How to Wrap Text around an image when the imgsrc is in a PHP variable!
Hi all,
Can someone tell me,
How to wrap text around an image when the image source is in a php variable?Here is my code.
<html>
<head>
<style type="text/css">
#divider {
background-color: #ccffff;
border: 1px solid #ccc;
padding: 0px;
margin: 1px 0px 0px 0px;
}
.fndslist{
position: absolute;
border: 2px solid #ccc;
margin: 1px 0px 0px 0px;
top: 111px;
left: 220px;
}
img.floatLeft
{ // I tried using float,align,valign but nothing worked.
float: left;
}
</style>
</head>
<body>
<?php
//database connections
echo '
<table width = "560" class="fndslist" BGCOLOR="white" border="0" cellpadding="4" cellspacing="4">';
mysql_select_db ('db1');
$result = mysql_query("SELECT * FROM table1")or die(mysql_error());
while($row = mysql_fetch_array( $result ))
{
echo '<tr id="divider"><td>'.$row['profileimg'].$row['username'].'</td></tr>';
}
echo '</table>';
?>
</body>
</html>
I have tried using valign,float,align but nothing is helping the text to wrap around the image. The text is simply getting displayed from the bottom-right corner of the image.
Can somebody help me resolve this issue.
Thanks in advance,
Kavitha Butchi
Junior Poster in Training
69 posts since May 2008
Reputation Points: 10
Solved Threads: 4
If alignment of the image wont work, why not make a barrier around the image. An example is
$imgwidth=64; //Image width
$imgheight=64; //Image height
echo '<tr id="divider"><td>
<table border=0 cellpadding=0 cellspacing=0 width=$imgwidth
height=$imgheight align=left>
<tr><td>
'.$row['profileimg'].'
</td></tr></table>
'.$row['username'].'</td></tr>';
The above is just from the top of my head so don't be too suprised if there is a bug or 2.
cwarn23
Occupation: Genius
3,033 posts since Sep 2007
Reputation Points: 413
Solved Threads: 259
You are such a Genius it just worked very fine.
Thanks a Ton for it!!! Thankyou for your time :)
Kavitha Butchi
Junior Poster in Training
69 posts since May 2008
Reputation Points: 10
Solved Threads: 4
