PHP MongoDB汇总$ match和$ group和$ addToSet [英] PHP MongoDB aggregate $match and $group and $addToSet

问题描述

在MongoDB中具有此文档结构:

Having this document structure in MongoDB :

{
"_id":<MongoDBID>,
"chatUser1ID": 2,
"chatUser2ID": 3
}

现在，我想从Mongo获得所有聊天伙伴，其中ID为2的聊天伙伴包含在任一"chatUser1"中.或"chatUser2".为此，我想使用$ match和$ group函数.

Now i want to get all chat partners from Mongo where the Chat Partner with the ID 2 is included in either "chatUser1" or "chatUser2". For that i want to use the $match and $group function.

$chatUserID = $_POST["chatUserID"]; // 2 in my example
$chatCursor = $chatCollection->aggregate([
            [
                '$match'  => 
                    [
                        '$or' => 
                            [
                                ["chatUser1ID" => $chatUserID],
                                ["chatUser2ID" => $chatUserID]
                            ]
                    ]
            ]
            ,[
                '$group' => 
                    [
                        '_id' => 0,
                        'chatUsers' => ['$addToSet' => '$chatUser1ID'],
                        'chatUsers' => ['$addToSet' => '$chatUser2ID'],
                        'chatUsers1' => ['$addToSet' => '$chatUser1ID'],
                        'chatUsers2' => ['$addToSet' => '$chatUser2ID'],
                    ]
            ]
        ]);

'chatUsers1' => ['$addToSet' => '$chatUser1ID'],

这会将字段chatUser1ID的所有ID放入chatUsers1集中，并且

This is putting all the ID's of the field chatUser1ID in the chatUsers1 set and

'chatUsers2' => ['$addToSet' => '$chatUser2ID']

将字段 chatUser2ID 的所有ID放入 chatUsers2 集中，其中 $ chatUserI D要么位于文档的chatUser1ID 或 chatUser2ID 字段.

is putting all all the ID's of the field chatUser2ID in the chatUsers2 set where the $chatUserID is either in the chatUser1ID or chatUser2ID field of the document.

之后，我想获取唯一的ID

After that i want want to get the unique ID's

$chatUserIDs = array();
foreach ($chatCursor as $counter => $document) {
    $bson = MongoDB\BSON\fromPHP($document);
    $value = MongoDB\BSON\toJSON($bson);
    $value = json_decode($value, true);
    $chatUserIDs = array_unique(array_merge($value['chatUsers1'], $value['chatUsers2']));
}
unset($chatUserIDs[array_search($chatUserID, $chatUserIDs)]);
array_unshift($chatUserIDs);

所以基本上可以正常工作，但是我想要一个解决方案，在该解决方案中，我可以直接从数据库中获取唯一ID的列表.本来我以为是线

So basically it's working but i want the solution where i get a list of unique ID's right away from the database. Originally i thought the lines

'chatUsers' => ['$addToSet' => '$chatUser1ID'],
'chatUsers' => ['$addToSet' => '$chatUser2ID'],

会将ID添加到集合chatUsers中，但是不幸的是，该集合在第二行中被覆盖.有没有一种方法可以附加"将ID替换为集合，而不是覆盖它们?也许是我可以排除$ chatUserID的一种方式，因为我只需要chatPartners.

would add the ID's to the set chatUsers but unfortunately the set is overwritten in the second line. Is there a way to "append" the ID's to the set instead of overwrite them ? And maybe a way where i can exclude the $chatUserID because i want only the chatPartners.

提前谢谢.

推荐答案

如果您不关心它们出现的顺序，则可以构建两个数组user1和user2，然后在稍后阶段将它们连接在一起.但是，这将无法处理重复数据删除.

If you don't care about the order they appear in, you can build two arrays of user1 and user2, then concat them together in a later stage. This won't handle deduplicating though.

$chatUserID = $_POST["chatUserID"]; // 2 in my example
$chatCursor = $chatCollection->aggregate([
    [
        '$match' => [
            '$or' =>[
                ["chatUser1ID" => $chatUserID],
                ["chatUser2ID" => $chatUserID]
            ]
        ]
    ], [
        '$group' => [
            '_id' => 0,
            'chatUsers1' => ['$addToSet' => '$chatUser1ID'],
            'chatUsers2' => ['$addToSet' => '$chatUser2ID'],
        ]
    ], [
        '$addFields' => [
            'chatUsers' => [
                 '$concatArrays' => [
                     '$chatUsers1',
                     '$chatUsers2'
                 ]
            ]
         ]
    ],
]);

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