MongoDB 对某列去重后，查询该列对应的另外一值是否都为 xx

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This topic created in 2047 days ago, the information mentioned may be changed or developed.

数据形如

[
{'status':'ok','message_id': '1', 'group_id':1},
{'status':'bad','message_id': '2', 'group_id':1},
{'status':'ok','message_id': '1', 'group_id':2}
],

需求是要判断当前库中同一 group 的 status 都为 ok，有一个 bad 则返回 False 。由于不太熟悉 MongoDB，使用了比较难看的做法：

group_ids = db.tasks.distinct('group_id')
for group_id in group_ids:
____status = db.tasks.find_one({'group_id': group_id, 'status': 'bad'})
____return False if task else True

请问大佬们，有什么好的办法可以完成这个需求呢

3 replies • 2020-10-28 23:02:10 +08:00

catinsides

Oct 28, 2020

```
aggregate([
{
$group: {
_id: '$group_id',
statusAry: {
$addToSet: '$status'
}
}
},
{
$addFields: {
len: {
$size: '$statusAry'
}
}
},
{
$addFields: {
results: {
$cond: { if: { $eq: [ "$len", 1 ] }, then: true, else: false }
}
}
}
])
```

catinsides

Oct 28, 2020

没判断全是 bad

lithiumii

Oct 28, 2020

你这个是 pymongo ？如果可以在 py 里做的话

group_ids = db.tasks.distinct('group_id')
bad_ids = db.tasks.find({'status':'bad'}).distinct('group_id')
good_ids = [x for x in group_ids if x not in bad_ids]

或者用 MongoDB 的 aggregation