Question

大家。在mongo组查询中，结果仅显示参数中的键。如何将每个组中的第一个文档保存为mysql查询组。例如：

-------------------------------------------------------------------------
|  name  | age  |  sex  | province |   city   |   area   |   address     |
-------------------------------------------------------------------------
| ddl1st | 22   | 纯爷们 |  BeiJing |  BeiJing | ChaoYang | QingNianLu    |
| ddl1st | 24   | 纯爷们 |  BeiJing |  BeiJing | XuHui    | ZhaoJiaBangLu |
|  24k   | 220  | ...   |  ....    |  ...     | ...      | ...           |
-------------------------------------------------------------------------



db.users.group({key: { name: 1},reduce: function ( curr, result ) { result.count ++ },initial: {count : 0 } })

结果：

[
{
    "name" : "ddl1st",
    "count" : 1
},
{
    "name" : "24k",
    "count" : 1
}
]

如何获得以下内容：

[
   {
   "name" : "ddl1st",
   "age" : 22,
   "sex" : "纯爷们",
   "province" : "BeiJing",
   "city" : "BeiJing",
   "area" : "ChaoYang",
   "address" : "QingNianLu",
   "count" : 1
   },
   {
   "name" : "24k",
   "age" : 220,
   "sex" : "...",
   "province" : "...",
   "city" : "...",
   "area" : "...",
   "address" : "...",
   "count" : 1
}
]

Answer 1

如果您想保留每个组的第一个匹配条目的信息，您可以尝试汇总，如：

db.test.aggregate({
  $group: {
    _id: '$name',
   name : { $first: '$name' }
   age : { $first: '$age' },
   sex : { $first: '$sex' },
   province : { $first: '$province' },
   city : { $first: '$city' },
   area : { $first: '$area' },
   address : { $first: '$address' },
   count: { $sum: 1 }
  }
}

Answer 2

如果具有多个字段的文档面临相同的问题，请快速更新。可以使用组合$replaceRoot流水线阶段和$mergeObjects流水线运算符的功能。

db.users.aggregate([
  {
    $group: {
      _id: '$name',
      user: { $first: '$$ROOT' },
      count: { $sum: 1 }
    },
  },
  {
    $replaceRoot: {
      newRoot: { $mergeObjects: [{ count: '$count' }, '$user'] }
    }
  }
])

Answer 3

在$$ROOT文档中使用$first，然后在第一个字段中使用$replaceRoot。

db.test.aggregate([
  { "$group": {
    "_id": "$name",
    "doc": { "$first": "$$ROOT" }
  }},
  { "$replaceRoot": { "newRoot": "$doc" }}
])

Answer 4

这就是我所做的，效果很好。

db.person.aggregate([
{
  $group: { _id: '$name'}, // pass the set of field to be grouped
   age : { $first: '$age' }, // retain remaining field
   count: { $sum: 1 } // count based on your group
},
{
  $project:{
       name:"$_id.name",
       age: "$age",
       count: "$count",
       _id:0 
  }
}])

Answer 5

我不了解.group帮助器，但如果您更喜欢使用Aggregation Framework，那么您必须指定要返回的字段。如果我错了，请纠正我，但在SQL中你无论如何都必须这样做。

嗯，这就是你用前面提到的聚合框架做的事情：

db.test.aggregate({
  $group: {
    _id: { name: "$name", city: "$city", fieldName: "$fieldName" },
    count: { $sum: 1 }
  }
})

Answer 6

顺便说一句，如果您不仅要保留第一个文档，还可以使用$addToSet 例如：

db.test.aggregate({
  $group: {
    _id: '$name',
    name : { $addToSet: '$name' }
    age : { $addToSet: '$age' },
    count: { $sum: 1 }
  }
}

Answer 7

我创建了此功能以概括逆转放松阶段的情况...让我知道你们是否遇到任何错误，但对我来说效果很好！

const createReverseUnwindStages = unwoundField => {
  const stages = [
    //
    // Group by the unwound field, pushing each unwound value into an array,
    //
    // Store the data from the first unwound document
    // (which should all be the same apart from the unwound field)
    // on a field called data.
    // This is important, since otherwise we have to specify every field we want to keep individually.
    //
    {
      $group: {
        _id: '$_id',
        data: {$first: '$$ROOT'},
        [unwoundField]: {$push: `$${unwoundField}`},
      },
    },

    //
    // Copy the array of unwound fields resulting from the group into the data object,
    // overwriting the singular unwound value
    //
    {
      $addFields: {[`data.${unwoundField}`]: `$${unwoundField}`},
    },

    //
    // Replace the root with our data object
    //
    {
      $replaceRoot: {
        newRoot: '$data',
      },
    },
  ]

  return stages
}

Answer 8

我来这里寻找答案，但对所选答案不满意（特别是考虑到年龄）。我发现this answer是更好的解决方案（已适应）：

db.test.aggregate({
  $group: {
    _id: '$name',
   person: { "$first": "$$ROOT" },
   count: { $sum: 1 }
  },
  {
    "$replaceRoot": { "newRoot": "$person" }
  }
}

Answer 9

您可以尝试一下

db.test.aggregate({
      { $group: 
            { _id: '$name',count: { $sum: 1 }, data: { $push: '$$ROOT' } } },
      {
        $project: {
          _id:0,
          data:1,
          count :1
        }
      }

}

Answer 10

如果要投影所有字段，请在查询下方使用此文件。

db.persons.aggregate({
      { $group: { _id: '$name', data: { $push: '$$ROOT' }, total: { $sum: 1 }} },
      {
        $project: {
          _id:0,
          data:1,
          total :1
        }
      }
}

Answer 11

我喜欢把将要与 $first 选项一起使用的所有内容放入字典中，以便在最后提取。

现在，只需复制字典，您就不再需要一次拖着所有这些信息 1！

{'$set': 
  {'collection_name':
    'collection_item1': '$collection_item1',
    'collection_item2': '$collection_item2',
    ...
  }
}

Answer 12

以下是答案＆gt;＆gt;＆gt;＆gt;

    $m = new \MongoDB\Driver\Manager();

    $command = new \MongoDB\Driver\Command([
        'aggregate' => 'mytestusers',
        'pipeline' => [
            ['$match' => ['name' => 'Pankaj Choudhary']],

            ['$unwind'=>'$skills'],
            ['$lookup' => array('from'=>'mytestskills','localField'=>'skills','foreignField'=>'_id','as'=>'sdfg')],
            ['$unwind'=>'$sdfg'],

            ['$group'=>array('_id'=>array('_id'=>'$_id','name'=>'$name','email'=>'$email'),'skills'=>array('$push'=>'$skills'),'sdfg'=>array('$push'=>'$sdfg'))],


        ],
        'cursor' => new \stdClass,
    ]);
    $cursor = $m->executeCommand('targetjob-plus', $command);
    $result = $cursor->toArray();

mongo group查询如何保留字段

12 个答案: