本文介绍了使用 Zend DB 编写子查询的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我在将下面的 SQL 转换为 Zend Db 查询时遇到了一些问题.
I am having some problems turning the SQL below into a Zend Db query.
$select = ' SELECT s.id, i.id as instance_id, i.reference, i.name, i.sic_code, i.start_date
FROM sles s
JOIN sle_instances i
ON s.id = i.sle_id
WHERE i.id = ( SELECT MAX(id)
FROM sle_instances
WHERE sle_id = s.id
)
ORDER BY i.name ASC';
我已经了解了之前的代码 - 但是 Zend Db 没有正确生成查询.谁能告诉我我错过了什么??
I have got as far as the code before - but Zend Db isn't producing the query correctly. Can any one show me what I missing??
$select = $db->select() ->from('sles', array( 'id',
'instance_id' => 'sle_instances.id',
'reference' => 'sle_instances.reference',
'name' => 'sle_instances.name',
'sic_code' => 'sle_instances.sic_code',
'start_date' => 'sle_instances.start_date'
)
)
->join('sle_instances', 'sles.id = sle_instances.sle_id')
->where('sles.id = (SELECT MAX(id) FROM sle_instances WHERE sle_id = sles.id)')
->order('sle_instances.name ASC');
顺便说一下,SQL 确实有效.我正在使用 Zend Db 重写它,因为我希望使用 Zend Paginator 功能.
The SQL does work by the way. I am rewriting it using Zend Db as I wish to use the Zend Paginator functionality.
非常感谢任何帮助.
PJ
推荐答案
这个:
$select = $db->select()->from(array("s" => "sles"), array("s.id","i.id as instanceid","i.reference","i.name","i.sic_code","i.start_date"))
->join(array('i' => "sle_instances"),"s.id = i.sle_id",array())
->where("i.id = (select max(id) from sle_instances where sle_id = s.id)")
->order('i.name asc');
给出这个:
"SELECT `s`.`id`, `i`.`id` AS `instanceid`, `i`.`reference`, `i`.`name`, `i`.`sic_code`, `i`.`start_date` FROM `sles` AS `s`
INNER JOIN `sle_instances` AS `i` ON s.id = i.sle_id WHERE (i.id = (select max(id) from sle_instances where sle_id = s.id)) ORDER BY `i`.`name` asc"
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