使用外部腳本從具有不同數據庫的多個 wordpress 站點獲取前綴

Question

我正在嘗試獲取具有不同前綴的 wordpress 安裝的多個數據庫的前綴。 但是沒有做到這一點，我只能得到所有文件的第一個表前綴：

文件夾結構（基本 - 包含所有必需的 wordpress 文件）：

home:
+---cpanel1
|   \---public_html
|       +---wp-load.php => $table_prefix = 'wp1_';
+---cpanel2
|   \---public_html
|       +---wp-load.php => $table_prefix = 'wp2_';
\---cpanel3
    \---public_html
        +---wp-load.php => $table_prefix = 'wp3_';
+---get_prefixes.php

腳本 get_prefixes.php (PHP)：

$myfile = '/wp-load.php';
class execute_sql_code {
    public function __construct($myfile) {
        $this->main_work($myfile);
    }
    public $cpanels = array(
            'cpanel1',
            'cpanel2',
            'cpanel3'
    );
    public function main_work($myfile) {
        foreach ($this->cpanels as $cpanel) {
            $the_file = $cpanel.'/public_html'.$myfile;
            var_dump($the_file);
            require($the_file);
            global $wpdb;
            var_dump($wpdb->prefix);
        }
    }
}
new execute_sql_code($myfile);

var_dump($the_file) => OUTPUTS正確的每條路徑。

var_dump($wpdb->prefix) => 每次迭代輸出錯誤wp1_ ：

電流輸出：

迭代 1 => cpanel1/public_html/wp-load.php => $wpdb->prefix = wp1_

迭代 2 => cpanel2/public_html/wp-load.php => $wpdb->prefix = wp1_

迭代 3 => cpanel3/public_html/wp-load.php => $wpdb->prefix = wp1_

預期產出：

迭代 1 => cpanel1/public_html/wp-load.php => $wpdb->prefix = wp1_

迭代 2 => cpanel2/public_html/wp-load.php => $wpdb->prefix = wp2_ // 正常

迭代 3 => cpanel3/public_html/wp-load.php => $wpdb->prefix = wp3_ // 正常

Answer 1

試試這個，你需要更改array("username", "password", "DB1", "localhost")等，配置以連接多個數據庫。 還要確保您每個文件夾的 cpanel wp-load.php文件加載成功。

使用new wpdb()類將創建一個新對象來使用當前的數據庫或 db 函數。

有關更多信息，我想建議wordpress stackexchange

$myfile = 'wp-load.php';

class execute_sql_code
{
    public function __construct($myfile)
    {
        $this->main_work($myfile);
    }

    public $cpanels = array(
       'cpanel1' => array("username", "password", "DB1", "localhost"),
       'cpanel2' => array("username", "password", "DB2", "localhost"),
       'cpanel3' => array("username", "password", "DB3", "localhost"),
    );


    public function main_work($myfile)
    {
        $i = 0;
        $cpanelKeys = array_keys($this->cpanels);
        foreach ($this->cpanels as $cpanel) {

            $path = $cpanelKeys[$i];
            $dbuser = ($cpanel[0]);
            $dbpassword = ($cpanel[1]);
            $dbname = ($cpanel[2]);
            $dbhost = ($cpanel[3]);

            $the_file = $_SERVER['DOCUMENT_ROOT'] . '/' . $path . '/public_html/' . $myfile;
            include_once($the_file);

            $wpdb = new wpdb($dbuser, $dbpassword, $dbname, $dbhost);
            $table_name = empty($wpdb->prefix) ? 'wp_' . "options" : $wpdb->prefix . "options";
            $sql = "SELECT option_value FROM " . $table_name . " WHERE option_name = 'siteurl';";
            $root_url = $wpdb->get_var($sql);
            var_dump($root_url);
            echo "<br>";
            unset($wpdb);
            $i++;
        }
    }
}

$obj = new execute_sql_code($myfile);

Answer 2

我懷疑 PHP 在全局命名空間中創建后無法覆蓋對象$wpdb 。 嘗試在var_dump($wpdb->prefix)之后unset($wpdb) var_dump($wpdb->prefix) ：

public function main_work($myfile) {
    foreach ($this->cpanels as $cpanel) {
        $the_file = $cpanel.'/public_html'.$myfile;
        var_dump($the_file);
        require($the_file);
        global $wpdb;
        var_dump($wpdb->prefix);
        unset($wpdb);
        global $table_prefix;
        unset($table_prefix);
    }
}

編輯：

可能創建了一些額外的全局變量，您必須在require($the_file);之前和之后檢查$GLOBALS require($the_file); 推斷要取消設置的內容。 我建議嘗試unset($table_prefix); 例如。