使用列表中的标签获取文档，按匹配的总数排序

本文介绍了使用列表中的标签获取文档，按匹配的总数排序的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 给定以下MongoDB文档集合： { title：'shirt one' tags：['shirt'，'cotton'，'t-shirt'，'black'] }， { title：'shirt two' tags：['shirt'，'white'，'button down collar'] }， { title：'shirt three'标签：['shirt'，'cotton'，'red' ] }， ... 检索与标签列表匹配的项目列表，按匹配标签的总数排序？例如，将此标签列表作为输入： ['shirt'，'cotton'，'black'] 我想要检索按匹配标签总数排序的项目： 项目总计匹配 -------- -------------- Shirt One 3（匹配衬衫+棉花+黑色）衬衫三2（匹配衬衫+棉花）衬衫2 1（匹配衬衫） 在关系模式中，标签将是一个单独的表，您可以对该表进行连接，计数匹配，并按计数排序。 但是，在Mongo ...？ 看起来这种方法可以工作， 将输入标签分成多个IN语句 通过OR将标签输入一起查询项目 ie例如，这将返回衬衫一的三个实例，衬衫一的两个实例，衬衫一的两个实例， Shirt Three等。 map / reduce输出 reduce：计算总出现次数_id > 但我不清楚如何将其实现为Mongo查询，这是最高效的方法。解决方案正如我在在MongoDB中搜索数组并按匹配数排序 这可能使用聚合框架。 假设 / p> 这种方法迫使你解开结果，并重新评估带有解卷结果的匹配谓词，所以它真的效率很低。 db.test_col.aggregate（ {$ match：{tags：{$ in：[ 衬衫，棉花，黑色]}}}， {$ unwind：$ tags}， {$ match：{tags：{$ in：[shirt， cotton，black]}}}， {$ group：{ _id：{_ id：1}， matches：{$ sum：1} }}， {$ sort：{matches：-1}} ）; 预期结果 {result：[ {_id：{_id：ObjectId（5051f1786a64bd2c54918b26） }，matches：3 }， {_id：{_id：ObjectId（5051f1726a64bd2c54918b24）}，matches：2 }， {_id：{_id：ObjectId（5051f1756a64bd2c54918b25）}，matches：1 } ]，ok：1 } Given the following MongoDB collection of documents :{ title : 'shirt one' tags : [ 'shirt', 'cotton', 't-shirt', 'black' ]},{ title : 'shirt two' tags : [ 'shirt', 'white', 'button down collar' ]},{ title : 'shirt three' tags : [ 'shirt', 'cotton', 'red' ]},...How do you retrieve a list of items matching a list of tags, ordered by the total number of matched tags? For example, given this list of tags as input:['shirt', 'cotton', 'black']I'd want to retrieve the items ranked in desc order by total number of matching tags:item total matches-------- --------------Shirt One 3 (matched shirt + cotton + black)Shirt Three 2 (matched shirt + cotton)Shirt Two 1 (matched shirt)In a relational schema, tags would be a separate table, and you could join against that table, count the matches, and order by the count.But, in Mongo... ?Seems this approach could work, break the input tags into multiple "IN" statementsquery for items by "OR"'ing together the tag inputsi.e. where ( 'shirt' IN items.tags ) OR ( 'cotton' IN items.tags )this would return, for example, three instances of "Shirt One", 2 instances of "Shirt Three", etcmap/reduce that outputmap: emit(this._id, {...});reduce: count total occurrences of _idfinalize: sort by counted totalBut I'm not clear on how to implement this as a Mongo query, or if this is even the most efficient approach. 解决方案 As i answered in In MongoDB search in an array and sort by number of matchesIt's possible using Aggregation Framework. Assumptionstags attribute is a set (no repeated elements)Query This approach forces you to unwind the results and reevaluate the match predicate with unwinded results, so its really inefficient.db.test_col.aggregate( {$match: {tags: {$in: ["shirt","cotton","black"]}}}, {$unwind: "$tags"}, {$match: {tags: {$in: ["shirt","cotton","black"]}}}, {$group: { _id:{"_id":1}, matches:{$sum:1} }}, {$sort:{matches:-1}});Expected Results{ "result" : [ { "_id" : { "_id" : ObjectId("5051f1786a64bd2c54918b26") }, "matches" : 3 }, { "_id" : { "_id" : ObjectId("5051f1726a64bd2c54918b24") }, "matches" : 2 }, { "_id" : { "_id" : ObjectId("5051f1756a64bd2c54918b25") }, "matches" : 1 } ], "ok" : 1} 这篇关于使用列表中的标签获取文档，按匹配的总数排序的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！