Unable to store each value of array as separate rows in database table

Question

My app sends off a body message with a user's information when they first create an account. One of the values is an "Interests" array, and my goal is to have my Node.js web service store each value of the "Interests" array as a separate row in one of my mysql database tables. However, the issue is that the last value of my "Interests" array is stored for each row. Below, you will find further details

req.body example:

{"firstName": "Mason",
 "lastName": "Rad",
 "age": 19,
 "userId": "radMas28",
 "userPassword": "fgjfh4534534",
 "interests": ["hockey", "baseball", "basketball"]
}

What I've tried:

var storeInterests = function(body) {
    for(var interest in body.interests) {
        database.sequelize.sync()
            .then(() => database.Interests.create( {
                userId: body.userId,
                interest: body.interests[interest]
            }));
    }
};

What's stored in my database:

I've also tried using a while loop that continues until the counter variable reaches the body.interests array .length property, however the same issue occurred.

Your help is very much appreciated.

kane · Accepted Answer · 2017-05-23 21:52:03Z

1

The issue is that database.sequelize.sync().then() is asynchronous and by the time each runs, the loop counter has iterated to the last value.

Try this

var storeInterests = function(body) {
  for(var i=0; i<body.interests.length; i++) {
    store(body.userId, body.interests[i])
  }
};

function store(userId, interest) {
    database.sequelize.sync()
        .then(() => database.Interests.create( {
            userId: userId,
            interest: body.interests[i]
        }));
}

edited May 23, 2017 at 21:52

answered May 23, 2017 at 21:29

kane

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3 Comments

Jared Hart Over a year ago

The first way causes the number "2" to be stored in each row for my interest column. The second way causes no information to be stored in the interest column of the table, just the userId is stored in three separate rows, but no interest value :(

kane Over a year ago

@JaredHart I updated the answer. please try the new solution

Jared Hart Over a year ago

Hi Kane, I arrived at the same solution at the same time! It works now! Thanks so much for the help, really appreciate it.

Jared Hart · Accepted Answer · 2017-05-23 21:50:10Z

0

So I was able to figure out a solution, however I'm worried that it might be inefficient. Could someone please take a look?

var storeInterests = function(body) {
    for(var counter in body.interests) {
        interestFunction(body.userId, body.interests, counter);
    }
};

var interestFunction = function(id, interests, counter) {
    database.sequelize.sync()
        .then(() => database.Interests.create( {
            userId: id,
            interest: interests[counter]
        }));
}

answered May 23, 2017 at 21:50

Jared Hart

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