Show div if PHP form is successfull

[soder]

Heya, I got this code

Code: Select all

function initShowHideDivs()
{
   var divs = document.getElementsByTagName('DIV');
   var divCounter = 1;
   for(var no=0;no<divs.length;no++){
      if(divs[no].className=='dhtmlgoodies_question'){
         divs[no].onclick = showHideContent;
         divs[no].id = 'dhtmlgoodies_q'+divCounter;
         var answer = divs[no].nextSibling;
         while(answer && answer.tagName!='DIV'){
            answer = answer.nextSibling;
         }
         answer.id = 'dhtmlgoodies_a'+divCounter;
         contentDiv = answer.getElementsByTagName('DIV')[0];
         contentDiv.style.top = 0 - contentDiv.offsetHeight + 'px';
         contentDiv.className='dhtmlgoodies_answer_content';
         contentDiv.id = 'dhtmlgoodies_ac' + divCounter;
         answer.style.display='none';
         answer.style.height='1px';
         divCounter++;
      }
   }
}
window.onload = initShowHideDivs;


This is to hide the div when the page loads.

and I got this PHP form:

Code: Select all

if(strlen($_POST['cat']) > 2)
{
  $sql="INSERT INTO categories (username, name) VALUES ('$name','$_POST[cat]')";
  mysql_query("$sql");
  echo "Category ".$_POST[cat]." has been created.<br>";
?>
<script type="text/javascript">
  initShowHideDivs();
</script>
<?php
}


How do I make the div show when the form loads successfully?

Regards,
Soder.

[viper94x]

to hide it on start : css-> visibility: hidden;
To show it, use scriptaculous or jQuery.
Thats by far the easiest and best way.

Sam

[JordanMRichards]

Personally I'd use PHP to handle this.

check to see if the form has been submitted then echo the div out.. like this.

Code: Select all

//cat is the form data past through, change cat to the name of the element that you want to check if it's been submitted..

if(isset($_POST['cat'])) {
    echo "
//Your div here
<div>
<p>
SUCCESS! ;)
</p>
</div>

";
}



Hope this helped

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14 Year Old Programmer & Graphic Artist, Confident and Courageous