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0 points
Submitted by Daniel Taggeselle
about 11 years

little problem

Hi folks,

i need a little help. Maybe its not a big thing but here is my code

Answer 5156ae4c5ad91283c3001517

2 votes

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points
Submitted by Rick
almost 11 years

2 comments

Rick almost 11 years

i Promotion by it

Rick almost 11 years

maybe it can help you

Answer 5151e21ef51a82cf640002e4

1 vote

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You forgot a semicolon after defining your variables and $start needs to be defined before $time.

points
Submitted by Roni
about 11 years

2 comments

Ari Ogoke almost 11 years

Thanks Aaron y!

I see another problem… What is the value of the $time variable after the first assignment statement? In other words, what was the value of the $start variable before it was assigned the value of 20 in the second statement.

Roni almost 11 years

If a variable isn’t defined, it’s NULL and it throws a “Notice: Undefined variable.” You can use isset($variable) to check if a variable is defined. If the $start would be defined before $time, both of them would be “20”, sorry if this doesn’t help, didn’t quite understand your question ;D