php function parameter is not retaining full string passed in - php

Question

I have a PHP function that in echoing out a string, is losing some of the characters passed in it. Any reason why this would be happening?
I pass: $vhkvdov#jqlydk#p*_L#1qrlws|ufqh#KWLZ#1hwdgsX
It returns: #jqlydk#p*_L#1qrlws|ufqh#KWLZ#1hwdgsX
This is my PHP code:
<?php
function display($str) {
echo $str;
}
display("$vhkvdov#jqlydk#p*_L#1qrlws|ufqh#KWLZ#1hwdgsX");
?>

Answer 1

If you have $ in you string use single quotes:
display('$vhkvdov#jqlydk#p*_L#1qrlws|ufqh#KWLZ#1hwdgsX');
Otherwise PHP would try to use $vhkvdov as a variable name and replace it by its contents, what will be nothing as I expect that the variable is not set
PHP reference on strings
Note: During development I would set error_reporting in php.ini to
error_reporting=E_ALL | E_NOTICE;
and
display_errors=On
or
log_errors=On
error_log=PATH/TO/YOUR/LOG.file ; make sure this is writable by the web server
To see an error message like :
PHP Notice: Undefined variable: vhkvdov in

Answer 2

The problem is that you are using double quotes, which make PHP interpret the inner string for PHP variables.
Consider the following example:
$friend = "foo";
echo "hello $friend";
// Output: hello foo
Use single quotes instead if you do not want PHP to interpret the string contents.
$friend = "foo";
echo 'hello $friend';
// Output: hello $friend

Answer 3

Take a read of this to help clarify why this is happening:
http://php.net/manual/en/language.types.string.php
Specifically:
When a string is specified in double quotes or with heredoc, variables
are parsed within it.
This is whats happening with your code:
If a dollar sign ($) is encountered, the parser will greedily take as
many tokens as possible to form a valid variable name. Enclose the
variable name in curly braces to explicitly specify the end of the
name.
Here is an example:
<?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.";
?>
The above example will output:
He drank some apple juice.
He drank some juice made of .
When you use single quotes, you will avoid this issue as the string taken litterally:
// Outputs: This will not expand: \n a newline
echo 'This will not expand: \n a newline';
// Outputs: Variables do not $expand $either
echo 'Variables do not $expand $either';