Question

我想要改变我数据库中的形象,以便用我的“智者”来检索它,并且为了做到这一点,我没有这样做。

Inside my button onTouch i have written this code

//create string contain url address for php file, the file name is phpFile.php,it receives parameter :name 
NSString *strURL=[NSString stringWithFormat:@"....my.php","hhhhh"];

//to execute php code
NSData *dataURL=[NSData dataWithContentsOfURL:[NSURL URLWithString:strURL]]; 

//to receive the returned Result 
NSString *strResult = [[[NSString alloc] initWithData:dataURL encoding:NSUTF8StringEncoding]self]; 
NSLog(@"%@",strResult);

Answer 1

There are different ways of doing this:

If you deployment target is iOS 5.0 and up you could use: NSJSonSerialization

if you want to support versions below 5.0 you could use one of many libraries to do that, personally I find SBJson quite good. Take a look at this snippet to get an idea :

NSString *dataString = [[NSString alloc] initWithData:self.receivedData encoding:NSUTF8StringEncoding];
NSDictionary *dic = [dataString JSONValue];
NSArray * element = [dic objectForKey:@"jsonKey"];
NSString * someSimpleElement = [dic objectForKey:@"jsonKey"];

Answer 2

如果你针对Si,或者后来,你可以使用NSJSONSerialization。

NSError& anError = nil;
id foo = [NSJSONSerialization JSONObjectWithData: dataURL options: 0 error: &anError];
if (foo == nil)
{
    // handle the error
}
else
{
    //play with the returned object
}

如果你需要瞄准较早版本的SOS,那么你就需要SBJSON或TJSON或类似,但原则是相同的。

NB 还请读到,现行代码将服务器的数据同步输入。如果这是主线,那么,在你获得数据或连接时间之前,调查股将予以冻结。

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