CODEWITHSUNDEEP
javascriptunderscore.js
Sai
Sai

Reputation: 113

Perform omit on the original object

I am very new to underscore js, I am trying to omit a certain property on an Object. What I did was

myObj = _.omit(myObj,name)
console.log(myObj);

Still the myObj seems to have the property name. Although if I do this it seemes to work

newMyObj= _.omit(myObj,name) 
console.log (newMyObj)

it seemed to work fine. What am I doing wrong, can someone help? Ok, so myObj looks like this

Angola: "4.134137685",Brunei: "2.532726835",Countries: "2004",Croatia: "1.717672961", keys: Array[11]

I am trying to omit "keys" which again is an array of objects

Thanks

Upvotes: 0

Views: 180

Answers (2)

stavarotti
stavarotti

Reputation: 582

I interpreted this question to mean you simply want to remove a property from an object using omit(). Note that omit() returns a copy of the object sans the specified property to remove. The method does not alter the object in place.

Given this premise, the code below, which matches what you have, works just fine:

var copy,
    obj = {
        Angola: "4.134137685",
        Brunei: "2.532726835",
        Countries: "2004",
        Croatia: "1.717672961", 
        keys: Array[11]
    },
    check = function (o) {
        _.each(o, function (value, key) {
            console.log("key: " + key + " value: " + value);
        });
    };


copy = _.omit(obj, "keys");
check(copy);

obj = _.omit(obj, "keys");
check(obj);

You will get the same result whether you are using a new variable or the existing one.

Upvotes: 0

user663031
user663031

Reputation:

There are these things called "debuggers". If you don't know what they are, then stop everything you're doing and learn about them now. Search Google for "Chrome devtools", for instance. Stop your code (put a breakpoint) at the point before the call to _.omit. In the console, type in myObj to see exactly what it contains, then also name. Or, you could use the 'Scope Variables" section of devtools to check the value of these variables. Now, make a single step (F10). See if or how the variables have changed, or type myObj again into the console to check its value.

In your particular case, you report that the deletion of the property occurs properly when you do 

newMyObj= _.omit(myObj,name)

but not with

myObj= _.omit(myObj,name)

In and of itself, that behavior is completely unexplainable. So there's something else going on that you're not telling us about. My guess is that you are doing something like this:

myObj = { keys: [] };
name = "keys";
delete_property();
console.log(myObj.keys); // []

function delete_property(myObj) {
    myObj = _.omit (myObj, name);
}

However, this does not do what you might think. The assignment to myObj within the function does nothing; it just reassigns the value of the function argument. It has no effect on the myObj outside the function.

To be sure, we'd need to see more of your actual code, but this is just a regular old debugging problem of the sort you will encounter thousands of times in your programming career, so you're better off learning to solve it yourself.

Upvotes: 2

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