mongodb查询玩家排名
摘要:
db.user.aggregatet([ { $sort:{score: -1} }, { $group:{ _id:null, users:${ $push:{nickname:"$nickname", score: "$score"} }, count: {$sum:1}, } }, { $pr 阅读全文
posted @ 2024-01-03 14:22 悉野 阅读(97) 评论(0) 推荐(0)
posted @ 2024-01-03 14:22 悉野 阅读(97) 评论(0) 推荐(0)