Question

我想在条件中添加一个连接到我的查询，但它返回了一个错误。我们如何在学说中做到这一点？有没有像： $ qb-＆gt; andWhere（＆＃39; p。$ gender =：g＆＃39;） - ＆gt; setParameter（＆＃39; g＆＃39;，$ gender）; 加入学说symfony ??

   $qb->select("p")->from("PfmSanadBundle:Person",  "p");

    if ((isset($province) && trim($province) !== '') && (isset($City) && trim($City) !== '')){
        $qb->join("p" ,"students", "ps")
            ->join("ps" ,"organization", "po")
            ->join("po" ,"cityProvince", "pc")
            ->join("pc" ,"province", "pp");
    }
   if ((isset($province) && trim($province) !== '') && !isset($City)) {
        $qb->join("p.students", "ps")
            ->join("ps.organization", "po");
    }
    $res = $qb->getQuery()->getArrayResult();

错误（邮递员）：

＆＃34;错误＆＃34;：{ ＆＃34;代码＆＃34;：500，＆＃34;消息＆＃34;：＆＃34;内部服务器错误＆＃34;，＆＃34;例外＆＃34;：[ { ＆＃34;消息＆＃34;：＆＃34; [语义错误]第0行，第49列附近的学生INNER＆＃39;：错误：课程＆＃39; p＆＃39;没有定义。＆＃34;，＆＃34; class＆＃34;：＆＃34; Doctrine \ ORM \ Query \ QueryException＆＃34;，＆＃34;追踪＆＃34;：[ {

Answer 1

加入方法用法如下

// Example - $qb->join('u.Group', 'g', Expr\Join::WITH, $qb->expr()->eq('u.status_id', '?1'))
// Example - $qb->join('u.Group', 'g', 'WITH', 'u.status = ?1')
// Example - $qb->join('u.Group', 'g', 'WITH', 'u.status = ?1', 'g.id')
public function join($join, $alias, $conditionType = null, $condition = null, $indexBy = null);

如下行

$qb->join("p" ,"students", "ps")
    ->join("ps" ,"organization", "po")
    ->join("po" ,"cityProvince", "pc")
    ->join("pc" ,"province", "pp");

更改为

$qb->join("p.students", "ps")
    ->join("ps.organization", "po")
    ->join("po.cityProvince", "pc")
    ->join("pc.province", "pp");

如何在学说中通过qb连接？

1 个答案: