[williamh26]

Hi guys i have this code but it does not give results

Code:

<?php

 $error = "Could not connect to the database";
 mysql_connect('localhost','root','')   or die($error);
 mysql_select_db('hairsalon') or die($error);
 //max displayed per page
 $per_page = 2;

 //get start varible

 $start = $_GET['start'];

 //count records

 $record_count = mysql_num_rows (mysql_query("SELECT * FROM products"));

 //count max pages

 $max_pages = $record_count / $per_page; // may come out as decimal

 if (!$start)
    $start =0;

 //display data

 $get = mysql_query("SELECT  prodid, name, description, picturename  FROM products ORDER BY name LIMIT $start, $per_page");

echo "<table  summary='Services, or Links box template' class='servicesT' cellspacing='0'>";
echo"<br>";
echo "<tr> <th align= left>Name</th>  <th align= left colspan='0'>Description</th> <th align= left colspan='0'>Picture</th>";
echo"<p>";
  while ($row = mysql_fetch_assoc($get))
  {
     // get data
        $prodid = $row['prodid'];
        $name = $row['name'];
        $description = $row['description'];
        $picturename = $row['picturename'];

      echo "<a href=\"index.php?&content=updatetutor&id=$prodid\">";
      echo strtoupper("<tr><td class='servBodL' width=25%><b><u>$name</td>\n");
      echo "<p>";
      echo strtoupper("<td class = 'servBodL' width= 30%>$description</td>");
      echo "<p>";
      echo strtoupper("<td class = 'servBodL' width= 30%>$picturename</td>");
      echo "<p>";





  }
    echo" </table> ";
   echo"<br>\n";
   //SETUP PREV AND NEXT VARIABLES

   $prev = $start - $per_page;

   $next = $start + $per_page;

   // show pre button

   if(!($start< 1))

   echo "<a href='index.php?content=ourproducts&start=$prev'>Prev</a>\n";
   //show page numbers

   //set variable for first page

   $i=1;

   for($x=0; $x<$record_count; $x=$x+$per_page)
   {

     if($start!=$x)

     echo "<a href='index.php?content=ourproducts&start=$x'>$i</a>\n";

     else

     echo "<a href='index.php?content=ourproducts&start=$x'>$i</b></a>\n";
     $i++;
   }


    if (!($start>=$record_count=$per_page))
   echo "<a href='index.php?content=ourproducts&start=$next'>Next</a>\n";

   echo "<br>";
?>

This is the error i recieve

Notice: Undefined index: start in C:\wamp\www\hairsalon2\includes\ourproducts.inc.ph p on line 11

and i have the pictures in a file called images, and it does not show the pictures



CAN ANY ONE HELP ME PLEASE!!!

[Wolf1994]

Notice: Undefined index: start in C:\wamp\www\hairsalon2\includes\ourproducts.inc.ph p on line 11

That's not the error. Use check like:

Code:

if(isset ($_GET['start']))
 $start = $_GET['start'];
else
 $start="";

to avoid this notice.

[williamh26]

thank you it works


but now it does not show the pictures can u help me with it plealse

[Wolf1994]

Did it show them before?

Try to add or die(mysql_error()); to each sql query, like:

Code:

mysql_query("SELECT * FROM products") or die (mysql_error());

to check if they all correct.

[williamh26]

the pictures images are on a file??

[Golith]

if picture name is held in a file then why is it being queried in sql
my suggestion would be to enter a file location in picture name field on db so u can pull it and use it or better still hold th e image on the db.

would make things easier