带有百分比符号的密码始终是错误的iPhone [英] password with percentage symbol always wrong iPhone

问题描述

我正在使用具有登录功能的Xcode开发iPhone应用程序，但是在使用以下代码时遇到了麻烦:

I'm developing an iPhone application in Xcode with a login function and I'm having trouble with the following code:

NSString *post = [NSString stringWithFormat:@"username=%@&password=%@",[NSString stringWithFormat:@"%@",username],password];
NSLog(post);

NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:url]];
NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding];
[request setHTTPMethod:@"POST"];
[request addValue:[NSString stringWithFormat:@"%@",username] forHTTPHeaderField:@"username"];
[request addValue:password forHTTPHeaderField:@"password"];
[request setHTTPBody:postData];
action = @"token";
NSURLConnection *connection;
connection = [[NSURLConnection alloc] initWithRequest:request delegate:self];

现在有一个成员，其密码为％32dzs3 *，并且该应用程序始终会给出密码不正确的错误.当我NSLog密码时，实际上不是我在文本字段中键入的密码.它看起来像这样:50883393zs3 *.我也尝试过这个:

Now there is a member with the password %32dzs3* and the app always gives an error that the password is incorrect. When I NSLog the password it's indeed not the password I typed in the textfield. it looked like this: 50883393zs3* . I also tried this:

for(int i = 0; i < [password length];i++){
    if([password characterAtIndex:i] == '%'){
        NSString *temporarypw = [password substringWithRange:NSMakeRange(0, i)];
        password = [NSString stringWithFormat:@"%@%%%%%@",temporarypw,[password substringWithRange:NSMakeRange((i+1), (password.length - i -1))]];
        break;
    }
}

这将以%%%% 32dzs3 *的形式返回密码，并且在我NSLogLog时密码正确.当我运行请求时，响应仍然是错误的密码响应.我该如何解决?

this returns the password as %%%%32dzs3* and the password is right when I NSLog it. still the response is a wrong password response when I run the request. How can I fix this?

推荐答案

%是HTTP URL中的特殊字符.您需要

% is a special character in HTTP URLs. You need to escape it before creating the URL, also :, /, #, ;, and @. Fortunately NSString has a method to do it.

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