[outputting Mysql results into tables]

Hi I want to create a photo gallery using PHP/MySQL. I'm having trouble comming up with the algorythm to put more than one record per line.

any input would be appreciated

[Allow Editable PHP Variables Dreamweaver Templates]

try putting the variables in an include file such as configure.php

[Layers Vs Field (PHP MYSQL) - Help]

can you show me the code for that field?

 

[php arithmic trouble]

I think what's happening is that $ans is an array and the user imputs are separate elements try to concatinate the user imput

 

try this

 

$ans = $_POST[first] .$_POST[calc] .$_POST[second];

 

note the periods

[Images with NULL value in MySql]

on one of my sites I use this;

 

if($row[4]==""){

echo"<td class=\"producttext\"><a href=\"details.php?id=$row[0]&storeID=$storeID&currency=$currency\"><img src=\"../images/savers/image_not_available.jpg\"></a></td>";

}else{

echo "<TD valign=\"top\" class=\"producttext\"><a href=\"details.php?id=$row[0]&storeID=$storeID&currency=$currency\"> <img src=\"../images/savers/$row[4]\"></a> </TD>";

}

 

where row[4] would be my image variable

[Images with NULL value in MySql]

if there is a null in the image varialbe it will not display the image but the HTML is still there so you get a image missing x on the browser. Try a n if statement to check that the image exits first if so display it other wise display a default image or don't sent any html.

 

if(row_Bilder[1}(

img src=<?php echo $row_Bilder['bilde1];

} else{

display either a default image or no image.

[Passing a variable to the browser to view a record]

Try to send the the record id which should be part of the databse as an autoincrement colum

like this page.php?id=$id

 

Then use this id to modify your query with; select * from tablename where id=$id.