如何通过PickerValue在URL中的iphone

Question

arrayNo = [[NSMutableArray alloc] init];    //pickerview radius array 
//[arrayNo addObject:@" Radius "]; 

[arrayNo addObject:@" 10 "]; 
[arrayNo addObject:@" 15000 "]; 
[arrayNo addObject:@" 20000 "]; 
[pickerView selectRow:1 inComponent:0 animated:NO]; 
[addressField resignFirstResponder]; 
radius.text= [arrayNo objectAtIndex:[pickerView selectedRowInComponent:0]]; 

我想通过上面的半径选择器在以下网址在半径

NSString *url = [NSString stringWithFormat:@"http://.............../phpsqlsearch_genxml.php?lat=%f&lng=%f&radius=%f",lati,longi,radius]; 

我用下面的按钮单击事件代码，但不工作尝试。

NSString * radius = [[[NSString alloc] initWithFormat:@"%f",[arrayNo objectAtIndex:[pickerView selectedRowInComponent:0]]]autorelease]; 

我应该写什么代码一样的...在此先感谢

Answer 1

你可以改变你的代码，

float radius = [arrayNo objectAtIndex:[pickerView selectedRowInComponent:0]] floatValue]; 
NSString *url = [NSString stringWithFormat:@"http://.............../phpsqlsearch_genxml.php?lat=%f&lng=%f&radius=%f",lati,longi,radius]; 

而且，你并不需要最后一行NSString * radius = ... 。