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基本上我想使用 php PDO 创建一个查询来检查我的数据库“test”中是否存在表“page”。我不知道该怎么做,我在这里得到了一些帮助。

我的代码运行完美......直到现在我把所有东西都放在课堂上......现在var_dump($r2)返回NULL,我不知道代码有什么问题。除了将它放入OOP之外,我没有改变任何东西......

谁能发现问题??因为我看不到。谢谢你

$r1 = $this->db->query('SHOW TABLES LIKE \'page\'');


 // Debbug
    $r2 = $r1->fetchAll;
    var_dump ($r2);

    if (count($r1->fetchAll()) > 0 ) {

        echo "The table PAGE exists";

    }

完整的类是以下一个

    class phase2 {

        function __construct () {

        $dbFile = 'dbconfig.php';
        $this->dbFile = $dbFile;

        require_once ("$dbFile");   


        $step = $_GET["step"];

        $username = $DB_USER;
        $password = $DB_PASS;
        $server = $DB_SERVER;
        $dbName = $DB_NAME;

        $this->step = $step;
        $this->dbFile = $dbFile;
        $this->username = $username;
        $this->password = $password;
        $this->server = $server;
        $this->dbName = $dbName;

        $db = new PDO ('mysql:host=' .$server.';dbname='.$this->dbName,$this->username,$this->password);

        $this->db = $db;

        if (empty ($_GET['fot']) ) { 

            $fOT = 'false'; 

        } elseif ($_GET['true']) { $fOT = 'true'; }

        $this->fOT = $fOT;

        $this->IDB = $this->handleDatabase( 1 );
        $this->IDB2 = $this->handleDatabase( 2 );
        $this->IDB3 = $this->handleDatabase( 3 );

        }



public function handleDatabase ($num = 1){

// Prepare SQL Statements

    $IDB1 = $this->db->prepare( 
         "CREATE TABLE pages (
          id int(11) NOT NULL auto_increment,
         subject_id int(11) NOT NULL,
          menu_name varchar(30) NOT NULL,
          position int(3) NOT NULL,
          visible tinyint(1) NOT NULL,
          content text NOT NULL,
          PRIMARY KEY  (id)
    )ENGINE=MyISAM AUTO_INCREMENT=7 DEFAULT CHARSET=utf8");

    $IDB2 = $this->db->prepare("
        CREATE TABLE subjects (
          id int(11) NOT NULL auto_increment,
          menu_name varchar(30) NOT NULL,
          position int(3) NOT NULL,
          visible tinyint(1) NOT NULL,
          PRIMARY KEY  (id)
    )ENGINE=MyISAM AUTO_INCREMENT=6 DEFAULT CHARSET=utf8");

    $IDB3 = $this->db->prepare("
        CREATE TABLE users (
          id int(11) NOT NULL auto_increment,
          username varchar(50) NOT NULL,
          hashed_password varchar(40) NOT NULL,
          PRIMARY KEY  (id)
    )ENGINE=MyISAM AUTO_INCREMENT=2 DEFAULT CHARSET=utf8");

    $name = "IDB".$num;
    return isset( $$name)?$$name:false;

}
//Set Option to True or False

function createTablePages ($fOT){


    $r1 = $this->db->query('SHOW TABLES LIKE \'page\'');

// Debbug
    $r2 = $r1->fetchAll;
    var_dump ($r2);


    if (count($r1->fetchAll()) > 0) {


        echo "The table PAGE exists";

    } elseif ($fOT == 'true') {

            echo "enteres";
            $this->IDB1->execute();
            $this->stepFunction (1,false);

    } 

}
function createTableSubjects ($fOT){

    $r2 = $this->db->query('SHOW TABLES LIKE \'subjects\'');

    if (count($r2->fetchAll()) > 0  && $fOT == 'false') {

            echo "The table SUBJECTS exists ";

    } elseif ($fOT == 'true') {

        $this->IDB2->execute();
        $this->stepFunction (2,false);

    }
}

function createTableUsers ($fOT){

    $r3 = $this->db->query('SHOW TABLES LIKE \'users\'');   

    if (count($r3->fetchAll()) > 0  && $fOT == 'false') {

            echo "The table USERS exists";

    } elseif ($fOT == 'true') {

            $this->IDB3->execute();
            echo "Would you like to populate all the tables?";
    }   
}


public function stepFunction ($fOT,$step){

switch ($step) {

    case 0: 
            $this->createTablePages ($fOT);
            break;
    case 1: 
            $this->createTableSubjects($fOT);
            break;
    case 2: $this->createTableUsers ($fOT);
            break;
    }


}

    }
4

3 回答 3

1

我可以看到的最大问题是您没有创建 $db - 仅在构造内部。尝试添加到此部分:

class phase2 {

    function __construct () {

添加此语句public $db;

class phase2 {

    public $db;

    function __construct () {

除非我弄错了,否则您不能在不先声明的情况下从方法中强制转换变量。您需要对需要从该类中的其他方法访问的任何其他变量执行相同的操作。阅读基础知识:http ://www.php.net/manual/en/language.oop5.basic.php

另外,我建议打开错误报告。

于 2012-08-29T23:00:50.753 回答
1

您的查询试图找到一个名为 的表page,但是,您CREATE TABLE创建了一个名为 的表pages

$IDB1 = $this->db->prepare( 
    "CREATE TABLE pages ("
...

$r1 = $this->db->query('SHOW TABLES LIKE \'page\'');

除非您实际上同时拥有两张表,否则错误出在这两个地方之一。

于 2012-08-29T23:06:03.637 回答
0

您在构造函数中使用 require_once() 。该类只会在第一次正确初始化。之后配置将不会加载,因此不会设置变量。

于 2012-08-30T00:39:58.553 回答