小编use*_*210的帖子

我有一个PHP脚本,用于连接MySQL数据库.通过mysql_connect连接工作完美,但在尝试使用PDO时,我收到以下错误:

SQLSTATE[HY000] [2005] Unknown MySQL server host 'hostname' (3)

我用来连接的代码如下:

    <?php  
    ini_set('display_errors', 1);
    error_reporting(E_ALL);

    $hostname_localhost ="hostname";  
    $database_localhost ="dbname";  
    $username_localhost ="user";  
    $password_localhost ="pass";  
    $user = $_GET['user'];  
    $pass = $_GET['pass'];

    try{
        $dbh = new PDO("mysql:host=$hostname_localhost;dbname=$database_localhost",$username_localhost,$password_localhost);
        echo 'Connected to DB';
        $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
        $stmt = $dbh->prepare("SELECT check_user_company(:user,:pass)");
        $stmt = $dbh->bindParam(':user',$user,PDO::PARAM_STR, 16);
        $stmt = $dbh->bindParam(':pass',$pass,PDO::PARAM_STR, 32);

        $stmt->execute();

        $result = $stmt->fetchAll();

        foreach($result as $row)
        {
            echo $row['company_id'].'<br />';
        }

        $dbh = null;
    }
    catch(PDOException $e)
    {
        echo $e->getMessage();
    } 
    ?>  

提前致谢

我在Android开发人员的摄像头上遇到了问题,这是我的代码:

// create Intent to take a picture and return control to the calling application
Intent intent = new Intent(MediaStore.ACTION_IMAGE_CAPTURE);
fileUri = getOutputMediaFileUri(MEDIA_TYPE_IMAGE);

// create a file to save the image
intent.putExtra(MediaStore.EXTRA_OUTPUT, fileUri); // set the image file name
// start the image capture Intent
startActivityForResult(intent, CAPTURE_IMAGE_ACTIVITY_REQUEST_CODE);

问题是'MEDIA_TYPE_IMAGE',它表示无法将其解析为变量.我将mediastore,camera和URI导入到我的项目中.提前致谢!