----- Original Message -----
From: "Jay Blanchard" <jblanchard@xxxxxxxxxx>
To: "Gustav Wiberg" <gustav@xxxxxxxxxxxxxx>; "PHP General"
<php-general@xxxxxxxxxxxxx>
Sent: Tuesday, March 28, 2006 10:00 PM
Subject: RE: Div-element at same vert. position?
[snip]
I have a code-snippet here:
while ($dbArray = mysql_fetch_array($querys)) {
$dbIDLevel1 = $dbArray["IDLevel1"];
$dbLevel1Name = $dbArray["level1Name"];
$dbFactsLevel1Name = $dbArray["factsLevel1Name"];
$dbFactsPictureLevel1Name = $dbArray["factsPictureLevel1Name"];
?>
<div id="factsmenu_<?php echo $dbIDLevel1;?>" name="factsmenu_<?php echo
$dbIDLevel1;?>" style="position:absolute; width:250px; height:300px;
z-index:<?php echo 200 + $dbIDLevel1;?>; left: <?php echo $xTextledins +
310;?>px; top: <?php echo $yRightmenu;?>px; overflow: auto; visibility:
hidden">
<p class="textstyle"><b><?php echo $dbLevel1Name;?></b><br>
<img src="pictures/level1/<?php echo $dbFactsPictureLevel1Name;?>"
align="top" width="200" height="150"><br>
<?php echo $dbFactsLevel1Name;?></p>
</div>
<div id="rightmenu_<?php echo $dbIDLevel1;?>" name="rightmenu_<?php echo
$dbIDLevel1;?>" style="position:absolute; width:150px; height:300px;
z-index:<?php echo 100 + $dbIDLevel1;?>; left: <?php echo $xTextledins +
130;?>px; top: <?php echo $yMenu;?>px; overflow: auto; visibility:
hidden">
<p class="textstyle">
<?php
$sql2 = "SELECT tblevel2catlevel1cat.ForIDLevel1Cat,
tblevel2catlevel1cat.ForIDLevel2Cat, tblevel2cat.IDLevel2,
tblevel2cat.level2Name FROM tblevel2catlevel1cat";
$sql2 .= " LEFT JOIN tblevel2cat ON (tblevel2cat.IDLevel2 =
tblevel2catlevel1cat.ForIDLevel2Cat)";
$sql2 .= " WHERE tblevel2catlevel1cat.ForIDLevel1Cat = " .
safeQuote($dbIDLevel1);
$querys2 = mysql_query($sql2);
$nrRows++;
while ($dbArray2 = mysql_fetch_array($querys2)) {
$dbIDLevel2 = $dbArray2["IDLevel2"];
$dbLevel2Name = $dbArray2["level2Name"];
?>
<a href="page_productarea2.php?ID=<?php echo $dbIDLevel2;?>"
title="<?php echo $dbLevel2Name;?>"><font color="#015730"><?php echo
$dbLevel2Name;?></font></a><br>
<?php
$nrRows++;
}
?>
</p>
</div>
<?php
//Here...
$yMenu +=(8*intval($nrRows));
}
If you don't bother about the database:
You'll see that two div-elements are created. I want "rightmenu_<?php
echo
$dbIDLevel1;?>" to be at the same vertical level as div-element:
"factsmenu_<?php echo $dbIDLevel1;?>"
Any ideas? (I tried with $yMenu +=(8*intval($nrRows)) with the rule that
any
row is 8 pixels, but this doesn't seem to be true)
[/snip]
A. The is a CSS question
2. You didn't provide a way to see it.
III. Using absolute positioning is a bad thing and the left div should
be floated to the left.
Ok, I got your message you all! But I really wanted to do the postitioning
in PHP with absolute positioning (just for the sakes cause ;-))
/G
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