To remove duplicates in MongoDB, you need to follow a series of steps:
First, you need to identify the duplicates. MongoDB doesn’t support multiple unique keys, so you have to use a traditional loop-lookup system. Run this on your MongoDB:
```
db.collection.aggregate([
{$group: {“_id”: “$duplicateField”, “count”: {“$sum”: 1}, dups: {“$addToSet”: “$_id”}}},
{$match: {“count”: {“$gte”: 2 }}}
]);
```
You replace `collection` and `duplicateField` by your collection’s and duplicate field’s name respectively.
This will gather all the documents that have at least a duplicate and output them in an array. `_id` becomes the duplicate value and `dups` is the array of documents `id`s.
In order to eliminate the duplicates, you need to write and execute a script. An example of this script:
```
var duplicates = db.collection.aggregate([
{$group: {“_id”: “$duplicateField”, “count”: {“$sum”: 1}, dups: {“$addToSet”: “$_id”}}},
{$match: {“count”: {“$gte”: 2 }}}
]);
duplicates.forEach(function(doc) {
doc.dups.shift(); // First one skip for deleting
doc.dups.forEach( function(dupId){
db.collection.remove({ _id : dupId }); // Delete remaining duplicates
})
});
```
This script is for deleting duplicate items based on a particular key (field) in the documents. The first item is preserved, subsequent duplicate items are deleted. Please replace `collection` and `duplicateField` with your collection’s and duplicate field’s name.
In order to not make any mistakes, make sure you have a backup, and first try this in the development environment before running it on production.
Also, to prevent further duplicates, you can use unique indexes:
`db.collection.createIndex( { “
