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Submitted by Ahmad Aiman
over 8 years

Ques : 26 Change Variable Value

I cannot solve this part. it says :

Oops, try again. It looks like you didn’t log your whole name to the console.

mine :

// On line 2, declare a variable myName and give it your name. var myName = ‘David’; // On line 4, use console.log to print out the myName variable. (console.log = myName); // On line 7, change the value of myName to be just the first 2 // letters of your name. myName = “David’s son”; // On line 9, use console.log to print out the myName variable. { (console.log = myName); }

can anyone help me?

Answer 55b867b8d3292f38f700086d

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// On line 2, declare a variable myName and give it your name. var myName = “David”; // On line 4, use console.log to print out the myName variable. console.log(myName); // On line 7, change the value of myName to be just the first 2 // letters of your name. myName = myName.substring(0,2); // On line 9, use console.log to print out the myName variable. console.log(myName);

points
over 8 years

Answer 55bdcd5ce39efe7e5a000445

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Did the same thing, the quotations around myName make it a string and not the variable.

points
Submitted by JustinET19
over 8 years

Answer 55fe37dad3292fa8c30003d9

-1 votes

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// On line 2, declare a variable myName and give it your name. var myName = “Ivan”;
// On line 4, use console.log to print out the myName variable. console.log (myName); // On line 7, change the value of myName to be just the first 2 // letters of your name. myName=”Ivan”.substring(0,2); // On line 9, use console.log to print out the myName variable. console.log (myName);

points
Submitted by signatural
over 8 years

1 comments

While this works it’s actually bad code. Once the variable has been declared and a value given to it you shouldn’t retype the value anywhere else in the code. Always manipulate the variable. myName=”Ivan”.substring(0,2); should be myName=myName.substring(0,2);