11.creating object



var meObj = new Object();
meObj["name"] ="ashish";
meObj["age"] =25;

Only we are able to see age 25 not the name

Replace this line with your code.


I am sure on of the mods can give a better answer, but there is no reason to worry.
You don't have a console.log method in your code and therefore nothing should get printed.
But on the "backend" there is probably a fail-safe method that console.log the objects age prop value.


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