Question

我正在从Joomla 2.5中的自定义类别表中加载值。这很好用，我可以显示每个类别的名称。

然后我想将具有特定id的类别添加到新数组中，但我无法使其工作。在C＃中，我确切地知道如何做到这一点，但不是在PHP中。

这是我的代码（用于修复代码格式的代码）：

<?php
// Get a db connection.
$db = JFactory::getDbo();

// Create a new query object.
$query = $db->getQuery(true);
$query->select('*');
$query->from('ind_category');
$query->order('Name ASC');

echo $query . "</br>";
// 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).
$categories = $db->loadObjectList();

// show the retrieved categories
if ($categories != null) {
    foreach ($categories as $cat) {
        echo "Name: " . $cat->Name . "</br>";
    }

} else {
    echo "<br> NULL results from " . $query;
}
echo "<br>";
echo "<br>";

// add objects from $categories with specific Id to $arr2
$arr2 = array();

foreach ($categories as $cat) {
    if ($cat->Id == 1) {
        $arr2[] = $cat;
    }
}
?>

Answer 1

根据joomla docs loadObjectList（）返回如下数组：

Array ( 
    [0] => stdClass Object ( [id] => 1 [name] => John Smith 
    [email] => johnsmith@domain.example [username] => johnsmith ) 
    [1] => stdClass Object ( [id] => 2 [name] => Magda Hellman 
    [email] => magda_h@domain.example [username] => magdah ) 
    [2] => stdClass Object ( [id] => 3 [name] => Yvonne de Gaulle 
    [email] => ydg@domain.example [username] => ydegaulle ) 
)

请尝试$cat->id（小写I）

在PHP查询结果中查找对象

1 个答案: