iphone - Xcode sdk 将参数传递给 NSURLRequest

Question

对于我的 iphone 应用程序，我需要使用 url 中的参数向服务器发出请求。我是开发 ios 应用程序的新手，我不知道如何将我设置的值传递到 url。

我已经尝试过这样的事情，但它说 URLWITHString 的“参数太多”。我如何以正确的方式写下来？

  NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:@"http://test.localhost/api/request.php?url=%@", passedValue] cachePolicy: NSURLRequestUseProtocolCachePolicy timeoutInterval:15.0];
NSURLConnection *connection = [[NSURLConnection alloc] initWithRequest:request delegate:self];

在此先感谢，尼克

Answer 1

NSString *urlString = [NSString stringWithFormat:@"http://test.localhost/api/request.php?url=%@", passedValue]; 
NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:urlString] cachePolicy: NSURLRequestUseProtocolCachePolicy timeoutInterval:15.0];

编辑 -

NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:[NSString stringWithFormat:@"http://test.localhost/api/request.php?url=%@", passedValue]] cachePolicy: NSURLRequestUseProtocolCachePolicy timeoutInterval:15.0];

Answer 2

试试这个：-

NSString *tempURL = [NSString stringWithFormat:@"http://test.localhost/api/request.php?url=%@", passedValue]; 
NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:tempURL] cachePolicy: NSURLRequestUseProtocolCachePolicy timeoutInterval:15.0];
NSURLConnection *connection = [[NSURLConnection alloc] initWithRequest:request delegate:self];

Answer 3

你必须给你的 urlwithstring 参数一个 NSString 对象

NSURLRequest *request = [NSURLRequest requestWithURL:[NSURL URLWithString:[NSString stringWithFormat:@"http://test.localhost/api/request.php?url=%@", passedValue]] cachePolicy: NSURLRequestUseProtocolCachePolicy timeoutInterval:15.0];
NSURLConnection *connection = [[NSURLConnection alloc] initWithRequest:request delegate:self];