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]];
最佳答案
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];
关于iphone - 我怎样才能在ios5中替换这个语句,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9306018/