[ACCEPTED]-How to pass parameters to PHP template rendered with 'include'?-include

Accepted answer
Score: 34

Consider including a PHP file as if you 5 were copy-pasting the code from the include 4 into the position where the include-statement 3 stands. This means that you inherit the 2 current scope.

So, in your case, $param is already 1 available in the given template.

Score: 25

$param should be already available inside 12 the template. When you include() a file 11 it should have the same scope as where it 10 was included.

from http://php.net/manual/en/function.include.php

When a file is included, the 9 code it contains inherits the variable 8 scope of the line on which the include occurs. Any 7 variables available at that line in the 6 calling file will be available within 5 the called file, from that point forward. However, all functions 4 and classes defined in the included file 3 have the global scope.

You could also do 2 something like:

print render("/templates/blog_entry.php", array('row'=>$row));

function render($template, $param){
   //extract everything in param into the current scope
   extract($param, EXTR_SKIP);

Then $row would be available, but 1 still called $row.

Score: 2

I use the following helper functions when 5 I work on simple websites:

function function_get_output($fn)
  $args = func_get_args();unset($args[0]);
  call_user_func_array($fn, $args);
  $output = ob_get_contents();
  return $output;

function display($template, $params = array())
  include $template;

function render($template, $params = array())
  return function_get_output('display', $template, $params);

display will output 4 the template to the screen directly. render 3 will return it as a string. It makes use 2 of ob_get_contents to return the printed 1 output of a function.

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