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Symfony Doctrine 查询构建器查找具有多对一关系的实体

[英]Symfony Doctrine query builder find entity with many to one relation

I'm building a website with Symfony for a project which will act like "booking.com" websites but much simplier and with some variations.我正在使用 Symfony 为一个项目构建一个网站,该网站的行为类似于“booking.com”网站,但要简单得多,并且有一些变化。

Some errors on the database fields but not really important for my issue.数据库字段上的一些错误,但对我的问题并不重要。

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As you can see, these tables concern two entities : apartment and visit.如您所见,这些表涉及两个实体:公寓和访问。 A customer can ask for a visit of an apartment.客户可以要求参观公寓。 It is a many to one relationship.这是多对一的关系。

I have a search form to search for apartments with criterias.我有一个搜索表格来搜索符合条件的公寓。 I want to only show apartments that don't have any visits between the arrival and departure dates that the user provided.我只想显示在用户提供的到达和离开日期之间没有任何访问的公寓。 So I ended up making a function in apartmentRepository to manage that and other cases.因此,我最终在 apartmentRepository 中创建了一个函数来管理该情况和其他情况。

Problem is: how can I get these apartments ?问题是:我怎样才能得到这些公寓?

Here is a draft of this function which is of course not finished neither perfect (if you have some comments to improve it, it would be great !).这是这个功能的草稿,当然还没有完成,也不完美(如果你有一些评论来改进它,那就太好了!)。

public function findByCustom($parameters): ?array
{
   $query =  $this->createQueryBuilder('a');

   foreach ($parameters as $key=> $parameter){
       if($key != 'keywords' and $key!= 'priceMin' & $key!='priceMax' and $key!="typeAp" and $key!="searchVisitDate") $query->orWhere('a.'.$key." = '".$parameter."'");

       if($key == "typeAp")
       {
           $typeApQuery = "";
           foreach ($parameters[$key] as $index => $type)
           {
               if($index !== count($parameters[$key])-1)
               {
                   $typeApQuery.=" a.typeAp = '".$type."' or";
               }
               else
               {
                   $typeApQuery.= " a.typeAp = '".$type."'";
               }
           }
           $query->andWhere($typeApQuery);
       }
   }

   $query->andWhere('a.rentPrice >='.$parameters['priceMin']." and a.rentPrice <= ".$parameters['priceMax']);
   $withoutInner = $query;
   $query
       ->join('App\Entity\Visit', 'v', Join::ON, 'a = v.apartment')
       ->where("v.date between '2020-03-15' and '2020-03-19'");
    $query->getQuery()->getResult();
    $sorted = $withoutInner->andWhere($this->createQueryBuilder('a')->expr()->notIn('a.id', $query));

   return array($sorted);

Of course apartment has a collection of visits and visit as a field named "apartment" which is related to the apartment object.当然,公寓有一个访问和访问的集合,作为一个名为“公寓”的字段,与公寓对象相关。

I really didn't find a proprer way to do it and I want to avoid doing SQL, to improve my understanding of Doctrine.我真的没有找到合适的方法来做到这一点,我想避免使用 SQL,以提高我对 Doctrine 的理解。

Thank you for your help because I'm stuck right now :/感谢您的帮助,因为我现在被困住了:/

EDIT 1: forgot to mention that I want to get apartments that don't have visits between required dates or that don't have any visits at all编辑 1:忘了提到我想要在规定的日期之间没有访问或根本没有访问的公寓

EDIT 2 :编辑 2:

public function findByCustom($parameters): ?array
{
   $query =  $this->createQueryBuilder('a');
    $withoutInner = $this->createQueryBuilder("a");

   foreach ($parameters as $key=> $parameter){
       if($key != 'keywords' and $key!= 'priceMin' & $key!='priceMax' and $key!="typeAp" and $key!="searchVisitDate")
       {
           $withoutInner->orWhere('a.'.$key." = '".$parameter."'");
           $query->orWhere('a.'.$key." = '".$parameter."'");
       }


       if($key == "typeAp")
       {
           $typeApQuery = "";
           foreach ($parameters[$key] as $index => $type)
           {
               if($index !== count($parameters[$key])-1)
               {
                   $typeApQuery.=" a.typeAp = '".$type."' or";
               }
               else
               {
                   $typeApQuery.= " a.typeAp = '".$type."'";
               }
           }
           $withoutInner->andWhere($typeApQuery);
           $query->andWhere($typeApQuery);
       }
   }

   $query->andWhere('a.rentPrice >='.$parameters['priceMin']." and a.rentPrice <= ".$parameters['priceMax']);
   $withoutInner->andWhere('a.rentPrice >='.$parameters['priceMin']." and a.rentPrice <= ".$parameters['priceMax']);
   $query
       ->join('App\Entity\Visit', 'v', Join::WITH, 'a.id = v.apartment')
       ->where("v.date between '2020-03-15' and '2020-03-19'");

