Question

我正在尝试从mySQL数据库中提取信息。我正在使用的代码是：

  <?php global $current_user;
  get_currentuserinfo();

  //echo 'Username: ' . $current_user->user_login . "\n";
  //echo 'User email: ' . $current_user->user_email . "\n";
  //echo 'User first name: ' . $current_user->user_firstname . "\n";
  //echo 'User last name: ' . $current_user->user_lastname . "\n";
  //echo 'User display name: ' . $current_user->display_name . "\n";
  echo  $current_user->display_name . "\n";
  //echo 'User ID: ' . $current_user->ID . "\n";
?>
<?php
            $members_settings = get_option('members_db');
        $host       = $members_settings['members_db_server']; // ''; //replace your IP or hostname
    $user       = $members_settings['members_db_user']; // ''; //database user
    $password   = $members_settings['members_db_pass']; //'';//database password
    $database   = $members_settings['members_db_database']; /*''; //database name
    $prefix     = ''; //prefix if any else just give any random value
    $driver     = 'mysql'; //here u can also have ms sql database driver, postgres, etc
    $options    = array ( 'driver' => $driver, 'host' => $host, 'user' => $user, 'password' => $password,  'database' => $database, 'prefix' => $prefix );

    //$db = JDatabase::getInstance($options);
    $con = mysql_connect($host,$user,$password);
            $db_selected = mysql_select_db($database, $con);
           // if (!$db_selected) { die ("Can\'t use ".$database." : " . mysql_error()); }


            $userid = $current_user->user_login;
    $sql = "SELECT * FROM  ibew_Members WHERE Card = $userid";
            //echo $sql;
    $result = mysql_query($sql);
            if ($result) {
                list($id) = mysql_fetch_array(mysql_query($sql));
                //echo $id;
            } else {
                echo "query failed";
            }

    if (mysqli_connect_errno($con))
    {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

?>

目前我收到的查询失败是我的错误，我不确定问题出在哪里，因为在这方面的专业知识方面，我仍然有点茫然。

对于如何解决这个问题的任何见解将不胜感激。

Answer 1

不推荐使用Mysql_和mysqli_，您应该考虑迁移到PDO框架。这是一个例子：

function createWindow() {
  var note_window = document.createElement("div");
  note_window.setAttribute("id", "note_window"); 
  note_window.setAttribute("class", "notification-window");
  note_window.style.visibility = 'visible';
  note_window.innerHTML = 'Text';
  var title = document.createElement("div");
  title.setAttribute("id", "title");
  title.setAttribute("class", "col-md-12");
  title.innerHTML = "More Text";
  note_window.appendChild(title);

var navbar = document.getElementById("navbar");
  navbar.appendChild(note_window);
}

createWindow();

这为您提供了更清晰的数据库事务，并有助于防止使用PDO对象进行注入。 Try / Catch还会报告错误并提供更详细的系统错误。

Answer 2

顺便说一下，我看到你正在评论JDatabase :: ...如果这是在joomla中，你仍然可以使用Joomla框架来获取你的查询数据：

define( '_JEXEC', 1 );
define( 'JPATH_BASE', '/Your local path to root/' );
require_once ( JPATH_BASE.'/includes/defines.php' );
require_once ( JPATH_BASE.'/includes/framework.php' );
$mainframe  = JFactory::getApplication('site');
$mainframe->initialise();
$user       = JFactory::getUser();
$config     =& JFactory::getConfig();
$session    =& JFactory::getSession();
// get the Joomla DBO config
$db         = JFactory::getDbo();
$query      = $db->getQuery(true);

$cardsQuery = "SELECT * FROM ibew_Members WHERE Card = ".$db->quote($user->id);
$db->setQuery($cardsQuery);
try {
    $results = $db->loadAssocList();
    if (count($results) > 0) { // we have results
        foreach($results as $result) { 
            echo "Relevant Info: ".$result['fieldname'];
        }
    } else {
        echo "No results found in the database for this user<br>\n";
    }
}
catch (Exception $e) {
    echo "Unable to complete Transaction: ".$e."<br>\n";
}

与我的其他答案基本相同，只是在Joomla PDO框架下。

接收查询失败错误

2 个答案: