Question

我正在该应用程序中开发一个应用程序我需要在NSURL中一次传递多个参数我的代码是

responseData = [[NSMutableData data] retain];
ArrData = [NSMutableArray array];
NSURL *url = [NSURL URLWithString:[NSString stringWithFormat:@"http://rate-exchange.appspot.com/currency?from=%@&to=%@&q=%@",strfrom,strto,strgo]];
NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:url]];
//NSURLRequest *request1 = [NSURLRequest requestWithURL:
//[NSURL URLWithString:@"http://rate-exchange.appspot.com/currency?from=%@&to=%@&q=1",strfrom,strto]];

上面的代码我需要动态传递多个参数。可能吗？如果是，那怎么样？谢谢＆amp;问候

Answer 1

在添加到

之类的URL之前尝试创建单独的字符串

 NSSString *strURL=[NSString stringWithFormat:@"http://rate-exchange.appspot.com/currency?from=%@&to=%@&q=%@"‌,strfrom,strto,strgo];

然后将此strURL添加到URL

NSURL *url = [NSURL URLWithString:strURL];

最后将其添加到请求中，您的代码错误，您在请求中添加了网址，URL不是字符串，而是URL，因此它应该是requestWithURL而不是URLWithString，它应该像这个

NSURLRequest *request = [NSURLRequest requestWithURL:url];

Answer 2

缺少许多这些答案的一件事是使用[NSString stringByAddingPercentEscapesUsingEncoding:]来避免在网址中使用无效字符：

NSString *baseURL = [NSString stringWithFormat:@"http://rate-exchange.appspot.com/currency?from=%@&to=%@&q=%@",strfrom,strto,strgo];
NSURL *url = [NSURL URLWithString:[baseURL stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
NSURLRequest *request = [NSURLRequest requestWithURL:url];

如何在iOS中传递NSURL字符串中的多个参数？

2 个答案: