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Displaying mysql records on same div container?


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Hello all,

 

I have a layout with 2 sides(left side and right side). I get from mysql the menus name and content text for every menu. I want when I click o a menu on the left side div to display in the right side div the content text for that menu.

Example:

LEFT SIDE div      |      RIGHT SIDE div

______________|_____________

Left menu1          |      menu1 content text

Left menu2          |      content text menu1

 

 

any ideas?

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For each menu item create a DIV for the content to be displayed. Each div should have a name in the format: "content_1", "content_2" etc.

 

In the menu items have an onclick event which calls a function and passes the number for each div. The function can iterate through all of the divs and set the display property for each div based upon the one which was selected.

 

Example:

<html>
<head>

<script type="text/javascript">

function showContent(menuID)
{
    var index = 1;
    var divObj;
    while(document.getElementById('content_' + index))
    {
        divObj = document.getElementById('content_'+index);
        divObj.style.display = (index==menuID) ? 'block' : 'none';
        index++;
    }
    return;
}
</script>
</head>

<body>

<div id="menu" style="border:1px solid black;float:left;">
<a onclick="showContent('1');">Menu 1</a><br />
<a onclick="showContent('2');">Menu 2</a><br />
<a onclick="showContent('3');">Menu 3</a><br />
</div>

<div id="content_1" style="border:1px solid black;display:none;">
Content for menu item 1<br />
Content for menu item 1<br />
Content for menu item 1<br />
</div>
<div id="content_2" style="border:1px solid black;display:none;">
Content for menu item 2<br />
Content for menu item 2<br />
Content for menu item 2<br />
</div>
<div id="content_3" style="border:1px solid black;display:none;">
Content for menu item 3<br />
Content for menu item 3<br />
Content for menu item 3<br />
</div>

</body>
</html>

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Many thanks for the replay

I've tested the example you provided and it works with default values but not to the ones I have(my right div has custom position)

 

<script type="text/javascript">
function showContent(menuID)
{
    var index = 1;
    var divObj;
    while(document.getElementById('content_' + index))
    {
        divObj = document.getElementById('content_'+index);
        divObj.style.display = (index==menuID) ? 'block' : 'none';
        index++;
    }
    return;
}
</script>
<div id="menu" style="border:1px solid black;float:left;">
<a onclick="showContent('1');">Menu 1</a><br />
<a onclick="showContent('2');">Menu 2</a><br />
<a onclick="showContent('3');">Menu 3</a><br />
</div>


<div id="content_1" style="	width: 260px;height: 490px;position:relative;right: 20px;top: -475px;overflow: hidden;float:right;display:none;">
Content for menu item 1<br />
Content for menu item 1<br />
Content for menu item 1<br />
</div>

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Seriously?

 

You're giving the DIV a relative position of "top: -475"!!! That is 475 pixels above the viewable area of the browser window. The DIV has a height of 490 pixels. So, IF the DIV had content to completely fill the height the user would only see the bottom 15 pixels of it.

 

The code is working perfectly - you just can't see the div because you placed it outside the viewable area.

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No it does not display OK in IE & FF. If you remove the "display:none" from the DIV style properties you still won't see it. Because it is outside the page boundries - at least using the code you provided. The position is relative. So, if you place the code for that DIV where the "inline" position would be at least 475 pixels down from the top, then it will be visible. Are you purposefully setting that value to -475 for a reason or are you setting it through trial and error. If the latter you may need to reevaluate your placement/styles as there seems to be a problem.

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