方法调用预期1的参数太多有3个

Question

请帮我找错,我的代码是

NSURL *url = [NSURL URLWithString: @"http://www.xxxx.com/xxxx_webservice/login.php?user=%@&pass=%@&format=json",Username,Password];

Answer 1

像这样做.首先制作一个正确的URL字符串,然后对其进行编码,最后制作一个NSURL.确保传递正确和正确的参数集.

NSString *link = [NSString stringWithFormat:@"http://www.example.com/xx_webservice/login.php?user=%@&pass=%@&format=json",Username,Password];
NSString *encdLink = [link stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
NSURL *url=[NSURL URLWithString:encdLink];