如何使用原生Joomla 2.5 php对象显示Joomla数据库表
我试图使用原生joomla PHP对象显示MySQL数据库表中的数据，但是我永远无法让它正常工作。它从不显示任何数据 我知道如何通过旧的方式连接到数据库来显示数据，如下所示如何使用原生Joomla 2.5 php对象显示Joomla数据库表,php,mysql,class,object,joomla,Php,Mysql,Class,Object,Joomla,我试图使用原生joomla PHP对象显示MySQL数据库表中的数据，但是我永远无法让它正常工作。它从不显示任何数据 我知道如何通过旧的方式连接到数据库来显示数据，如下所示 $con = mysqli_connect("localhost","xxxxxx_user10","$dbpass","xxxxxxx_final"); // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: "
$con = mysqli_connect("localhost","xxxxxx_user10","$dbpass","xxxxxxx_final");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
} else {
//What is the base model
$sku = $finish;
// Select The Correct Table
$result = mysqli_query($con,"SELECT * FROM BAH_finish WHERE color_value='$sku' GROUP BY uniqueID ORDER BY uniqueID ASC");
$row = mysqli_fetch_array($result);
$finishname = $row['color_name'];
$finishtype = $row['color_type'];
| ID | Last | First | Address | City | State | ZIP
<?php
// Get a db connection.
$db = JFactory::getDbo();
// Create a new query object.
$query = $db->getQuery(true);
// Select all records from the user profile table where key begins with "custom.".
// Order it by the ordering field.
$query->select($db->quoteName(array('Last', 'First', 'Address', 'City', 'State', 'ZIP')));
$query->from($db->quoteName('111testnames'));
$query->where($db->quoteName('Last') . $db->quote('buffet'));
// Reset the query using our newly populated query object.
$db->setQuery($query);
// Load the results as a list of stdClass objects (see later for more options on retrieving data).
$results = $db->loadObjectList();
echo $results;
?>