How to define objects with a function in javascript?

Question

I have three identical objects that I'd like to define in three slightly different ways. The only difference is that I'm accessing a different property within my source data (via Bracket Notation) for each object as shown below

The object I'm using is not a function. For example Object1 is defined by

var Object1  = dc.lineChart("#chart-line-hitsperday"); 


var D = "Dimension1"
Object1
    .width(200).height(200)
    .dimension(SourceData.dimension(function (d) {return d[D];}))
    .group(SourceData.dimension(function (d) {return d[D];}).group().reduceSum(function (d) {return d.Amount;}))


var D = "Dimension2"
Object2
    .width(200).height(200)
    .dimension(SourceData.dimension(function (d) {return d[D];}))
    .group(SourceData.dimension(function (d) {return d[D];}).group().reduceSum(function (d) {return d.Amount;}))


var D = "Dimension3"
Object3
    .width(200).height(200)
    .dimension(SourceData.dimension(function (d) {return d[D];}))
    .group(SourceData.dimension(function (d) {return d[D];}).group().reduceSum(function (d) {return d.Amount;}))

Is there a way that I can define each with 1 line of code as oppose to three lines of code. In other words, I'd like to convert above into the following

Object1.definition("Dimension1")
Object2.definition("Dimension2")
Object3.definition("Dimension3")

by defining 'definition' as something like :

definition(D) =         
    .width(200).height(200)
    .dimension(SourceData.dimension(function (d) {return d[D];}))
    .group(SourceData.dimension(function (d) {return d[D];}).group().reduceSum(function (d) {return d.Amount;}))

Is this possible?

See JSFiddle Here: http://jsfiddle.net/chrisguzman/juhaoem2/

I've tried the following with no luck:

var definition = function(D){     
    this.width(200).height(200)
    .dimension(SourceData.dimension(function (d) {return d[D];}))
    .group(SourceData.dimension(function (d) {return d[D];}).group().reduceSum(function (d) {return d.Amount;}))
}


Object1.definition("Dimension1")

Second Try

Object1.prototype.definition = function(dim){
    this.width(200).height(200)
        .dimension(SourceData.dimension(function (d) {return d[dim];}))
        .group(SourceData.dimension(function (d) {return d[dim];}).group().reduceSum(function (d) {return d.Amount;}))
};

Your code as posted won't work anyway because of the way you re-use "D". — Pointy
– Pointy, Commented Aug 16, 2014 at 14:10
Right, that's kind of the point. This is the code that I wish could work, but doesnt work. But do you get what I'm trying to do? Is there a way to re-use D so that it works? — Chris
– Chris, Commented Aug 16, 2014 at 14:11
No, your second try had some mistakes, please see my updated answer — frogatto
– frogatto, Commented Aug 16, 2014 at 14:27
Code after I've tried the following with no luck would work if you call it the following way: Object1.definition.call(Object1,"Dimension1"); as in Vinz243's answer, I've updated it with how to use it. — HMR
– HMR, Commented Aug 16, 2014 at 17:16

frogatto · Accepted Answer · 2014-08-16 14:28:15Z

2

Try this:

SuperObject.prototype.definition = function(dim){
    this.width(200).height(200)
        .dimension(SourceData.dimension(function (d) {return d[dim];}))
        .group(SourceData.dimension(function (d) {return d[dim];}).group().reduceSum(function (d) {return d.Amount;}))
};

Also you should define the following functions as well:

SuperObject.prototype.width()
SuperObject.prototype.height()
SuperObject.prototype.dimenstion()
SuperObject.prototype.group()

Or put them in the SuperObject prototype chain.

Note that SuperObject is the constructor of Object1, Object2,...

UPDATE #1

Usage:

var obj = new SuperObject();
obj.definition("Dimension1");

edited Aug 16, 2014 at 14:28

answered Aug 16, 2014 at 14:14

frogatto

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8 Comments

Pointy Over a year ago

This will only make sense if "Object1" is a function, and there's no indication that that's the case in the OP.

Rudie Over a year ago

I am prototypr! Resistence is futile!

Chris Over a year ago

The object I'm using is not a function, so prototype doesnt seem to be working for me. var Object1 = dc.lineChart("#chart-line-hitsperday");

frogatto Over a year ago

@Chris No, prototype is available for all JavaScript objects, your objects is not going to be SuperObject. rather that SuperObject is a factory/constructor of your objects. type of your objects and what they should to do is specified by you in the SuperObject definiftion

frogatto Over a year ago

@Chris for clarification, could you please create a jsfiddle.net?

|

HMR · Accepted Answer · 2014-08-16 17:21:37Z

1

Using this?

definition = function(D){     
    this.width(200).height(200)
    .dimension(SourceData.dimension(function (d) {return d[D];}))
    .group(SourceData.dimension(function (d) {return d[D];}).group().reduceSum(function (d) {return d.Amount;}))
}

Use:

definition.call(Object1,"Dimension1");

edited Aug 16, 2014 at 17:21

HMR

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answered Aug 16, 2014 at 14:12

Vinz243

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Comments

HMR · Accepted Answer · 2014-08-16 16:53:03Z

The code as you posted is highly unreadable so I'll try with a simpler example:

var o1={},o2={},o3={}
,dimensions = ['Dimension1','Dimension2','Dimension3']
,objects=[o1,o2,o3]
,len=dimensions.length,i=-1;
while(++i<len){
  //function returning function creating a closure
  //  in a while loop
  objects[i].somefn=(function(d){
    return function(){
      console.log(d);
    }
  }(dimensions[i]));
  objects[i].sameForAll='anything';
}
o1.somefn();
o2.somefn();
o3.somefn();

When a variable name starts with a capital it usually means it's a constructor so I'd advice not to use capitals unless they are all caps to indicate they are constants.

[update]

Here is what it looks like applied to your code:

//wrap in IIFE
(function(dimensions,objects){
  var i=-1,len=dimensions.length;
  //for every object and dimension
  while(++i<len){
    //wrap in IIFE for closure
    (function(object,D){//see later what object and D is
      object
        .width(200).height(200)
        .dimension(
          SourceData.dimension(
            function (d) {
              return d[D];
            }
          )
        )
        .group(
          SourceData.dimension(
            function (d) {
              return d[D];
            }
          )
          .group()
          .reduceSum(
            function (d) {
              return d.Amount;
            }
          )
        );
    }(objects[i],dimensions[i]));//pass object and D
  }
}(['Dimension1','Dimension2','Dimension3'],[Object1,Object2,Object3]))

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