如何在不使用eval（）的情况下评估PHP表达式

Question

I want to display a variable or an array by passing the variable name to a function and display it without having to use the dangerous eval() function. I am having trouble making it happen. Is it possible to do?

Here's the code:

show( '$_SESSION', 'a' ); // it does not work
show( '_SESSION', 'a' ); // it does not work

function show( $showWhat = null, $showType = null ) {
  echo '<pre>';
  if( strtolower( $showType ) == 'a' ) { // 'a' represents array()'s
    print_r( '$' . $showWhat ); // it does not work
    print_r( $showWhat ); // it does not work
  }
  else { // 'v' represents variables
    echo $showWhat;
  }
  echo '</pre>';
  exit;
}

Answer 1

After a lot of trials and tests I got it to work this way:

show( $_SESSION );

/**
  * halts processing and displays arrays & variables
  */
  function show( $showWhat = null ) {

    echo '<pre>';

    if( is_array( $showWhat ) ) {
      echo 'Array: ';
      print_r( $showWhat );
    }
    else {
      echo 'Variable: ';
      echo $showWhat;
    }

    echo '</pre>';
    exit;

  } /* show() */