javascript object/array for loop to assigns keys

Question

I have an object that looks like this:

var obj = {
    "cat" : [
        {val: 'res', string: "Residential & Single Family Homes"},
        {val: 'Dup', string: "Duplexes & Apartments"},
        {val: 'Con', string: "Condominiums"},
        {val: 'Lot', string: "Lots, Land & Farms"},
        {val: 'Com', string: "Commercial"},
        {val: 'Mob', string: "Mobile Homes"},
    ],
    "bdrms" : [
        {val: "1", string: "1 Bedroom"},
        {val: "2", string: "2 Bedrooms"},
        {val: "3", string: "3 Bedrooms"},
        {val: "4", string: "4 Bedrooms"},
        {val: "5", string: "5 Bedrooms"},
    ],
    "bthrms" : [
        {val: "1", string: "1 Bathroom"},
        {val: "1.1", string: "1 1/2 Bathrooms"},
        {val: "2", string: "2 Bathrooms"},
        {val: "2.1", string: "2 1/2 Bathrooms"},
        {val: "3", string: "3 Bathrooms"},
        {val: "3.1", string: "3 1/2 Bathrooms"},
        {val: "4", string: "4 Bathrooms"},
        {val: "4.1", string: "4 1/2 Bathrooms"},
        {val: "5", string: "5 Bathrooms"},
    ],
    "zip" : [

    ]
};

and what i would like to do is dynamically generate an array of zip codes inside "zip" using a for loop. what I have so far isnt working as I am getting a bunch of errors such as unexpected token ect...

im assuming that i am doing this totally wron, but this is what i am attempting to do:

"zip" : [
    for(i=43000; i<=45999;){
        {val: i, string: i},
        i++
    };
]

Just as an FYI, you may get some weird behavior with a property named string. I believe that's typically a reserved word in JavaScript... — War10ck
– War10ck, Commented Jul 16, 2015 at 18:02

T.J. Crowder · Accepted Answer · 2015-07-16 18:04:26Z

4

The simplest thing is just to do it after the object initializer:

for(i=43000; i<=45999;i++){
    obj.zip.push({val: i, string: i});
}

(Note that I moved the i++, it has no business being in the body of the loop.) (Also note that unless you have i declared somewhere, that was falling prey to The Horror of Implicit Globals.) (And for blocks don't have a ; after them. :-) )

If it were really important that it be part of the initializer (which is unlikely), you could use a temporary function you call immediately:

"zip" : function() {
    var a = [];
    for(var i=43000; i<=45999;i++){
        a.push({val: i, string: i});
    }
    return a;
}()

...but I wouldn't without a good reason. Though I'm not sure why not...

edited Jul 16, 2015 at 18:04

answered Jul 16, 2015 at 18:01

T.J. Crowder

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5 Comments

Destination Designs Over a year ago

by "after the object initialize" do you mean outside of var obj = {...}; ?

xyz Over a year ago

for closure +1, even though only use is redability! And it's not a bad practice to initialise this way, so long as each initialisation does not contain any logic a model shouldn't have

xyz Over a year ago

@War10ck no ] should be needed can you elaborate

T.J. Crowder Over a year ago

@prakharsingh95: He was commenting on an earlier version of the answer that had a typo.

T.J. Crowder Over a year ago

@DestinationDesigns: Yes. Specifically, after it.

