create array using push method

Question

 var members = [
     ['Fred G. Aandahl', '1951-1953', 'North Dakota'],
     ['Watkins Moorman Abbitt', '1948-1973', 'Virginia'],
 ];

I need to create like this dynamically, I am using the following code to do that:

var data = new array();
var members = [];
$.each(data, function (i, elem) {
    data.push(elem["p_age"], elem["p_name"], elem["p_date"]);
    members.push(data);
});
console.log(members);
}

I need to print this values, for that.

for(var x = 0; x < members.length; x++) {
    console.log(members[i][0]);
    console.log(members[i][1]);
    console.log(members[i][2]);
}

so when i try this i get following.

  [object][object][object][object][object][object]

var members = [ ['Fred G. Aandahl', '1951-1953', 'North Dakota'], ['Watkins Moorman Abbitt', '1948-1973', 'Virginia'], ]; need to print the array like this. so i can use it further. — Dilshi
– Dilshi, Commented Aug 6, 2013 at 3:56
Your example starts with an empty array. Can you paste the actual code that creates var data? — Ross Allen
– Ross Allen, Commented Aug 6, 2013 at 4:00
Are you sure a multi-dimensional array like this is actually the best structure? An array of objects seems better suited to your problem. — Nick Tomlin
– Nick Tomlin, Commented Aug 6, 2013 at 4:00
How do you expect us to help you? You're just adding elements to your array, but who knows what's in them as you didn't provide any of the code. You're also not adding arrays to your array. — Brad
– Brad, Commented Aug 6, 2013 at 4:00

Shubh · Accepted Answer · 2013-08-06 05:33:07Z

1

I am not sure how is your code working! It has some error's if your already aware of.

Your code should work fine after you change to:-

var data = new Array();//Was an Error in your code
var members = [];
$.each(temp, function (i, elem) {
    data.push(elem["p_age"], elem["p_name"], elem["p_date"]);
    members.push(data);
});

console.log(members);
for (var x = 0; x < members.length; x++) {
    console.log(members[x][0]);//Was an Error in your code
    console.log(members[x][1]);
    console.log(members[x][2]);
}

Secondly, how does data.push(elem["p_age"], elem["p_name"], elem["p_date"]); works for you? It should give you undefined.

Just to get myself clear I wrote down your code to a fiddle. Have a look.

answered Aug 6, 2013 at 5:33

Shubh

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Comments

Arun P Johny · Accepted Answer · 2013-08-06 04:00:51Z

1

Try

var members=[];
$.each(data, function(i, elem) {
    members.push([elem["p_date"],elem["p_name"],elem["p_date"]]);
});
console.log(JSON.stringify(members))

answered Aug 6, 2013 at 4:00

Arun P Johny

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Comments

user1864610 · Accepted Answer · 2013-08-06 04:05:08Z

1

This looks suspect:

$.each(data, function(i, elem) {
  data.push(elem["p_date"],elem["p_name"],elem["p_date"]);

It looks like you're trying to iterate over data, pushing the elements back on to data. I imagine that the $.each() needs to iterate over something else.

I also question why you're pushing elem['p_date'] onto an array twice.

answered Aug 6, 2013 at 4:05

user1864610

Comments

Shiva Avula · Accepted Answer · 2013-08-06 04:47:51Z

1

Because it is treating them as objects. Try using toString() method.

answered Aug 6, 2013 at 4:47

Shiva Avula

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Sharad · Accepted Answer · 2013-08-06 04:51:09Z

1

Hi use x instead of i for loop.

for(var x=0;x<members.length;x++){
                      console.log(members[x][0]);
                      console.log(members[x][1]);
                       console.log(members[x][2]);
                 }

It will work.

answered Aug 6, 2013 at 4:51

Sharad

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Comments

c-smile · Accepted Answer · 2013-08-06 04:51:32Z

1

Not

var data=new array();

but

var data=new Array();

Array's class name is Array, but not 'array'.

answered Aug 6, 2013 at 4:51

c-smile

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