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Steven Cousineau
2,563 Pointsif statement will not execute!
Hello everyone, I'm kinda confused at the moment why my code will not execute the if statement, I've tryed the code in another editor and the same result. I have it logging the randomNumber in the console to make sure i can get the right answer to test as you can see. I would appreciate any input and I thank you!
_My workspace code here->https://w.trhou.se/9pmqp7wbwg
2 Answers
Michael Lambert
6,286 PointsHia,
It seems that the random number can't return a number to a variable and print a number to the console in one statement. The variable randomNumber never receives the random number which is why the 'if' part isn't working. Don't know how you could achieve both at the same time in this manner.
Steven Cousineau
2,563 PointsThank you for your reply. You both are right it seems as i rewrote the whole code wiithout the console.log and it worked. Which is weird because the first time I tried executing the original code didn't work, which is why I implemented the console.log to test the code more efficently. That was the idea anyways, regardless thank for your time guys!
Gavin Sevareid
3,121 PointsHello, Steven.
You're problem seems to be: var randomNumber = console.log(Math.floor(Math.random() * 6) + 1); The correct way is to just make it this: (Math.floor(Math.random() * 6) + 1);
Gavin Sevareid
3,121 PointsGavin Sevareid
3,121 PointsHello, Steven.
You're problem seems to be: var randomNumber = console.log(Math.floor(Math.random() * 6) + 1); The correct way is to just make it this: (Math.floor(Math.random() * 6) + 1);