Andy · Accepted Answer · 2015-07-16 18:02:30Z

2

Just have an empty array zip: [] and then use a loop to push objects containing the values into it, making sure you var your loop variables:

for (var i = 43000; i <= 45999; i++) {
    obj.zip.push({ val: i, string: i });
};

DEMO

answered Jul 16, 2015 at 18:02

Andy

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Comments

Sushanth -- · Accepted Answer · 2015-07-16 18:07:58Z

1

"zip" : [ <---- Having a loop inside an array will throw an error for(i=43000; i<=45999;){ {val: i, string: i}, i++ }; ]

Supposed to look like

"zip": function () { //<--- Need to write up a function
    var arr = [];    // that will return an array instead
    for (var i = 43000; i <= 45999; i++) {
        var oo = {   // 
            val: i,
            string: i
        };
        arr.push(oo);
    };

    return arr;

}()

So basically it is a function that computes the array and returns the zip codes.

edited Jul 16, 2015 at 18:07

answered Jul 16, 2015 at 18:02

Sushanth --

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Comments

Nathan Hornby · Accepted Answer · 2015-07-16 18:07:17Z

0

var zip = [];
for (var i = 43000; i <= 45999; i++) {
    zip.push({val: i, string: i});
}
console.log(zip);

Will output the zip array containing the objects you desire.

Edit: Beaten to the punch by Andy!

answered Jul 16, 2015 at 18:07

Nathan Hornby

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Comments

Nitesh Patel · Accepted Answer · 2015-07-16 18:17:19Z

0

Try this as a simple readable solution:

"zip": getZipCodes()

function getZipCodes() {
    var codes = [];
    for(var i = 43000; i <= 45999; i++){
        codes.push({val: i, string: i});
    }

    return codes;
};

That way you do not modify the object after creation.

answered Jul 16, 2015 at 18:17

Nitesh Patel

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Comments

dgavian · Accepted Answer · 2015-07-16 18:09:00Z

-2

You can't run your loop inside of the array literal. This will work:

   var obj = {
        "cat" : [
            {val: 'res', string: "Residential & Single Family Homes"},
            {val: 'Dup', string: "Duplexes & Apartments"},
            {val: 'Con', string: "Condominiums"},
            {val: 'Lot', string: "Lots, Land & Farms"},
            {val: 'Com', string: "Commercial"},
            {val: 'Mob', string: "Mobile Homes"},
        ],
        "bdrms" : [
            {val: "1", string: "1 Bedroom"},
            {val: "2", string: "2 Bedrooms"},
            {val: "3", string: "3 Bedrooms"},
            {val: "4", string: "4 Bedrooms"},
            {val: "5", string: "5 Bedrooms"},
        ],
        "bthrms" : [
            {val: "1", string: "1 Bathroom"},
            {val: "1.1", string: "1 1/2 Bathrooms"},
            {val: "2", string: "2 Bathrooms"},
            {val: "2.1", string: "2 1/2 Bathrooms"},
            {val: "3", string: "3 Bathrooms"},
            {val: "3.1", string: "3 1/2 Bathrooms"},
            {val: "4", string: "4 Bathrooms"},
            {val: "4.1", string: "4 1/2 Bathrooms"},
            {val: "5", string: "5 Bathrooms"},
        ],
        "zip" : [

        ]
    },
    zips = obj.zip,
    i,
    counter = 0,
    j;
    for(i=43000; i<=45999; i++){
        zips[counter] = { val: i, string: i };
        counter++
    };

answered Jul 16, 2015 at 18:09

dgavian

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5 Comments

xyz Over a year ago

counter = i always. Js engine will auto optimise this, but it's bad for redability

dgavian Over a year ago

I agree it would have been better to use push, and I also agree that using "i' for a counter is pretty standard, but I haven't heard the "counter = i always" rule. Do you have a link to any documentation on this that I can read up on?

War10ck Over a year ago

@prakharsingh95 counter = i is usually true if you're starting with i = 0. In this case, though, that doesn't hold true. counter should be 0 and i should be 43000. It's not a clean solution by any accounts but this would definitely work

xyz Over a year ago

@dgavian sorry I meant counter is just an offset from i. You are incrementing both when you only need to increment one. you could do zips[counter-43000]

xyz Over a year ago

@War10ck I clarify. If two variables have an unchanging relationship, and mathematically trivial keep only one. For ex, js engine or a compiler will automatically do this, and it's easier to read with one variable.

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