I am trying to pass a variable to a popup window, to do a search in a
database, and display an image based on the results... Here is what I have
messed up so far (this is the 1st time I have used php/mysql.... no
userID/Pswd on local host)
The script that passes the variable (variable does pass to 2nd html doc)
<script language="Javascript"><!--
function popImg1() {
window.open('../../assets/img760/img.php?getimg=alfstag1','img','toolbar=0,l
ocation=0,directories=0,status=1,menubar=0,scrollbars=0,resizable=0,copyhist
ory=0,width=800,height=600,left=0,top=0')
//-->
</script>
The php/mysql connect & search (have tested the connect and it works db
name=img760 table=img)
<?
$link = mysql_connect('localhost')or die ("Could not connect");
mysql_select_db('img760')or die ("Could not select database");
$query = "select ('imglocation','height','width') from img where
imgname=$getimg";
$imglocation = 'imglocation';
$height = 'height';
$width = 'width';
?>
The display of the image
<td align="center"><a href="javascript:void(0)" <img src="<? echo
$imglocation ?>" width="<? echo $width ?>" height="<? echo $height ?>"
border="0"></a></td>
I have been able to pass these variables directly in the url and display the
image..... trying to do it with a DB look up all I get in the code is the
literal not the variable (<img scr=imglocation width=width....etc) ...
Where am I going wrong? I have fiddled and changed until I have about
exhausted the possibilities... (Apache server on Win2k if it makes any
difference.... should not...)
What is a good reference book on PHP/MySQL... ??? I have one but obviously
it was a bad choice (title/author to remain unspoken..... forever)
Thanks
Mike
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