电子邮件地址中的"+"号

Question

我需要提交一个带有" +"符号的电子邮件地址,并在服务器上进行验证.但服务器接收" aaa+bbb@mail.com"为" aaa bbb@mail.com"的电子邮件.

我使用以下代码将所有数据作为POST请求发送

NSURL* url = [NSURL URLWithString:[NSString stringWithFormat:@"%@%@", url, @"/signUp"]];

NSString *post = [NSString stringWithFormat:@"&email=%@&userName=%@&password=%@",
                      user.email, 
                      user.userName, 
                      user.password];

NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:NO];

NSData* data = [self sendRequest:url postData:postData];

编码前的post变量有值

&email=aaa+bbb@gmail.coma&userName=Asdfasdfadsfadsf&password=sdfasdf

编码后它是一样的

&email=aaa+bbb@gmail.coma&userName=Asdfasdfadsfadsf&password=sdfasdf

我用来发送请求的方法如下代码:

-(id) sendRequest:(NSURL*) url postData:(NSData*)postData {
    // Create request
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];

    NSString *postLength = [NSString stringWithFormat:@"%d",[postData length]];

    [request setURL:url];
    [request setHTTPMethod:@"POST"];
    [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Current-Type"];
    [request setValue:postLength forHTTPHeaderField:@"Content-Length"];
    [request setHTTPBody:postData];

    NSURLResponse *urlResponse;

    NSData *data = [NSURLConnection sendSynchronousRequest:request returningResponse:&urlResponse error:nil];

    [request release];

    return data;
}

Answer 1

电子邮件,用户名和密码需要转义-stringByAddingPercentEscapesUsingEncoding:.

NSString *post = [NSString stringWithFormat:@"&email=%@&userName=%@&password=%@",
                  [user.email stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding],
                  ...

但是,这不会逃脱,+因为它是一个有效的URL字符.您可以使用更复杂的CFURLCreateStringByAddingPercentEscapes,或者为了简单起见,只需将以下内容替换+为%2B:

NSString *post = [NSString stringWithFormat:@"&email=%@&userName=%@&password=%@",
                  [[user.email stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]
                   stringByReplacingOccurrencesOfString:@"+" withString:@"%2B"], ...