在MongoDB中更新整个文档(所有字段)的最快方法是什么? [英] What is the fastest way to update the whole document (all fields) in MongoDB?

问题描述

假设我要更新整个文档并覆盖所有字段，但_id除外.就资源消耗而言，三种方法中哪一种是最好的:

Let's say I want to update the whole document and override all fields, except _id. What of the three methods is the best in terms of resource consumption:

1.将完整的文档设置为更新参数，以便传递所有字段

示例:

collection.update({_id: docId}, {$set:updateDoc});

2.计算原始文档和updateDoc之间的增量文档

示例:

const originalDoc = collection.findOne(docId);
const deltaDoc = calculateDeltaFct(originalDoc, updateDoc); //get changed fields
collection.update({_id: docId}, {$set:deltaDoc});

3.使用 Mongo 3.2. replaceOne函数

3. Use the Mongo 3.2. replaceOne function

示例:

collection.replaceOne({_id: docId}, {$set:updateDoc});

我对每种方法的优缺点都有一个假设，但我想确定选择什么以及为什么选择.我不确定如何精确测量它，所以也许有人可以帮忙.

I have an assumption of the pros and the cons of each methods but I want to be sure what to choose and why. I am not sure how to measure it precisely, so maybe someone can help.

背景:

我有一个度量标准集合，其中许多文档经常更新，但是要更新的字段变化很大，因此很难为每个字段编写更新方法.相反，我打算只放入所有数据并更新所有字段，因此我仅使用一种用于所有更新的更新方法来保持代码整洁.

I have a metrics collection where many documents are updated often, but the fields to be updated vary a lot, so it is hard to write an update method for each field. Instead I intend to just throw all data in and update all fields, so I keep my code clean with only one update method for all updates.

更新:

在我的设置中，文档结构中没有嵌入任何子文档.我的(dev)设置中也没有分片和复制功能.

In my setup, there are no subdocuments embedded in the document structure. I also have no sharding and replication in my (dev) setup.

此外，我找到了一些方法( collection.explain )，我也会用它来研究该主题.不过，非常感谢您的帮助或提示.

Furthermore I found some method (collection.explain) which I will use to research on that topic, too. Nevertheless, any help or hint is much appreciated.

推荐答案

这实际上取决于您在更新之前是否需要旧信息.如果您要覆盖信息，甚至覆盖一个键->值对，那么我将使用update或replaceOne.差异可能在时间上有所不同，具体取决于集合(数据集)的大小.如果这是值得关注的基准差异.就我个人而言，我倾向于replaceOne，但这只是基于经验和我处理过的馆藏.

It really depends if you require the old information before the update. If you are overwriting the info and even for one key - > value pair then I would use either update or replaceOne. The difference may be in time depending on the size of your collection (dataset). If that is of concern benchmark the difference. Personally I would lean towards replaceOne, but that is just based on experience and the collections I deal with.

对于您所解释的内容，我认为您的第二个选择要么不是高效的，要么是精通内存的.听起来好像不需要这样的计算就可以简单地更新数据，而无需担心覆盖先前的信息.

For what you have explained I don't think your second choice is either efficient or being memory savvy. It does not sound like you need such a calculation for simply updating data where there is no concern over overwriting previous information.

这篇关于在MongoDB中更新整个文档(所有字段)的最快方法是什么?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IT屋！