php - Xcode 中使用 php 和 mysql 的登录程序无法正常工作

Question

嗨，我正在我的应用程序中执行登录页面。我在控制台中收到错误和无效消息。帮助我得到罚款结果。提前致谢。我的 Xcode 在这里

- (IBAction)login:(id)sender {

NSInteger success = 0;
@try {

    if([[self.username text] isEqualToString:@""] || [[self.password text] isEqualToString:@""] ) {

        [self alertStatus:@"Please enter Email and Password" :@"Sign in Failed!" :0];

    } else {
        NSString *post =[[NSString alloc] initWithFormat:@"username=%@&password=%@",[self.username text],[self.password text]];
        NSLog(@"PostData: %@",post);

        NSURL *url=[NSURL URLWithString:@"http://192.168.1.5:81/priya/login.php"];

        NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];

        NSString *postLength = [NSString stringWithFormat:@"%lu", (unsigned long)[postData length]];

        NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
        [request setURL:url];
        [request setHTTPMethod:@"POST"];
        [request setValue:postLength forHTTPHeaderField:@"Content-Length"];
        [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
        [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
        [request setHTTPBody:postData];

        //[NSURLRequest setAllowsAnyHTTPSCertificate:YES forHost:[url host]];

        NSError *error = [[NSError alloc] init];
        NSHTTPURLResponse *response = nil;
        NSData *urlData=[NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];

        NSLog(@"Response code: %ld", (long)[response statusCode]);

        if ([response statusCode] >= 200 && [response statusCode] < 300)
        {
            NSString *responseData = [[NSString alloc]initWithData:urlData encoding:NSUTF8StringEncoding];
            NSLog(@"Response ==> %@", responseData);

            NSError *error = nil;
            NSDictionary *jsonData = [NSJSONSerialization
                                      JSONObjectWithData:urlData
                                      options:NSJSONReadingMutableContainers
                                      error:&error];

            success = [jsonData[@"success"] integerValue];
            NSLog(@"Success: %ld",(long)success);

            if(success == 1)
            {
                NSLog(@"Login SUCCESS");
            } else {

                NSString *error_msg = (NSString *) jsonData[@"error_message"];
                [self alertStatus:error_msg :@"Sign in Failed!" :0];
            }

        } else {
            //if (error) NSLog(@"Error: %@", error);
            [self alertStatus:@"Connection Failed" :@"Sign in Failed!" :0];
        }
    }
}
@catch (NSException * e) {
    NSLog(@"Exception: %@", e);
    [self alertStatus:@"Sign in Failed." :@"Error!" :0];
}
if (success) {
    [self performSegueWithIdentifier:@"login_success" sender:self];
}}

和我的 php 编码

<?php

$host = "localhost"; 
$user = "xcode"; 
$pass = "xcode"; 
$db="xcode";

$r = mysql_connect($host, $user, $pass);

if (!$r) {
    echo "Could not connect to server\n";
    trigger_error(mysql_error(), E_USER_ERROR);
} else {
    echo "Connection established\n"; 
}

echo mysql_get_server_info() . "\n"; 
$r2 = mysql_select_db($db);

if (!$r2) {
    echo "Cannot select database\n";
     trigger_error(mysql_error(), E_USER_ERROR);
} else {
    echo "database selected\n"; 
}
$u=$_POST['username'];
$pw=$_POST['password'];
$result=mysql_query("select username,password from login where username='$u' and      password='$pw'");

$row=mysql_fetch_array($result);
if($row["username"]==$u && $row["password"]==$pw)
{
echo '{"success":1}';
}
else
{
 '{"success":0,"error_message":"Username and/or password is invalid."}';
 }

  mysql_close();
  ?>

我在控制台中得到了结果

2014-02-22 13:44:48.172 school apps[1605:a0b] PostData: username=admin&password=xcode
2014-02-22 13:44:49.252 school apps[1605:a0b] Response code: 200
2014-02-22 13:44:49.252 school apps[1605:a0b] Response ==> Connection established
5.5.34
database selected
{"success":1}
2014-02-22 13:44:49.253 school apps[1605:a0b] Success: 0

如果有任何教程，请给我 url。我对 Xcode 很陌生。

Answer 1

您需要将mysql数据库输出转换为json格式。

   //set the mime type
   $document->setMimeEncoding('application/json');
   echo json_encode($yourdata);

转换后，您需要

    jexit();

排除源代码中的 html。