Firebase function - How to return the object in response as JSON after saving the data to database?

Question

I am using firebase function to write the data into the database. The function I am using is triggered by HTTP call (only POST). I was able to save the data to the database, however, couldn't return the data in proper format. I am getting, the following error

TypeError: snapshot.val is not a function

export const saveOrder = functions.https.onRequest(((request, response) => {

    if (request.method == "POST") {

        const data = JSON.stringify(request.body);
        let jsonData = data.replace(/\r?\n\t?/g, '');

        let object = {order: JSON.parse(jsonData), status: "pending"};

        return admin.database()
            .ref("orders")
            .push(object)
            .then(function (snapshot) {
                return response.send(200, snapshot.val());
            });
    } else {
        response.contentType("application/json");
        response.status(400).send('{"message":"Invalid method"}');
        return;
    }
}));

Hristo Eftimov · Accepted Answer · 2018-11-08 15:10:54Z

2

The .push() method doesn't return a Snapshot data, it returns a Reference.

You can get the key of the new reference:

    .then((reference) => { return response.send(200, reference.key) });

To return the data in a proper format, you can return the object that you are passing to Firebase:

return admin.database() ref("orders").push(object)
        .then(function(reference) {
            return response.send(200, {
                  key: reference.key,
                  data: object,
            });
        });

answered Nov 8, 2018 at 15:10

Hristo Eftimov

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