Closed zeroSal closed 5 months ago
Hi, are you sure that double $ is an error? The problem in your code is the scope of the variables.
Hi, this is not error at ALL! It's a language syntax, read PHP doc: PHP.net: Variable variables
According to PHP.net:
Sometimes it is convenient to be able to have variable variable names. That is, a variable name which can be set and used dynamically. A normal variable is set with a statement such as:
<?php
$a = 'hello';
?>
A variable variable takes the value of a variable and treats that as the name of a variable. In the above example, hello, can be used as the name of a variable by using two dollar signs. i.e.
<?php
$$a = 'world';
?>
At this point two variables have been defined and stored in the PHP symbol tree: $a with contents "hello" and $hello with contents "world". Therefore, this statement:
<?php
echo "$a {$$a}";
?>
produces the exact same output as:
<?php
echo "$a $hello";
?>
This is not an error. Thanks @hosni and @Xring-git for explaining me.
Describe the bug Intelephese does not recognize as error a variable with double
$
even if it is a function parameter.To Reproduce
Expected behavior Intelephese marks as error the line where the variable is references with double
$
Screenshots
Platform and version