mongodb type change to array

Question

I have a field in mongodb that's a string. {"field": "some text"}, I want to convert them all into arrays. {"field": ["some text"]}

I know I can just loop through all the documents, get the field, then update, but I'm wondering if there's a cleaner way.

Thanks.

Answer 1

Nitin Garg's answer above almost works, except his example converts from a string to a hash, NOT a string to an array.

Taking into account Joel Harris's comments, the proper solution would look like:

db.jobs.find( { "jobLocationCity" : { $type : 2 } } ).snapshot().forEach( function (x) {
    x.jobLocationCity = [ jobLocationCity ];
    db.jobs.save(x);
});

Or if using db.eval:

function f() {
    db.jobs.find( { "jobLocationCity" : { $type : 2 } } ).snapshot().forEach( function (x) {
        x.jobLocationCity = [ jobLocationCity ];
        db.jobs.save(x);
    });
}
db.eval(f);

Answer 2

Starting in Mongo 4.2, db.collection.update() can accept an aggregation pipeline, finally allowing the update of a field based on its current value:

// { field: "some text" }
db.collection.update(
  {},
  [{ $set: { field: ["$field"] } }],
  { multi: true }
)
// { field: [ "some text" ] }

• The first part {} is the match query, filtering which documents to update (in this case all documents).

• The second part [{ $set: { field: { ["$field"] } } }] is the update aggregation pipeline (note the squared brackets signifying the use of an aggregation pipeline). $set (alias of $addFields) is a new aggregation operator which in this case replaces the field's value (simply wrapping it into an array). Note how field is modified directly based on its own value ($field).

• Don't forget { multi: true } (or to use updateMany), otherwise only the first matching document will be updated.

Answer 3

Actually, the find( { "jobLocationCity" : { $type : 2 } } ) will not work properly, because if you'll run update script next time, it will treat ['mystring'] elements again as string type.

You should use something like this to prevent it:

db.message_info.find( { "jobLocationCity" : { $type : 2 } }  ).snapshot().forEach(
  function (x) {
    if (!Array.isArray(x.jobLocationCity)){
        x.jobLocationCity = [ x.jobLocationCity  ];
        db.jobs.save(x);
    }
  }
)

see http://docs.mongodb.org/manual/reference/operators/