Zend Framework 2 使用别名多次加入相同的 table
Zend Framework 2 join the same table multiple times using alias
我正在努力将多个左联接实施到 ZF2 中。我有第一个工作,但是当我添加另一个时,它不起作用。
这是我应该在 zf2 中实施的工作 SQL 查询:
SELECT
ac.ctr_id AS ctr_id,
ac.ctr_no AS ctr_no,
ac.ctr_marketer AS marketer,
ac.ctr_manager AS manager,
ac.ctr_recruiter AS recruiter,
l1.emp_realname AS marketer,
l2.emp_realname AS co_recruiter_manager,
l3.emp_realname AS recruiter
FROM
allcontracts AS ac
JOIN
lstemployees AS le ON ac.ctr_recruiter = le.emp_id
LEFT JOIN
lstemployees AS l2 ON ac.ctr_manager = l2.emp_id
LEFT JOIN
lstemployees AS l3 ON ac.ctr_recruiter = l3.emp_id
LEFT JOIN
lstemployees AS l1 ON ac.ctr_marketer = l1.emp_id
来自我的模型:
.....
$where = new Where();
$this->table='allcontracts';
$select = new Select($this->table);
$select->columns(array('*')); // TODO add columns from allcontracts table
// This one works
$select->join('lstemployees', 'allcontracts.ctr_recruiter = lstemployees.emp_id');
// When I add this one below it doesn't work
$select->join(array('l2' => 'lstemployees'), 'allcontracts.ctr_manager = l2.emp_id', array('*'), 'left');
$where->like('ctr_no', '%LT');
if($id!='' && $id > 0)
$where->equalTo('ctr_id', $id);
$select->where($where);
$resultSet = $this->selectWith($select);
......
有什么想法吗?
这是我的建议:
<?php
use Zend\Db\Sql\Select;
$select = new Select();
$select->columns([Select::SQL_STAR])
->from(['ac' => 'allcontracts '])
->join(['le' => 'lstemployees'], 'ac.ctr_recruiter = le.emp_id', [])
->join(['l1' => 'lstemployees'], 'ac.ctr_marketer = l1.emp_id', ['marketer' => 'emp_realname'], Select::JOIN_LEFT)
->join(['l2' => 'lstemployees'], 'ac.ctr_manager = l2.emp_id', ['co_recruiter_manager' => 'emp_realname'], Select::JOIN_LEFT)
->join(['l3' => 'lstemployees'], 'ac.ctr_recruiter = l3.emp_id', ['recruiter' => 'emp_realname'], Select::JOIN_LEFT);
// to debug your query
die($select->getSqlString($dbAdapter->getPlatform()));
// if you haven't $dbAdapter, replace by null but the result will be quoted.
我正在努力将多个左联接实施到 ZF2 中。我有第一个工作,但是当我添加另一个时,它不起作用。
这是我应该在 zf2 中实施的工作 SQL 查询:
SELECT
ac.ctr_id AS ctr_id,
ac.ctr_no AS ctr_no,
ac.ctr_marketer AS marketer,
ac.ctr_manager AS manager,
ac.ctr_recruiter AS recruiter,
l1.emp_realname AS marketer,
l2.emp_realname AS co_recruiter_manager,
l3.emp_realname AS recruiter
FROM
allcontracts AS ac
JOIN
lstemployees AS le ON ac.ctr_recruiter = le.emp_id
LEFT JOIN
lstemployees AS l2 ON ac.ctr_manager = l2.emp_id
LEFT JOIN
lstemployees AS l3 ON ac.ctr_recruiter = l3.emp_id
LEFT JOIN
lstemployees AS l1 ON ac.ctr_marketer = l1.emp_id
来自我的模型:
.....
$where = new Where();
$this->table='allcontracts';
$select = new Select($this->table);
$select->columns(array('*')); // TODO add columns from allcontracts table
// This one works
$select->join('lstemployees', 'allcontracts.ctr_recruiter = lstemployees.emp_id');
// When I add this one below it doesn't work
$select->join(array('l2' => 'lstemployees'), 'allcontracts.ctr_manager = l2.emp_id', array('*'), 'left');
$where->like('ctr_no', '%LT');
if($id!='' && $id > 0)
$where->equalTo('ctr_id', $id);
$select->where($where);
$resultSet = $this->selectWith($select);
......
有什么想法吗?
这是我的建议:
<?php
use Zend\Db\Sql\Select;
$select = new Select();
$select->columns([Select::SQL_STAR])
->from(['ac' => 'allcontracts '])
->join(['le' => 'lstemployees'], 'ac.ctr_recruiter = le.emp_id', [])
->join(['l1' => 'lstemployees'], 'ac.ctr_marketer = l1.emp_id', ['marketer' => 'emp_realname'], Select::JOIN_LEFT)
->join(['l2' => 'lstemployees'], 'ac.ctr_manager = l2.emp_id', ['co_recruiter_manager' => 'emp_realname'], Select::JOIN_LEFT)
->join(['l3' => 'lstemployees'], 'ac.ctr_recruiter = l3.emp_id', ['recruiter' => 'emp_realname'], Select::JOIN_LEFT);
// to debug your query
die($select->getSqlString($dbAdapter->getPlatform()));
// if you haven't $dbAdapter, replace by null but the result will be quoted.