obj = mysql_fetch_assoc($rs)) {
    $arr[] = $obj['valoracionEmpresa'];
}
echo json_encode($arr);

Answer 1

你必须创建一个 NSURLConnection 的对象而不是 NSData 作为响应，你将在 NSURLConnection 的委托(delegate)方法中获取数据。

NSURLConnection *connection=[[NSURLConnection alloc]initWithRequest:request delegate:self];     

#pragma NSURLConnection Delegate Methods
- (void)connection:(NSURLConnection *)connection didReceiveResponse:(NSURLResponse *)response
{
    //buffer is the object Of NSMutableData and it is global,so declare it in .h file
    buffer = [NSMutableData data];
}

- (void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data 
{
    [buffer appendData:data];
}

- (void)connectionDidFinishLoading:(NSURLConnection *)connection
{
    //Here You will get the whole data 
    NSArray *array = [NSJSONSerialization JSONObjectWithData:buffer options:0 error:&jsonParsingError];
    //And you can used this array
}

- (void)connection:(NSURLConnection *)connection didFailWithError:(NSError *)error
{
    //if any error occurs..
}