Utilize the push method to form a new array

Question

Utilize the push method to form a new array

 var teamMembers = [
     ['John D. Adams', '1959-1967', 'Ohio'],
     ['Dawson Mathis', '1971-1981', 'Georgia'],
 ];

To generate this dynamically, I am implementing the code below:

var data = new Array();
var members = [];
$.each(data, function (i, elem) {
    data.push(elem["person_age"], elem["person_name"], elem["person_date"]);
    members.push(data);
});
console.log(members);
}

I want to display these values as follows.

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

However, when I attempt this, I receive the following output.

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

javascript arrays

Answer 1

Answer №1

I'm a bit confused about how your code is functioning. There seem to be some errors that you may already be aware of.

If you make the following changes, your code should work fine:

var data = new Array();// There 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]);// There was an error in your code
    console.log(members[x][1]);
    console.log(members[x][2]);
}

Also, can you explain how

data.push(elem["p_age"], elem["p_name"], elem["p_date"]);

is working for you? It seems like it should return undefined.

To clarify things, I recreated your code in a fiddle. Take a look here.

Answer 2

I'm a bit confused about how your code is functioning. There seem to be some errors that you may already be aware of.

If you make the following changes, your code should work fine:

var data = new Array();// There 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]);// There was an error in your code
    console.log(members[x][1]);
    console.log(members[x][2]);
}

Also, can you explain how

data.push(elem["p_age"], elem["p_name"], elem["p_date"]);

is working for you? It seems like it should return undefined.

To clarify things, I recreated your code in a fiddle. Take a look here.

Answer 3

Answer №2

Attempt

let team=[];
$.each(data, function(index, element) {
    team.push([element["date"],element["name"],element["date"]]);
});
console.log(JSON.stringify(team))

Answer 4

Attempt

let team=[];
$.each(data, function(index, element) {
    team.push([element["date"],element["name"],element["date"]]);
});
console.log(JSON.stringify(team))

Answer 5

Answer №3

This seems questionable:

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

It appears as though you are attempting to loop through data, adding the elements back to data. Perhaps the $.each() function should be used to iterate over a different object.

I am also uncertain why you are duplicating elem['p_date'] in the array.

Answer 6

This seems questionable:

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

It appears as though you are attempting to loop through data, adding the elements back to data. Perhaps the $.each() function should be used to iterate over a different object.

I am also uncertain why you are duplicating elem['p_date'] in the array.

Answer 7

Answer №4

When working with these entities, remember to treat them as objects rather than simple data types. You can achieve this by utilizing the toString() method.

Answer 8

When working with these entities, remember to treat them as objects rather than simple data types. You can achieve this by utilizing the toString() method.

Answer 9

Answer №5

Hey, why not try using x instead of i in your loop?

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

This change should make it work perfectly!

Answer 10

Hey, why not try using x instead of i in your loop?

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

This change should make it work perfectly!

Answer 11

Answer №6

Do not

let info = new array()

Instead, use

let info = new Array()

The correct class name for arrays is Array, not 'array'.

Answer 12

Do not

let info = new array()

Instead, use

let info = new Array()

The correct class name for arrays is Array, not 'array'.

Utilize the push method to form a new array

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

Answer №6

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