mongodb: remove subdocument with where clause that includes document & subdocument

mongodb

Solution

The syntax of your `$pull` object is off. Try this instead:

db.workers.update({"type":"Manager","employees.id":101},
    {$pull : {"employees" : {"id" : 103}}},false,true)

To confirm they were removed:

db.workers.find({
    type: "Manager", 
    $and: [{'employees.id': 101}, {'employees.id': 103}]
})

Problem

Here is my collection (workers): ``` "type" : "Manager", "employees" : [{ "name" : "bob" "id" : 101 },{ "name" : "phil" "id" : 102 },{ "name" : "bob" "id" : 103 }] ``` First: this is NOT an array so $pullAll will not work or other array commands. All I want to do is: (1) search the collection for id 101 in ALL subdocuments with type Manager. (2) If 101 exists in a "Manager" subdocument, I want to remove item 103. I have been pouring over the interwebs for two days on this issue and cannot figure it out. I've tried this (and many other variations): ``` db.workers.update( {"type":"Manager","employees.id":101},{$pull : {"employees.id" : {"id" : 103}}},false,true) ```

Original source