ASIHTTPRequest简单帖子失败

时间:2012-06-17 17:57:28

标签: iphone objective-c ios post asihttprequest

我是objective-c的新手,我现在正试图在iPhone上编写一个简单的用户登录程序。我的想法是使用ASIHTTPrequest中的类将信息发布到localhost中的一个php文件,该文件链接到db。我设置了一个按钮,如果你按下它,它会发出一个帖子请求。但是,没有任何内容插入到db。 这是我的代码:

.h文件:

@interface ViewController : UIViewController
-(IBAction)post;

.m文件:

#import "ViewController.h"
#import "ASIHTTPRequest.h"
#import "ASIFormDataRequest.h"

@implementation ViewController

-(IBAction)post {
    NSURL *url = [NSURL URLWithString:@"localhost:8888/index.php"];
    ASIFormDataRequest *request = [ASIFormDataRequest requestWithURL:url];
    [request setPostValue:@"paul" forKey:@"user"];
    [request setPostValue:@"12345" forKey:@"pw"];
}

index.php文件:

<?php
include('database.php');
session_start();
$link = connect_mysql();
mysql_select_db("test",$link);

if (($_POST["user"]=="")||($_POST["pw"]=="")) { header('Location:error.html'); exit(); }
      $sql = 'select * from userinfor where name="'.$_POST["user"].'";';
      $result = mysql_query($sql, $link);
      $pass= mysql_fetch_object($result);
      if ($pass) { header('Location:again.html'); exit();}
      if ($_POST["pw"]!=$_POST["pw2"]) { header('Location:error2.html'); exit();}
      $sql = 'INSERT INTO userinfor (name, password) values ("'.$_POST["user"].'","'.$_POST["pw"].'")';
      $result = mysql_query($sql, $link);
      if (!$result) {
        echo "DB Error, could not query the database\n";
        echo 'MySQL Error: '. mysql_error();
        exit();
      } else {
        echo 'Sucessfully added entry<br/>';
        echo '<a href="login.html">Click to login</a>';
      }
?> 

database.php文件:

<?php
function connect_mysql() {
  //Connect to database (Please specify the password field if you needed
    $link = mysql_connect('localhost', 'root', '12345');
    if (!$link) {
      die('Could not connect to mysql');
    }

    $db = mysql_select_db('test', $link);
    if (!$db) {
      die('Could not select database');
    }

   return $link;

}

?>

1 个答案:

答案 0 :(得分:0)

问题是您没有说出要求启动请求。添加

[request startAsynchronous];

[request startSynchronous];

此外,由于您必须成为请求的委托,请确保添加方法(成为委托人:[request setDelegate:self];):

- (void)requestFinished:(ASIHTTPRequest *)request;

- (void)requestFailed:(ASIHTTPRequest *)request;