blocks|key|342917|text|就像MySQL+f.e.一样，MongoDB没有办法返回它，因为它不会将文档按顺序保存在数据库中。不命名行号。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|342918|jhonkola的ObjectID技巧只有在只有一个客户端创建新元素时才有效，因为ObjectID是在客户端生成的，第一部分是时间戳。如果不同的客户端与同一台服务器通信，则无法保证顺序。尽管如此，我不会依赖于这一点。|342919|我也不太明白你在尝试做什么，所以也许在你的问题中提到这一点？然后我就可以更新答案了。|342920|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

There is no way that MongoDB can return this as it does not keep documents in order in the database, just like MySQL f.e. doesn't name row numbers. 

The ObjectID trick from jhonkola will only work if only one client creates new elements, as the ObjectIDs are generated on the client side, with the first part being a timestamp. There is no guaranteed order if different clients talk to the same server. Still, I would not rely on this.

I also don't quite understand what you are trying to do though, so perhaps mention that in your question? I can then update the answer.

blocks|key|559004|text|如果您的要求是查找文档的位置而不考虑顺序，那么这是不可能的，因为MongoDb不会按特定顺序存储文档。但是，如果您想知道基于某个字段的索引，例如_id，则可以使用此方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|559005|如果您严格遵循_id字段中的自动递增。您可以计算所有值小于该_id的文档，比方说n，那么n%2B1将是基于_id的文档索引。|559006|n+=+db.myCollection.find({"id":+{+"$lt"+:+12345}}).count()+;|code-block|syntax|javascript|559007|如果从集合中删除文档，这也是有效的。|559008|entityMap^0|20|3|0|7|3|U|3|1F|3|0|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|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|10|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|11|8|@]|D|@]|E|$]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|P|$]]

If your requirement is to find the position of the document irrespective of any order, that is not 
possible as MongoDb does not store the documents in specific order.
However,if you want to know the index based on some field, say <code>_id</code> , you can use this method.

If you are strictly following auto increments in your <code>_id</code> field. You can count all the documents
that have value less than that <code>_id</code>, say n , then n + 1 would be index of the document based on <code>_id</code>.

<pre><code>n = db.myCollection.find({"id": { "$lt" : 12345}}).count() ;
</code></pre>

This would also be valid if documents are deleted from the collection.

blocks|key|343024|text|这可能是非常慢但简单的方法。在这里，您可以像往常一样传递查询。我只是循环所有的文档，并检查if条件是否与记录匹配。在这里，我正在检查_id字段。您可以使用任何其他单个字段或多个字段来检查它。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|343025|var+docIndex+=+0;
db.url_list.find({},{"_id":1}).forEach(function(doc){
++docIndex%2B%2B;
++if("5801ed58a8242ba30e8b46fa"==doc["_id"]){
++++print('document+position+is...'+%2B+docIndex);
++++return+false;
++}
});|code-block|syntax|javascript|343026|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This might be very slow but straightforward method. Here you can pass as usual query. Just I am looping all the documents and checking if condition to match the record. Here I am checking with _id field. You can use any other single field or multiple fields to check it.

<pre><code>var docIndex = 0;
db.url_list.find({},{"_id":1}).forEach(function(doc){
 docIndex++;
 if("5801ed58a8242ba30e8b46fa"==doc["_id"]){
 print('document position is...' + docIndex);
 return false;
 }
});
</code></pre>

blocks|key|343075|text|在使用所需位置作为查询搜索数据库集合(MyCollection)时，重新构造集合以包括任何条目的位置，即{'id'：12340，'name'：'G'，'+position+'：1}|type|unstyled|depth|inlineStyleRanges|entityRanges|data|343076|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Restructure your collection to include the position of any entry i.e {'id': 12340, 'name': 'G', 'position': 1} then when searching the database collection(myCollection) using the desired position as a query

blocks|key|343100|text|我使用的返回整个集合的查询都使用sort来获得可重复的顺序，find.sort.forEach使用上面的脚本来获得正确的索引。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|343101|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

The queries I use that return the entire collection all use sort to get a reproducible order, find.sort.forEach works with the script above to get the correct index.

How to get position (index) of selected document in mongo collection? 

E.g.
this document: <code>db.myCollection.find({"id":12345})</code>
has index <code>3</code> in <code>myCollection</code>

<pre><code>myCollection:
 id: 12340, name: 'G'
 id: 12343, name: 'V'
 id: 12345, name: 'A'
 id: 12348, name: 'N'
</code></pre>

Get position of selected document in collection [mongoDB]

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如何获取选中文档在mongo集合中的位置(索引)？例如，这个文档：db.myCollection.find({"id":12345})在myCollection中有索引3myCollection:    id: 12340, name: 'G'    id: 12343, name: 'V'    id: 12345,...

问获取所选文档在集合中的位置[mongoDB]
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