CODEWITHSUNDEEP
javascriptarrays
user2693135
user2693135

Reputation: 1306

Array push behavior

Although I have some basic JavaScript background, I stumbled upon this code that I wrote:

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[email protected]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[email protected]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28}]
var tempArr=[];
var table=[];
var dataArr = Object.keys(data).map(function(k) { return data[k] });
dataArr.forEach(function(user) {
  tempArr[0]=user.fName;
  tempArr[1]=user.lName;
  tempArr[2]=user.email;
  tempArr[3]=user.age;
  table.push(tempArr);
  console.log('table'+JSON.stringify(table));
 });

In the final loop, I expected table to contain the arrays for Tom, Pat, and Sam . Instead, this is what I got:

table[["Tom","Moody","[email protected]",30]]
table[["Pat","Smith","[email protected]",32],["Pat","Smith","[email protected]",32]]
table[["Sam","Dawn","[email protected]",28],["Sam","Dawn","[email protected]",28],["Sam","Dawn","[email protected]",28]]
table[["Sam","Dawn","[email protected]",28],["Sam","Dawn","[email protected]",28],["Sam","Dawn","[email protected]",28],["Sam","Dawn","[email protected]",28]]

Why is push() replacing the previous entry in table? Any help will be highly appreciated.

Upvotes: 4

Views: 321

Answers (6)

str
str

Reputation: 44969

The others already pointed out problems in your code.

However, you also make things more complicated than necessary. You can just do this:

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[email protected]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[email protected]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28}];

var table = data.map(function(user) {
    return [
      user.fName,
      user.lName,
      user.email,
      user.age,
    ];
});
      
console.log(table);

Or if you use ES6:

var table = data.map(user => [ user.fName, user.lName, user.email, user.age ];

Upvotes: 6

Yosvel Quintero
Yosvel Quintero

Reputation: 19070

var data = [{"_id": "57b3e7ec9b209674f1459f36","fName": "Tom","lName": "Moody","email": "[email protected]","age": 30}, {"_id": "57b3e8079b209674f1459f37","fName": "Pat","lName": "Smith","email": "[email protected]","age": 32}, {"_id": "57b3e8209b209674f1459f38","fName": "Sam","lName": "Dawn","email": "[email protected]","age": 28}, {"_id": "57b3e8219b209674f1459f39","fName": "Sam","lName": "Dawn","email": "[email protected]","age": 28}],
    table = [];

data.forEach(function(user) {
    table.push([user.fName, user.lName, user.email, user.age]);
});

console.log(table);

Upvotes: 0

Redu
Redu

Reputation: 26161

Obviously map isthe way to go for the sake of functional approach however if you like imperative styles one simplistic way could be using for of loop as follows.

var data = [{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[email protected]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[email protected]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28}],
   table = [];
for (var user of data) table.push([user.fName,user.lName,user.email,user.age]);
console.log(table);

Upvotes: 1

Yury Tarabanko
Yury Tarabanko

Reputation: 45121

You don't need to write all the boilerplate code by hand. Use a proper array iterator (map in your case).

var table = data.map(function(user) {
   return [user.fName, user.lName, user.email, user.age];
});

Upvotes: 2

Gareth Parker
Gareth Parker

Reputation: 5062

Well, unlike a lot of other languages, JavaScript has passes everything by reference. That means that when you table.push(tempArr);, you're not actually pushing the values of tempArr, you're pushing a reference to tempArr. So, if you do this:

var a = 'a';

var table = [];
table.push(a);
a = 'b';
console.log(table[0]);

You'll get b as your output. What you want to do is define a new variable to push, like this

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[email protected]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[email protected]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28}]

var table=[];
var dataArr = Object.keys(data).map(function(k) { return data[k] });
dataArr.forEach(function(user) {
    var tempArr=[];
    tempArr[0]=user.fName;
    tempArr[1]=user.lName;
    tempArr[2]=user.email;
    tempArr[3]=user.age;
    table.push(tempArr);
 });

 console.log('table'+JSON.stringify(table));

Upvotes: 0

Himanshu Pandey
Himanshu Pandey

Reputation: 736

Problem here is not with push. Variable in javascript store reference to array. And in table you are pushing reference to same array tempArr. You need to create new array before pushing it or create deep copy of array before pushing it.

Example 

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[email protected]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[email protected]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[email protected]","age":28}]
var table=[];
var dataArr = Object.keys(data).map(function(k) { return data[k] });
dataArr.forEach(function(user) {

  var tempArr=[];
  tempArr[0]=user.fName;
  tempArr[1]=user.lName;
  tempArr[2]=user.email;
  tempArr[3]=user.age;
  table.push(tempArr);
  console.log('table'+JSON.stringify(table));
 });

Upvotes: 0

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