iphone - 我怎样才能在ios5中替换这个语句

Question

当我写“stringWithContentsOfURL”时，我收到错误“stringWithContentsOfURL 已弃用”。

    -(CLLocationCoordinate2D) addressLocation 
{
        NSString *urlString = [NSString stringWithFormat:@"http://maps.google.com/maps/geo?q=%@&output=csv", 
      [addressField.text stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
        NSString *locationString = [NSString stringWithContentsOfURL:[NSURL URLWithString:urlString]];
        NSArray *listItems = [locationString componentsSeparatedByString:@","];

        double latitude = 0.0;
        double longitude = 0.0;

        if([listItems count] >= 4 && [[listItems objectAtIndex:0] isEqualToString:@"200"]) {
            latitude = [[listItems objectAtIndex:2] doubleValue];
            longitude = [[listItems objectAtIndex:3] doubleValue];
        }
        else {
            //Show error
        }
        CLLocationCoordinate2D location;
        location.latitude = latitude;
        location.longitude = longitude;

        return location;
    }

我怎样才能替换这条线。谢谢!

NSString *locationString = [NSString stringWithContentsOfURL:[NSURL URLWithString:urlString]];

Answer 1

已替换为

stringWithContentsOfURL:encoding:error: 

或

stringWithContentsOfURL:usedEncoding:error:

示例代码

NSError* error = nil;
NSString* locationString = [NSString stringWithContentsOfURL:[NSURL URLWithString:urlString] encoding:NSASCIIStringEncoding error:&error];