MongoDB: How to use a field of an array as a mutual exclusive flag - mongodb

Question

Here below is again my hypothetical Users collection where more than one address is allowed:
{
"firstName": "Joe",
"lastName": "Grey",
...
"addresses":
[
{
"name": "Default",
"street": "...",
...,
"isDefault": true
},
{
"name": "Home",
"street": "...",
...,
"isDefault": false
},
{
"name": "Office",
"street": "...",
...,
"isDefault": false
}
]
}
This time I've added the isDefault flag, which should be mutual exclusive. That is, when I update an address and set isDefault to true, I should ensure this flag in the other elements of the array are set to false. Is there a way to do that in one step without performing a find-and-update?

Answer 1

You will generally find this impossible at present as there is no way to update multiple fields in an array at once, let alone with different values. Consider even the following operation:
db.collection.update(
{ "addresses.isDefault": false },
{ "addresses.isDefault.$": true }
)
Now that is the reverse of what you want to do (well part of), but to illustrate my point, that will match the first item in the array that meets the query condition. Using the positional $ operator in the update, only the second item in the array would actually be set. The third element would be left alone because this operation does not work that way. The documentation covers this.
In order to set all the fields at once you must retrieve the entire document via find, flip your values in code, and then do a separate update while replacing the whole array.
Now the "link" in the comments says, you can do this all in one stroke. But even so you will be constructing the query and update for each element in the array. But it will be more efficient.