Just trying to declare a functino: why am I getting "Expected: variable Unexpected: variable" warning? Tirengarfio Created February 06, 2022 20:53 Hi, I'm on a PHP 7.4 configuration.
Hi there,
1. So what are you trying to write there? A function?
2. You said "I'm on a PHP 7.4"... but your code fragment is a broken PHP...
What is that? If it's a function declaration then where is the "function" keyword? And why you have both "private" and "public" at the same time?..
Change "public" to "function" and check again.