Give variable name but not its value to a function

Question

To remove tedious code i would like to create a function to simplify it.
So to change the settings that i stored in an object, I created this.

var settings = {
 color : 'red'
 //one of many settings
};
alert(settings.color); // returns red
function changeValue(property, value){
 settings.property = value;
}
changeValue('color', 'blue');
alert(settings.color); // still return red

it seems like its treated as a string instead.
And if i try this:

function changeValue(object, value){
 object = value;
}
changeValue(settings.color, 'blue');

it would only give the function the value which is 'red'.
and this:

changeValue('settings.color', 'blue');

would not work obviously.

So how do i solve this ? How do i pass down a variable to a function but not its value so it can be altered ?

Note: the function is simplified for now, but would contain a lot more magic when it works.

Answer 1

Pass the object and the name of the property that you want to change as separate arguments:

function changeValue(obj, prop, value) {
 obj[prop] = value;
}
changeValue(settings, "color", "blue");

Assuming var prop = "color" then obj[prop] is equivalent to obj.color. The first notation allows you to dynamically specify the property that should be changed.

• Einar Löve: Cheers. New to javascript so i had no idea of this method.

Answer 2

when you do settings.color you get a value, rather than object. You'd have to do something like

function changeValue(object, field, value){
 object[field] = value;
}

used like

changeValue(settings, 'color', 'blue')

which may not actually reduce time for you at all..

---

EDIT:: I have made this function for you. But please.. Don't ever use it.

var settings = {
 color : "blue"
}
function setField(str, value){
 var args = str.split(".").reverse();
 var obj = window[args.pop()];
 while(args.length - 1 > 0) 
 obj = obj[args.pop()]
 obj[args.pop()] = value;
}
alert(settings.color);
setField("settings.color", "red");
alert(settings.color);

• Einar Löve: since i wanted to "remove tedious code", this would definitely go against it.
• AlanFoster: @EinarLöve 'tedious code'? My 'solution' does exactly what you want. setField("settings.color", "red") works as you wanted. I also gave the 'correct' way of doing it. Although your original requirement is wrong, and both solutions should not be used (including lwburk's code (which is identical to mine))
• Einar Löve: You first function is definitely just what im looking for. my previous comment was a respond to your edit.