'MongoDB Replace specific array values

In MongoDB, I have a movie collection that has an array of languages , e.g.
languages: [0:USA, 1: German, 2: French, ...etc]

The array values are not in any particular order.

How can I now update an array value based on some specific value? Let's say I want to update all "French" and replace it with "Francais" for the entire collection. How can I do that?

arraysmongodb


Solution 1:[1]

Use the positional $ operator which identifies the element in the languages array to update without explicitly specifying its position in the array i.e. instead of knowing the position in advance and updating the element as:

db.movies.updateMany(
    { "languages": "French" }, 
    { "$set": { "languages.2": "Francais" } }
)

you can just use the $ operator as:

db.movies.updateMany(
    { "languages": "French" }, 
    { "$set": { "languages.$": "Francais" } }
)

Alternatively using the aggregation pipeline for update operations:

db.movies.updateMany(
    { "languages": "French" }, 
    [
        { "$set": { 
            "languages": {
                "$map": {
                    "input": "$languages",
                    "in": {
                        "$cond": [
                            { "$eq": ["$$this", "French"] }, 
                            "Francais", 
                            "$$this"
                        ]
                    }
                }
            }
        } }
    ]
)

Solution 2:[2]

In case you have duplicate entries in your array like this:

"languages" : [ 
    "USA", 
    "German", 
    "Francais",
    "Francais"
]

You can use the following syntax to replace all occurrences starting with MongoDB v3.5.12 (documentation)

db.movies.updateMany(
    { "languages": "French" },
    { "$set": { "languages.$[filter]": "Francais" } },
    { "arrayFilters": [ { "filter": "French" } ] }
)

Strictly speaking, the { "languages": "French" } filter is not needed. It will, however, speed up things by leveraging an index on "languages" in case there is one. If that's not needed/wanted, passing a simple {} value as the first parameter works as well.

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2 dnickless

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