Question

89

How to increment a value inside an object variable

rated 0 times [ 95] [ 6] / answers: 1 / hits: 22927 / 10 Years ago, tue, september 16, 2014, 12:00:00

I have the following.

var dataset = {val1 : 0, val2 : 0, val3 : 0};

var person = [];

var totalPeople = 10;



for(var i = 0; i <totalPeople; i++) {

   person[i] = dataset; 

}

Why i chose this approach, click here
.

I'm trying to make one of the values auto increment inside another for loop.

I've tried the following approaches to no avail.

person[1]{val1 : 0,

          val2 : 0,

          val3 : val3 + 1};



person[1]{val1 : 0,

          val2 : 0,

          val3 : person[1].val3 + 1};



person[1].val3 = person[1].val3 + 1;

any ideas?

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ammonderekm

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1 Year ago

answered 10 Years ago quentinaveryb · Accepted Answer

Totally sorry. The solution that you're referring to here was posted by me and is incorrect. I just updated my answer in that post.

Don't use the array initialization style that I originally posted:

var dataset = {val1 : 0, val2 : 0, val3 : 0};

var person = [];

var totalPeople = 10;



for(var i = 0; i < totalPeople; i++) {

  person[i] = dataset; // this assigns the *same* object reference to every 

                       // member of the person array.



}

This is the correct way to initialize your person array:

var person = [];

var totalPeople = 10;



for(var i = 0; i < totalPeople; i++) {

  person[i] = {val1 : 0, val2 : 0, val3 : 0}; // do this to create a *unique* object 

                                              // for every person array element

}

If you use the correct array initializtion shown directly above, then you can increment val3 like this with each loop iteration:

var person = [];

var totalPeople = 10;



for(var i = 0; i < totalPeople; i++) {

  person[i] = {val1 : 0, val2 : 0, val3 : 0};

  person[i]['val3'] = i;

}

Sorry again for the bad information that I provided in the other post. (All other info is correct. Just the array initialization code was bad.) I hope this updated information helps.