iOS中未使用格式字符串的數據參數

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

你已經錯過了格式說明嘗試使用此

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

Answer 2

您的格式字符串只有2個格式說明符，但您傳遞的是3個參數。讓他們匹配。

檢查：

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

Answer 3

試用確認passord這個..你已經錯過了格式說明(%@)。

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

更新：嘗試網址與http://

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

入住這也..

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