如何更新一个 MongoDB 文档的 _id？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/4012855/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How update the _id of one MongoDB Document?
提问by shingara
I want update an _idfield of one document. I know it's not a really good pratice. But with some technical reason, I need update it. If I try to update it I get:
我想更新_id一个文档的字段。我知道这不是一个很好的做法。但由于某些技术原因，我需要更新它。如果我尝试更新它，我会得到：
> db.clients.update({ _id: ObjectId("123")}, { $set: { _id: ObjectId("456")}})
Performing an update on the path '_id' would modify the immutable field '_id'
And the update is not made. How I can update it?
并且没有进行更新。我怎样才能更新它？
回答by Niels van der Rest
You cannot update it. You'll have to save the document using a new _id, and then remove the old document.
你不能更新它。您必须使用 new 保存文档_id，然后删除旧文档。
// store the document in a variable
doc = db.clients.findOne({_id: ObjectId("4cc45467c55f4d2d2a000002")})
// set a new _id on the document
doc._id = ObjectId("4c8a331bda76c559ef000004")
// insert the document, using the new _id
db.clients.insert(doc)
// remove the document with the old _id
db.clients.remove({_id: ObjectId("4cc45467c55f4d2d2a000002")})
回答by Patrick Wolf
To do it for your whole collection you can also use a loop (based on Niels example):
要为您的整个集合执行此操作，您还可以使用循环（基于 Niels 示例）：
db.status.find().forEach(function(doc){
doc._id=doc.UserId; db.status_new.insert(doc);
});
db.status_new.renameCollection("status", true);
In this case UserId was the new ID I wanted to use
在这种情况下 UserId 是我想使用的新 ID
回答by Mark
In case, you want to rename _id in same collection (for instance, if you want to prefix some _ids):
如果您想在同一集合中重命名 _id（例如，如果您想为某些 _id 加上前缀）：
db.someCollection.find().snapshot().forEach(function(doc) {
if (doc._id.indexOf("2019:") != 0) {
print("Processing: " + doc._id);
var oldDocId = doc._id;
doc._id = "2019:" + doc._id;
db.someCollection.insert(doc);
db.someCollection.remove({_id: oldDocId});
}
});
if (doc._id.indexOf("2019:") != 0) {...needed to prevent infinite loop, since forEach picks the inserted docs, even throught .snapshot()method used.
if (doc._id.indexOf("2019:") != 0) {...需要防止无限循环，因为 forEach 选择插入的文档，甚至通过使用.snapshot()方法。
回答by Florent Arlandis
Here I have a solution that avoid multiple requests, for loops and old document removal.
在这里，我有一个解决方案，可以避免多个请求、循环和旧文档删除。
You can easily create a new idea manually using something like:_id:ObjectId()But knowing Mongo will automatically assign an _id if missing, you can use aggregate to create a $projectcontaining all the fields of your document, but omit the field _id. You can then save it with $out
您可以使用以下内容轻松手动创建一个新想法：_id:ObjectId()但是知道 Mongo 会在丢失时自动分配一个 _id，您可以使用聚合创建一个$project包含文档所有字段的字段，但省略字段 _id。然后你可以保存它$out
So if your document is:
因此，如果您的文件是：
{
"_id":ObjectId("5b5ed345cfbce6787588e480"),
"title": "foo",
"description": "bar"
}
Then your query will be:
那么您的查询将是：
db.getCollection('myCollection').aggregate([
{$match:
{_id: ObjectId("5b5ed345cfbce6787588e480")}
}
{$project:
{
title: '$title',
description: '$description'
}
},
{$out: 'myCollection'}
])
