How to use javascript in PHP class functions?

Question

As you see I have class whose name is table; and I have many functions on it "mainface" is one of them; these function is like my interface there are some check box and combobox on it; I want to write some JavaScript on "All" checkbox -> onclick=" ".
I want that when the user click on the "All" checkbox and option will disappear on the screen but I cannot find a way to do it.
Is there any way to do it ? Can I add call the JavaScript function to onclick?

<script type="text/javascript" >
function hide(){
            if(document.form.All.checked){
                 document.getElementById('option').style.visibility="hidden";
            }
}
</script>

</head>
<body> 

<?php

class table{

   public function   __construct() {
       echo table:: mainface();              
   }

   public function mainface()
   {
      $arayuz=" ";
      $arayuz.="<form name=form action=index.php method=post>";
      $arayuz.="<fieldset>";

    $arayuz.="<input type=checkbox name=All value=All onclick='SOME JAVA SCRIPT FUNCTION'>;

}
?>

Although this may look like a duplicate of stackoverflow.com/questions/6545167/…, it's not, the older question was edited after it got closed. — user456814
– user456814, Commented Jul 1, 2011 at 8:51
Yes it is duplicate my older one is seems to very confuse ;now ı make clear my questions and make commmnet on it tobe understandable — bluedoor
– bluedoor, Commented Jul 1, 2011 at 8:55
LAmureTJ: Write your HTML and Javascript first: this is what the browser sees. Then, write your PHP around it: this is pre-processed and helps you to automatically generate your HTML and Javascript; at this stage, do not touch your HTML or Javascript. — Lightness Races in Orbit
– Lightness Races in Orbit, Commented Jul 1, 2011 at 9:30

Daniel Gruszczyk · Accepted Answer · 2011-07-01 09:18:02Z

3

php file:

public function mainface()
{
    $output = "<script type=\"text/javascript\" src=\"/path_to_your_js/table.js\">-</script>";
    $output.= "<form name =\"form\" action=\"index.php\" method=\"POST\">";
    $output.= "<fieldset>";
    $output.= "<input type=\"checkbox\" name=\"ALL\" value=\"ALL\" onclick=\"checkAll(this)\">";
}

And then in table.js:

function checkAll( myCheckBox )
{
    //do some stuff you need here
    myCheckBox.style.display = "none";
}

I hope that helps...

answered Jul 1, 2011 at 9:18

Daniel Gruszczyk

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Comments

qbert220 · Accepted Answer · 2011-07-01 09:01:00Z

You are mixing server side and client side script languages. PHP runs server side (on the web server), javascript runs client side (within the web browser). Onclick is javascript, so runs client side. You cannot call PHP functions directly from an Onclick action.

http://en.wikipedia.org/wiki/Client-side_scripting

http://en.wikipedia.org/wiki/Server-side

To achieve what you want you don't need any PHP scripting. It can all be done in Javascript. You'll need to add an ID to your form like this:

$arayuz.="<form name='form' id='myform' action='index.php' method='post'>";

Then use the following as the onclick action:

$arayuz.="<input type=checkbox name=All value=All onclick='document.getElementById(\"myform\").style.visibility = \"hidden\";'>"

Henry · Accepted Answer · 2011-07-01 09:12:21Z

0

That would work but printing out in-line JavaScript like that is hideous, I would really recommend you look into unobtrusive JavaScript and use something like jQuery. Jquery example that would work for all checkbox's:

$(":checkbox").live('click' , function () { $("#IdOfForm").hide(); });

answered Jul 1, 2011 at 9:12

Henry

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Comments

jeni · Accepted Answer · 2011-07-01 09:47:11Z

0

form id is used in onclick

  $arayuz.="<form name='form' id='form1' action='index.php' method='post'>";

this will be your onclick

  $arayuz.="<input type=checkbox name=All value=All onclick='document.getElementById(\"form1\").style.visibility = \"hidden\";'>"

answered Jul 1, 2011 at 9:47

jeni

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