wildteen88 Posted November 23, 2007 Share Posted November 23, 2007 Because variables act differently when used within functions. If you create a variable outside of a function, then that function will not be able to access variables outside of it and vice versa. In order to use a variable within a function that has been created outside of it you have two options: - Either define the variable as global within the function - Or pass that variable as a parameter to the function In order to understand this more you'll need to read up on variable scope Quote Link to comment https://forums.phpfreaks.com/topic/78440-array-error/page/2/#findComment-397632 Share on other sites More sharing options...
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.