Variables that will be expanded in double quoted strings are not parsed as variables #34
Description
http://php.net/manual/en/language.types.string.php#language.types.string.parsing
Program:
<?php
$juice = "apple";
echo "He drank some $juice juice.".PHP_EOL;
// Invalid. "s" is a valid character for a variable name, but the variable is $juice.
echo "He drank some juice made of $juices.";
// Valid. Explicitly specify the end of the variable name by enclosing it in braces:
echo "He drank some juice made of ${juice}s.";
?>Output:
He drank some apple juice.
He drank some juice made of .
He drank some juice made of apples.
Parser:
All of these variables are parsed as string:
text.html.php
program
echo_statement
string
Version:
0.13.0