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JavaScript

Joel K
Joel K
3,845 Points
Posted by Joel K
Joel K
3,845 Points

Question on Objects Section of Intro to Programming

I have the following code...

var me = {
    first_name: "Adam",
    last_name: "Smith",
    "Employee Number": 1
}

console.log(me.first_name);
console.log(me.last_name);
console.log(me["Employee Number"]);


var key = "first_name"
console.log(me[key]);

http://jsbin.com/opIDejoZ/1/edit

So i get the whole concept of the first block. The object me has properties and i can console.log those to display the key values. I'm just wondering about the last part with the var key="first name". How does the JS know that first_name is looking for the first key/value pair in the me object. Why does it not see "first_name" as a string and therefore perhaps different than the key first_name in the object above it?

I hope i was clear. This is probably a simple concept i'm missing.

Thanks.

Adam Sackfield
Adam Sackfield
Courses Plus Student 19,663 Points

In the final console.log you are passing it the variable object "me"

4 Answers

eirikvaa
eirikvaa
18,015 Points

Doing 

console.log(me.first_name);

is exactly the same as doing

console.log(me["first_name"]);

The latter is convenient when you have spaces in your key. You couldn't access the Employee Number like below. 

console.log(me."Employee Number");
James Barnett
James Barnett
39,199 Points
James Barnett
James Barnett
39,199 Points

What's going on there is JS is looking at the string first_name and then checking the object me and seeing if there is a key with the name first_name and returning it's value adam.

I hope I explained that clearly enough.

http://jsbin.com/opIDejoZ/1/edit

Ben Rubin
PLUS
Ben Rubin
Courses Plus Student 14,658 Points
Ben Rubin
Ben Rubin
Courses Plus Student 14,658 Points

It actually does see first_name as a string. key is a variable that you are setting to the string "first_name". That's what the double quotes mean. The line me[key] tells JS to go to the me object and get the property named first_name (since key contains the string "first_name").

You can do a test to see that. Try these two lines.

console.log(me["first_name"]);
console.log(me[key]);

You'll see that those both give the same result. Then try this line.

console.log(me[first_name]);

You see that that produces an error since first_name is neither a variable nor a string value.

Joel K
Joel K
3,845 Points
Joel K
Joel K
3,845 Points

Thanks everyone. Strange concept to wrap my head around but think i understand it.