how can i check if an element is visible in php?

Question

i have a couple of variables that i want to save or not save to a database depending on whether or not i have set them as hidden/visible in a javascript function, i have searched for a while but cant find anything. Thanks!

score 5 · Accepted Answer · answered Jul 11 '12 at 19:38

5

That's absolutely impossible since PHP runs on the server while JavaScript and CSS are client-side.

The only thing you can do is checking if the element is visible using JavaScript and sending that data to your PHP script, e.g. via a hidden <input> field.

answered Jul 11 '12 at 19:38

ThiefMaster

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is there any code that i can use to completely remove an element in the javascript based on its id? – user1445218 Jul 11 '12 at 19:47
Yes, with `elem` being the element: `elem.parentNode.removeChild(elem)` – ThiefMaster Jul 11 '12 at 19:51
for some reason this doesnt work, i saw the exact same code that you posted somewhere else and already tried it – user1445218 Jul 11 '12 at 20:01

score 3 · Answer 2 · edited Dec 28 '18 at 21:03

You can also have this alternative. NOTE that the page has to refresh before this take effect (i.e. before PHP has knowledge of what was going on)

//Your JavaScript
function setHidden()
{

   document.getElementById('elementForVar1').visibility = 'hidden';

   //use this to indicate field/variable is hidden. PHP will use this later
   document.getElementById('elementForVar1HiddenField').value = 1;
}

declare hidden fields in your form to store the states of the variables

<form name="xxx">
    <input type="hidden" id="elementForVar1HiddenField" name="elementForVar1HiddenField" value="0" />
</form>

Your javascript simply set the value of the hidden field to 1 indicating var1 is hidden

<?php
  if($_POST['elementForVar1HiddenField'] == 1)
     //variable was hidden

?>

how can i check if an element is visible in php?

2 Answers2