Array output behavior

Question

Although I have a basic JavaScript background, I came across this code that I wrote:

var data=[{"_id":"57b3e7ec9b209674f1459f36","fName":"Tom","lName":"Moody","email":"[e-mail]","age":30},{"_id":"57b3e8079b209674f1459f37","fName":"Pat","lName":"Smith","email":"[e-mail]","age":32},{"_id":"57b3e8209b209674f1459f38","fName":"Sam","lName":"Dawn","email":"[e-mail]","age":28},{"_id":"57b3e8219b209674f1459f39","fName":"Sam","lName":"Dawn","email":"[e-mail]","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 last loop, I expected the table to contain arrays for Tom, Pat, and Sam. Instead, I got:

table[["Tom","Moody","[e-mail]",30]]
table[["Pat","Smith","[e-mail]",32],["Pat","Smith","[e-mail]",32]]
table[["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28]]
table[["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28],["Sam","Dawn","[e-mail]",28]]

Why does push () replace the previous record in the table? Any help would be greatly appreciated.

Answer 1

Others have already indicated problems in your code.

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

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

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

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Or if you are using ES6:

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

Answer 2

You do not need to write all the boilerplate code manually. Use the correct array iterator ( mapin your case).

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

Answer 3

, - , , for of .

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

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Answer 4

. javascript . tempArr. , , .

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

Answer 5

, , JavaScript . , table.push(tempArr);, tempArr, tempArr. , :

var a = 'a';

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

b. , , , ,

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

Answer 6

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

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

console.log(table);

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