如何在 iphone 的 URL 中传递 PickerValue

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]];

我想在半径的以下网址中传递上述半径选择器

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

我尝试在按钮单击事件上使用以下代码，但无法正常工作。

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

我应该为相同的代码编写什么代码...在此先感谢

Answer 1

您可以将代码更改为，

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，你不需要最后一行 NSString * radius = ...。