table output to multiple pages?

[eraxian]

You know when you click on a link on some website and it gives you a list or a table, w/ say 50 results, and at the bottom will show you <next page> 1, 2, 3,...7 <last> and so on. So these other pages are created dynamically to show the next 50 results. Is there a way to do this w/ php and mysql?

 

I have a database w/ names, and a table w/ the letters of the alphabet, and when a user clicks a letter it shows a table w/ all the names starting w/ that letter, but i dont want the table to be a mile long, so i wanted it to create other pages, is there a way to do this w/ php, if not what do i need to do it?

 

 

Also, perhaps an alternative to that, i have this simple code to create a dynamic html table from a mysql query, but it only puts the data into 1 column, what if i wanted to create a table w/ 2 or 3 columns? Perhaps some nested loop is needed, but i am very noobish, so any help would be appreciated, thanks.

 

<?php
$con = mysql_connect("server","username","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("bands", $con);

$result = mysql_query("SELECT * FROM list WHERE name LIKE 'B%' ORDER BY name");

echo "<table border='1'>
<tr>
<th>Bands</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . "<a href=\"/bands/list/" . $row['link'] . ".html\">" . $row['name'] .  "</a>" . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>

[pocobueno1388]

To separate your results into pages your need to use pagination. Just Google "PHP MySQL pagination" and you should get plenty of results.

 

As for displaying your data in columns, take a look at this post (tutorial):

http://www.phpfreaks.com/forums/index.php/topic,95426.0.html

[eraxian]

thank you!!!

[tabatsoy]

try this one:

echo '<tr>
				<td width = "20" align = "left" valign = "top"><font class = "questions">'.$row['link'].'.</span><td>';
				echo '<td width = "280" align = "left" valign = "top" ><font class = "questions">'.$row['name'].'</span></td>';		

			'</tr>';

[eraxian]

Ok here is what i got so far, based off of the tutorial you linked me to, however i keep getting the error "no database selected" when clearly i have selected the database...

 

<?php
$con = mysql_connect("server","username","password");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("list", $con);
$query = "SELECT * FROM name WHERE name LIKE \'A%\' ORDER BY name";
$result = mysql_query($query) or die("There was a problem with the SQL query: " . mysql_error()); 
if($result && mysql_num_rows($result) > 0)
{
    $i = 0;
    $max_columns = 3;
    while($row = mysql_fetch_array($result))        
   {
       // make the variables easy to deal with
       extract($row);

       // open row if counter is zero
       if($i == 0)
          echo "<tr>";

       // make sure we have a valid product
       if($name != "" && $name != null)
          echo "<td>" . "<a href=\"/bands/list/" . $row['link'] . ".html\">" . $name .  "</a>" . "</td>";
    
       // increment counter - if counter = max columns, reset counter and close row
       if(++$i == $max_columns) 
       {
           echo "</tr>";
           $i=0;
       }  // end if 
   } // end while
} // end if results

// clean up table - makes your code valid!
if($i < $max_columns)
{
    for($j=$i; $j<$max_columns;$j++)
        echo "<td> </td>";
}
?>
</tr>
</table>

 

is my syntax wrong or something?

[pocobueno1388]

Try doing this:

mysql_select_db("list", $con)or die(mysql_error());

[eraxian]

You rule paco, thank you! There was also an error in my select_db string, i mistyped the name of the DB

[pocobueno1388]

Awesome, so did you get everything working okay?

[eraxian]

i did just now, got the columns and the pages to work together. they look like crap but i'll make it pretty later . Thanks for your help paco