Just trying to declare a functino: why am I getting "Expected: variable Unexpected: variable" warning?

 

Hi,

I'm on a PHP 7.4 configuration.

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1 comment

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...

private public loadData($a)

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.

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