blocks|key|2498190|text|这可以通过使用$set用name子文档中的值更新所有文档来实现：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2498191|db.collection.find({+"name":+{"$exists":+1+}+}).forEach(function(doc)+{
++++var+setName+=+{};
++++for+(+var+k+in+doc.name+)+{
++++++++setName[k]+=+doc.name[k];
++++}
++++db.collection.update(
++++++++{+"_id":+doc._id+},
++++++++{+"$set":+setName,+"$unset":+"name"+}
++++);
})|code-block|syntax|javascript|2498192|虽然我建议您使用$project，因为它比这个解决方案性能更好，但我可以理解您不想使用$project的原因。|2498193|entityMap^0|7|4|C|4|0|0|8|8|17|8|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

This may be achieved by using <code>$set</code> to update all documents with the values in the <code>name</code> sub-document:

<pre><code>db.collection.find({ "name": {"$exists": 1 } }).forEach(function(doc) {
 var setName = {};
 for ( var k in doc.name ) {
 setName[k] = doc.name[k];
 }
 db.collection.update(
 { "_id": doc._id },
 { "$set": setName, "$unset": "name" }
 );
})
</code></pre>

While I'll recommend you use <code>$project</code> because it would be more performant than this solution, I can understand why you wouldn't want to use <code>$project</code>.

blocks|key|2708100|text|以下是使用JavaScript变量的解决方案。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2708101|#+Set+Object+for+what+to+project
var+projectWhat+=+{'_id'+:+0};

#+Fill+Object+with+keys
Object.keys(db.coll.findOne().name).forEach(function(x){
++++projectWhat[x]+=+"$name."+%2B+x;
});

#+Do+Aggregate
db.coll.aggregate([{$project+:+projectWhat}])|code-block|syntax|javascript|2708102|输出结果将是|2708103|{+"firstName"+:+"John",+"lastname"+:+"Peters"+}
{+"firstName"+:+"Mary",+"lastname"+:+"Jones"+}|2708104|希望这能有所帮助。|2708105|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here is the solution which uses JavaScript variable.

<pre><code># Set Object for what to project
var projectWhat = {'_id' : 0};

# Fill Object with keys
Object.keys(db.coll.findOne().name).forEach(function(x){
 projectWhat[x] = "$name." + x;
});

# Do Aggregate
db.coll.aggregate([{$project : projectWhat}])
</code></pre>

And the output will be

<pre><code>{ "firstName" : "John", "lastname" : "Peters" }
{ "firstName" : "Mary", "lastname" : "Jones" }
</code></pre>

Hope this helps.

blocks|key|2708123|text|MongoDB+3.4在聚合管道中有了一个新阶段--+$replaceRoot，它做的正是我们所要求的。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2708124|https://docs.mongodb.com/manual/reference/operator/aggregation/replaceRoot/|2708125|entityMap|0|LINK|mutability|MUTABLE|url^0|R|C|0|0|23|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@$9|S|A|T|1|U]]|E|$]]|$1|H|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|G]]]]

MongoDB 3.4 has the new stage in aggregation pipeline - <code>$replaceRoot</code>, which does exactly what was asked.

<a href="https://docs.mongodb.com/manual/reference/operator/aggregation/replaceRoot/" rel="noreferrer">https://docs.mongodb.com/manual/reference/operator/aggregation/replaceRoot/</a>

blocks|key|2498260|text|从Mongo+4.2开始，可以使用$replaceWith聚合操作符将一个文档替换为另一个文档(在我们的示例中，替换为一个子文档)：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2498261|//+{+_id:+"1",+name:+{+firstName:+"John",+lastname:+"Peters"+}+}
//+{+_id:+"2",+name:+{+firstName:+"Mary",+lastname:+"Jones"++}+}
db.collection.aggregate({+$replaceWith:+"$name"+})
//+{+firstName:+"John",+lastname:+"Peters"+}
//+{+firstName:+"Mary",+lastname:+"Jones"+}|code-block|syntax|javascript|2498262|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.mongodb.com/master/reference/operator/aggregation/replaceWith/#pipe._S_replaceWith^0|1|9|H|C|H|C|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@$9|X|A|Y|1|Z]]|E|$]]|$1|F|3|G|5|H|7|10|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

Starting <code>Mongo 4.2</code>, the <a href="https://docs.mongodb.com/master/reference/operator/aggregation/replaceWith/#pipe._S_replaceWith" rel="nofollow noreferrer"><code>$replaceWith</code></a> aggregation operator can be used to replace a document by another (in our case by a sub-document):

<pre><code>// { _id: "1", name: { firstName: "John", lastname: "Peters" } }
// { _id: "2", name: { firstName: "Mary", lastname: "Jones" } }
db.collection.aggregate({ $replaceWith: "$name" })
// { firstName: "John", lastname: "Peters" }
// { firstName: "Mary", lastname: "Jones" }
</code></pre>

blocks|key|33500|text|您可以像这样使用$replaceRoot：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|33501|db.collection.aggregate(
++++[
++++++++{
++++++++++++$replaceRoot+:+{
++++++++++++++++newRoot:+{"$name"}
++++++++++++}
++++++++}
++++]
)|code-block|syntax|javascript|33502|此外，如果您希望保留根文档中的某个字段，则可以使用$mergeObjects将其与嵌入的对象组合：|33503|db.collection.aggregate(
++++[
++++++++{
++++++++++++$replaceRoot+:+{
++++++++++++++++newRoot:+{
+++++++++++++++++++$mergeObjects:+[
++++++++++++++++++++++{"_id":+"$_id"},
++++++++++++++++++++++"$name"
+++++++++++++++++++]
++++++++++++++++}
++++++++++++}
++++++++}
++++]
)|33504|entityMap^0|8|C|0|0|P|D|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|P|$]]

You can use <code>$replaceRoot</code> like this:
<pre><code>db.collection.aggregate(
 [
 {
 $replaceRoot : {
 newRoot: {&quot;$name&quot;}
 }
 }
 ]
)
</code></pre>
Also if you have a field in the root document you want to retain you can use a <code>$mergeObjects</code> to combine it with your embedded object:
<pre><code>db.collection.aggregate(
 [
 {
 $replaceRoot : {
 newRoot: {
 $mergeObjects: [
 {&quot;_id&quot;: &quot;$_id&quot;},
 &quot;$name&quot;
 ]
 }
 }
 }
 ]
)
</code></pre>

Using the aggregate pipeline, I am trying to project an embedded document to the root level WITHOUT projecting each field individually.

For example, I want to project <code>name</code> from this collection to the root level:

<pre><code>[
 {
 _id: "1",
 name: {
 firstName: "John",
 lastname: "Peters"
 }
 },
 {
 _id: "2",
 name: {
 firstName: "Mary",
 lastname: "Jones"
 }
 }
]
</code></pre>

This is what I am looking for:

<pre><code>[
 {
 firstName: "John",
 lastname: "Peters"
 },
 {
 firstName: "Mary",
 lastname: "Jones"
 }
]
</code></pre>

Is there a way to do this without projecting each field individually? I don't want to have to do this:

<pre><code>db.collection.aggregate(
 [
 {
 $project : {
 "_id" : 0,
 "firstName" : "$name.firstName",
 "lastName" : "$name.lastName"
 }
 }
 ]
</code></pre>

MongoDB $project embedded document to root level

MongoDB

使用聚合管道，我尝试将嵌入的文档投影到根级别的，而不是单独投影每个字段。例如，我希望将name从这个集合投影到根级别：[    {        _id: "1",        name: {            firstName: "John",            lastname: "Peters"   ...

问将MongoDB $project嵌入文档添加到根级别
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