    $query->getQuery()->getResult();
    $sorted = $withoutInner->andWhere($this->createQueryBuilder('a')->expr()->notIn('a.id', $query));

With this function I get doctrine error :有了这个功能,我得到了学说错误:

Error: Method Doctrine\Common\Collections\ArrayCollection::__toString() must not throw an exception, caught ErrorException: Catchable Fatal Error: Object of class Doctrine\ORM\EntityManager could not be converted to string

I haven't tested this myself, but something like this should work:我自己还没有测试过,但是这样的事情应该可行:

$query
    ->leftJoin('App\Entity\Visit', 'v', Join::WITH, 'a = v.apartment AND v.date between "2020-03-15" and "2020-03-19"')
    ->where('a.visits IS EMPTY');

The idea here is to use leftJoin and select only those result that do not have a corresponding entry in the visit table.这里的想法是使用leftJoin并仅选择那些在visit表中没有相应条目的结果。

I found a solution to get all apartments that have no rents (where the join is null).我找到了一个解决方案来获取所有没有租金的公寓(连接为空)。 Yes I changed visits for rents because that was a mistake in my main question.是的,我改变了租金的访问,因为这是我的主要问题中的一个错误。

 $query->leftJoin('a.rents', 'r',Join::WITH,'a = r.apartment')
        ->where(
            $query->expr()->andX($query->expr()->isNull('r')))
        ->orWhere($query->expr()-> **);

I have an attribute in my apartment entity which has all rents of the current apartment.我的公寓实体中有一个属性,其中包含当前公寓的所有租金。 So I use it to make the join.所以我用它来进行连接。 With the expression isNull I get it to work for the apartments with no visits.使用表达式 isNull 我让它在没有访问的情况下为公寓工作。

** : I want to be able to get the apartments that have no locations where user entered required arrival date is not between r.arrival and r.departure. ** :我希望能够获得没有用户输入所需到达日期不在 r.arrival 和 r.departure 之间的位置的公寓。 This would give me all apartments without any rents and those which are free to book.这将给我所有没有租金的公寓和那些可以免费预订的公寓。

I thought about doing another query and do a notIn but I don't know how to do it neither.我想过做另一个查询并做一个 notIn 但我也不知道该怎么做。

Thank you.谢谢你。

EDIT and solution :编辑和解决方案:

I found out by myself how to do it.我自己发现了如何做到这一点。 This is absolutly not the best approach neither the best solution but I'm out of time for my projet so I needed to get this done.这绝对不是最好的方法,也不是最好的解决方案,但我的项目没时间了,所以我需要完成这项工作。

$rents = $this->getEntityManager()->createQueryBuilder()
        ->select("a.id")->from('App:Rent', 'r')
        ->andWhere('r.arrival between :arrival and :departure')
        ->leftJoin('r.apartment', 'a')
        ->setParameters(array("arrival"=>\DateTime::createFromFormat('Y-m-d', "2020-04-20")->setTime(00,00,00), "departure"=>\DateTime::createFromFormat('Y-m-d',"2020-04-31")->setTime(00,00,00)))
        ->getQuery()->getArrayResult();


    $rentsSorted = array();
    foreach ($rents as $rent)
    {
        if(!in_array( $rent['id'],$rentsSorted))
        {
            $rentsSorted[] = $rent['id'];
        }
    }
    if(count($rentsSorted)>0)
    {
        $withRentsNotBetweenDates->andWhere('a1.rentPrice >=' . $parameters['priceMin'] . " and a1.rentPrice <= " . $parameters['priceMax'])
            ->andWhere($withRentsNotBetweenDates->expr()->notIn('a1.id', $rentsSorted));
    }
    else
    {
        $withRentsNotBetweenDates->andWhere('a1.rentPrice >=' . $parameters['priceMin'] . " and a1.rentPrice <= " . $parameters['priceMax']);
    }

Of course I'm going to change the parameters to the required dates given by my form.当然,我要将参数更改为我的表单给出的所需日期。

This code allows me to get all rents that exists within these required dates.此代码允许我获得这些所需日期内存在的所有租金。 I then store each apartment id in an array that I will pass in the not in to exclude these apartments because they have a rent at this period.然后我将每个公寓 id 存储在一个数组中,我将传入 not in 以排除这些公寓,因为它们在此期间有租金。

If you have any improvents or a better way to do this don't be shy xD.如果您有任何改进或更好的方法来做到这一点,请不要害羞 xD。

Have a good day.祝你有美好的一天。

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