Mouseover in PHP

[DSTR3]

I am trying to do a mouseover/mouseout on an image in a table. Right now I have it working with Javascript. You click on the camera and the image comes up in a new window. However; I want to do it with a mouseover like this.

 

http://www.menuhead.net/PopUP/mouseover-popup/over.php

 

I know mouseover is client side so how would I do this with php.

 

function newPopup(url) {
popupWindow = window.open(
 url,'popUpWindow','height=300,width=300,left=10,top=10,resizable=no,scrollbars=no,toolbar=no,menubar=no,location=no,directories=no,status=no'  )
}

 

<td><a href=Javascript:newPopup('$row[LocationPix]');><img src='http://www.menuhead.net/Images/Buttons/PShotClear.png'</td>

 

Thank you.

[Jessica]

I know mouseover is client side so how would I do this with php.

 

Since PHP is server side, and you know what you need to do is client side, what could you possibly be asking??

 

 

I think my favorite part is that you posted this in the MySQL help forum.

Edited February 19, 2013 by Jessica

[DSTR3]

Useless. You shouldn't let your dog answer the questions.

Edited February 19, 2013 by DSTR3

[DSTR3]

Closer, but still not the image.....

http://www.menuhead.net/PopUP/mouseover-popup/overthis.php

 

<td><a href ='#' id='trigger'('$row[LocationPix]');><img src='http://www.menuhead.net/Images/Buttons/PShotClear.png'</td>

 

$(function() {
 var moveLeft = 20;
 var moveDown = 10;
 $('a#trigger').hover(function(e) {
   $('div#pop-up').show();
 }, function() {
   $('div#pop-up').hide();
 });
 $('a#trigger').mousemove(function(e) {
   $("div#pop-up").css('top', e.pageY + moveDown).css('left', e.pageX + moveLeft);
 });
});

 

Now how to get the picture?

[Mikey]

The HTML you have posted is not valid unfortunately. For one you're not closing your IMG tag...

 

Unless you perhaps need something like this:

 

<td><a href='#' onclick='trigger("{$row[LocationPix]}");'><img src='http://www.menuhead.net/Images/Buttons/PShotClear.png' /></td>

 

Otherwise, as Jessica said I'm not entirely sure what you're asking.

[cyberRobot]

I know mouseover is client side so how would I do this with php.

 

 

To populate the anchor tag with a PHP variable, the variable needs to be enclosed in PHP tags:

 

<td><a href="#" onmouseover="newPopup('<?php print $row['LocationPix']; ?>');"><img src='http://www.menuhead.net/Images/Buttons/PShotClear.png' /></a></td>

 

Note that I moved the pop-up code to the "onmouseover" attribute.