1

好的，我真的无法按照我需要的方式进行这项工作。我有 2 张桌子：

gerencia_clientes和gerencia_vendas

第一个保存我的客户，第二个保存发票

我需要使用这样的查询列出它们

$db = JFactory::getDBO();
$query = "select * from jos_gerencia_vendas where id_edicao = ".$iddaedicao." order by id_cliente";

$db->setQuery($query);
$list = $db->loadObjectList();
foreach ($list as $item)

但不是按 id_cliente 排序，我需要正确的语法来使用另一个表中按字母顺序排列的客户进行排序，

例子

$query = "select * from jos_gerencia_vendas where id_edicao = ".$iddaedicao." order by (select nomecliente from jos_gerencia_clientes)";

有什么帮助吗？

4

3 回答 3

1 
$query = "select * from jos_gerencia_vendas AS a 
          LEFT JOIN jos_gerencia_clientes AS b ON b.id_cliente=a.id_cliente
          where a.id_edicao = ".$iddaedicao." 
          order by b.nome_cliente ASC"
于 2012-08-22T00:18:03.840 回答
1

还有一种方法可以使用 Joomla 平台 API 执行包含子查询的查询。如何使用子查询的基本思想是基于gunjanpatel。

这个答案也发布在Stackexchange for Joomla上。

这是在嵌套集模型上执行查询的示例：

SQL查询：

-- Find the Immediate Subordinates of a Node
SELECT node.title, (COUNT(parent.id) - (sub_tree.depth + 1)) AS depth
FROM lubd3_usergroups AS node,
        lubd3_usergroups AS parent,
        lubd3_usergroups AS sub_parent,
        (
                SELECT node.id, (COUNT(parent.id) - 1) AS depth
                FROM lubd3_usergroups AS node,
                        lubd3_usergroups AS parent
                WHERE node.lft BETWEEN parent.lft AND parent.rgt
                        AND node.id = 1
                GROUP BY node.id
                ORDER BY node.lft
        )AS sub_tree
WHERE node.lft BETWEEN parent.lft AND parent.rgt
        AND node.lft BETWEEN sub_parent.lft AND sub_parent.rgt
        AND sub_parent.id = sub_tree.id
GROUP BY node.id
-- not showing the parent node
HAVING depth = 1
-- showing the parent node
-- HAVING depth <= 1
ORDER BY node.lft;

以及由 Joomla 执行的转换后的查询：

// Create the subQuery select statement.
// Nested Set Queries: http://mikehillyer.com/articles/managing-hierarchical-data-in-mysql/
// CROSS JOIN: http://www.informit.com/articles/article.aspx?p=30875&seqNum=5
$subQuery->select(array('node.id', '(COUNT(parent.id) - 1) AS depth'))
    ->from($db->quoteName('#__usergroups') . 'node')
    ->join('CROSS', $db->quoteName('#__usergroups', 'parent'))
    ->where($db->quoteName('node.lft') . ' BETWEEN  ' . $db->quoteName('parent.lft') . ' AND ' . $db->quoteName('parent.rgt') . ' AND ' . $db->quoteName('node.id') . ' = ' . $db->quote('1'))
    ->group($db->quoteName('node.id'))
    ->order($db->quoteName('node.lft'));

// Create the base select statement.
$query->select(array('node.title', '(COUNT(parent.id) - (sub_tree.depth + 1)) AS depth'))
    ->from($db->quoteName('#__usergroups') . 'node')
    ->join('CROSS', $db->quoteName('#__usergroups', 'parent'))
    ->join('CROSS', $db->quoteName('#__usergroups', 'sub_parent'))
    ->join('CROSS', '(' . $subQuery .') AS sub_tree')
    ->where($db->quoteName('node.lft') . ' BETWEEN  ' . $db->quoteName('parent.lft') . ' AND ' . $db->quoteName('parent.rgt')
    . ' AND ' . $db->quoteName('node.lft') . ' BETWEEN  ' . $db->quoteName('sub_parent.lft') . ' AND ' . $db->quoteName('sub_parent.rgt')
    . ' AND ' . $db->quoteName('sub_parent.id') . ' = ' . $db->quoteName('sub_tree.id'))
    ->group($db->quoteName('node.id'))
    ->having($db->quoteName('depth') . ' = ' . $db->quote('1'))
    ->order($db->quoteName('node.lft'));

// Set the query and load the result.
$db->setQuery($query);
$rowList = $db->loadAssocList();

echo "<pre>";
print_r($rowList);
echo "</pre>";
于 2015-09-01T08:33:33.787 回答
0

不确定您用于数据库表的列，但这可能有效：

$query = "select * from jos_gerencia_vendas where id_edicao = ".$iddaedicao." order by names ASC";

根据您的表格列更改“名称”。

希望这可以帮助

于 2012-08-21T18:00:36.920 回答