mongodb使用forEach更新所有文档的密钥

本文介绍了mongodb使用forEach更新所有文档的密钥的处理方法，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

我想在Mongo中将所有文档的订单"字段更新为1..2..3..4 .... 34.

I want to update in Mongo the 'order' field to all of my documents so they will be 1..2..3..4....34.

运行此命令后，它们都具有"order":"34". 我在做什么错了?

After running this, they all have "order": "34". What am I doing wrong?

var i = 1; db.images.find().forEach(function() { db.images.update( {}, { "$set": {"order": NumberInt(i)} }, { multi: true } ); i++; })

推荐答案

multi : true表示所有与查询匹配的文档都将被更新.您的查询是{}，它与所有文档匹配.因此，基本上，您是在每次迭代中更新所有文档的order.

multi : true means all documents matching the query will be updated. And your query is {}, which matches all the documents. So, basically you are updating the order of all the documents in every iteration.

此外，必须启用 snapshot 模式光标上，以确保不会多次返回同一文档.

Also, snapshot mode has to be enabled on the cursor to ensure that the same document isn't returned more than once.

您可以尝试以下方法:

var i = 1; db.images.find().snapshot().forEach(function(image) { db.images.update( {"_id" : image._id}, { "$set": {"order": NumberInt(i)} } ); i++; })

从性能的角度来看，最好使用批量API. 批量写入

From a performance standpoint, it is better to use the bulk APIs. bulkwrite