iOS中格式字符串未使用的数据参数

Question

NSURL *url = [NSURL URLWithString: @"www.bikrionline.com/database/iqry_mbr_mst.php"];
NSString *body = [NSString stringWithFormat: @"txtemail=%@&txtpwd=%@&txtcpwd", email, password,confirmpassword];
NSMutableURLRequest *request = [[NSMutableURLRequest alloc]initWithURL: url];
[request setHTTPMethod: @"POST"];
[request setHTTPBody: [body dataUsingEncoding: NSUTF8StringEncoding]];
[_webView loadRequest: request];

Answer 1

您的格式字符串只有2个格式说明符，但您传递的是3个参数。让它们匹配。

检查一下：

[NSString stringWithFormat: @"txtemail=%@&txtpwd=%@&txtcpwd=%@", email, password,confirmpassword];

Answer 2

试试这个..你错过了格式说明符(%@)来确认密码。

NSString *body = [NSString stringWithFormat: @"txtemail=%@&txtpwd=%@&txtcpwd=%@", email, password,confirmpassword];

更新：使用http://

尝试网址

NSURL *url = [NSURL URLWithString: @"http://www.bikrionline.com/database/iqry_mbr_mst.php"];

同时检查..

UIWebView *webView=[[UIWebView alloc] initWithFrame:CGRectMake(0,0,320,480);
[self.view addSubView:webView];

Answer 3

您错过了格式说明符，请尝试使用此

NSString *body = [NSString stringWithFormat: @"txtemail=%@&txtpwd=%@&txtcpwd=%@", email, password,confirmpassword];