Question

我正在将mongo查询转换为symfony4学说ODM查询。在这种情况下，学说将不接受多个分组依据字段。如何解决以下用例？

Mongo查询：

db.Article.aggregate([
    {
        $group: {
            _id: { department: "$department", status : "$status" },
            MIN: { $min: "$salary" },
            MAX: { $max: "$salary" }
        }
    }
]);

Mongo查询结果：

{ "_id" : { "department" : "sales", "status" : 0 }, "MIN" : 100, "MAX" : 5000 }
{ "_id" : { "department" : "sales", "status" : 1 }, "MIN" : 10000, "MAX" : 16500 }
{ "_id" : { "department" : "IT", "status" : 1 }, "MIN" : 5000, "MAX" : 5000 }
{ "_id" : { "department" : "DOCTOR", "status" : 1 }, "MIN" : 20000000, "MAX" : 20000000 }
{ "_id" : { "department" : "IAS", "status" : 1 }, "MIN" : 120000, "MAX" : 120000 }

Symfony4与教义odm：

$qb = $this->createAggregationBuilder($documentClass);
$qb
    ->group()
        ->field('id')
        ->expression('$department')
        ->field('department')
        ->first('$department')
        ->field('status')
        ->first('$status')
        ->field('lowestValue')
        ->min('$salary')
        ->field('highestValue')
        ->max('$salary')
        ->field('totalValue')
        ->sum(1);

仅按部门领域考虑。我想按部门和基于状态的分组结果分组。

Answer 1

此解决方案适用于Symfony3。在ArticleRepository.php文件中，编写以下行：

$qb = $this->createAggregationBuilder('Document\Article');
$group_expression = $qb->expr()
                       ->field('departement')
                       ->expression('$department')
                       ->field('max')
                       ->expression('$max')
                       ->field('min')
                       ->expression('$min');
$qbf->group()
    ->field('id')
    ->expression($group_expression)
    ->field('department')
    ->first('$department')
    ->field('status')
    ->first('$status')
    ->field('lowestValue')
    ->min('$salary')
    ->field('highestValue')
    ->max('$salary')
    ->field('totalValue')
    ->sum(1);

symfony4与doctrine-odm：如何在聚合构建器中添加一个或多个组

1 个答案: