Iphone sdk，如何将数据发布到网络服务器？有人有例子吗？

Question

伙计们，我想将数据发布到网络服务器，例如用户名和密码，在网络服务器上，我使用PHP来回显是或否，我附加了所有代码。谁能帮我看看代码有什么问题吗？因为它总是说密码或用户名不正确。我尝试用里面的用户名和密码测试 php 代码，它工作正常。所以请帮助我。谢谢。

头文件

@interface kiksyloginViewController : UIViewController {
    IBOutlet UITextField *usernameField;
    IBOutlet UITextField *passwordField;
    IBOutlet UIButton *loginButton;
}

@property (nonatomic, retain) UITextField *usernameField;
@property (nonatomic, retain) UITextField *passwordField;
@property (nonatomic, retain) UIButton *loginButton;

- (IBAction) login: (id) sender;

@end

实现文件

#import "kiksyloginViewController.h"

@implementation kiksyloginViewController

@synthesize usernameField;
@synthesize passwordField;
@synthesize loginButton;

- (IBAction) login: (id) sender
{

    NSString *post =[NSString stringWithFormat:@"username=%@&password=%@",usernameField.text, passwordField.text];

    NSString *hostStr = @"http://localhost/userlogin.php";
    hostStr = [hostStr stringByAppendingString:post];
    NSData *dataURL =  [NSData dataWithContentsOfURL: [ NSURL URLWithString: hostStr ]];    
    NSString *serverOutput = [[NSString alloc] initWithData:dataURL encoding: NSASCIIStringEncoding];


    if([serverOutput isEqualToString:@"Yes"]){

        UIAlertView *alertsuccess = [[UIAlertView alloc] initWithTitle:@"Congrats" message:@"You are authorized "
                                                              delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];

        [alertsuccess show];
        [alertsuccess release];

    } else {
        UIAlertView *alertsuccess = [[UIAlertView alloc] initWithTitle:@"Error" message:@"Username or Password Incorrect"
                                                              delegate:self cancelButtonTitle:@"OK" otherButtonTitles:nil, nil];
        [alertsuccess show];
        [alertsuccess release];

    }
}

- (void)didReceiveMemoryWarning {
    // Releases the view if it doesn’t have a superview.
    [super didReceiveMemoryWarning];

    // Release any cached data, images, etc that aren’t in use.
}

- (void)viewDidUnload {
    // Release any retained subviews of the main view.
    // e.g. self.myOutlet = nil;
}


- (void)dealloc {
    [super dealloc];
}

@end

PHP 文件

<?php 
//$Container = new Container; 

$u = $_GET['username']; 
$pw =$_GET['password']; 
$check = "select username, password from user where username='$u' and password='$pw'"; 
function myDbconn() 

{ 
    mysql_connect("localhost", "root","") or die(); 
    mysql_select_db("test") or die (mysql_error()); 
} 

myDbconn(); 


$login = mysql_query($check) or die (mysql_error()); 
//$run = DB()->select($check); 



if (mysql_num_rows($login)==1){ 
    //$row = DB()->fetch($run); 
    //$row = DB()->fetch($login); 
    $row = mysql_fetch_assoc($login); 
    echo 'yes'; 
    exit;         
} 
else { 
    echo 'No'; 
    exit; 
} 

?>

Answer 1

我看到的第一个问题可能是最大的问题是没有“？”任何地方。具体来说，是这样的:

@"username=%@&password=%@"

应该看起来更像这样:

@"?username=%@&password=%@"

因为您实际上并未发布到服务器，所以您只是使用 GET 请求和 URL 参数。

我不知道您是否计划在您的设备上后台运行 Web 服务器和 PHP，但在某些时候我假设您会更改 http://localhost更改为其他内容，希望您在让人们导航到您的网页并在 URL 查询中输入他们想要的任何内容之前先清理您的 SQL 查询。