ianmitchell
03-04-2007, 06:02 PM
I'm new to PHP so i'm not sure how to do this. What I'm trying to do is to return the results of a database, and display each row in seperate defined div tags, so that each row can potentially be displayed at independent locations on the page using CSS.
For example
.div {
position:absolute;
left: 100px;}
.div2 {postion:absolute;
left: 100px;
top: 500px;}
<div class="div1">(1st row of data)</div>
<div class="div2">(2nd row of data)</div>
Heres my PHP code:
<?php
$hostname_connPHP = "localhost";
$database_connPHP = "scores";
$username_connPHP = "root";
$password_connPHP = "*****";
$Connection1 = mysql_pconnect($hostname_connPHP, $username_connPHP, $password_connPHP) or die(mysql_error());
mysql_select_db("scores") or die(mysql_error());
$result = mysql_query("SELECT * FROM scores")
or die(mysql_error());
$rowsPerPage = 12;
// by default we show first page
$pageNum = 1;
// if $_GET['page'] defined, use it as page number
if(isset($_GET['page']))
{
$pageNum = $_GET['page'];
}
// counting the offset
$offset = ($pageNum - 1) * $rowsPerPage;
$query = "SELECT name, school, score
FROM scores
order by score desc
LIMIT $offset, $rowsPerPage";
$result = mysql_query($query) or die('Error, query failed');
// print the student info in table
echo '<table cellpading="0" cellspacing="0">
<tr>
<td>Name</td>
<td>School</td>
<td>Score</td>
</tr>';
while(list($name, $school, $score) = mysql_fetch_array($result))
{
echo "<tr><td>$name</td><td>$school</td><td>$score</td></tr>";
}
echo '</table>';
echo '<br>';
$query = "SELECT *
FROM scores ";
$pagingQuery = "LIMIT $offset, $rowsPerPage";
$result = mysql_query($query . $pagingQuery) or die('Error, query failed');
// ... the code that prints the result in a table
// how many rows we have in database
$result = mysql_query($query) or die('Error, query failed');
$numrows = mysql_num_rows($result);
?>
Any help would be great, thanks :thumbsup:
For example
.div {
position:absolute;
left: 100px;}
.div2 {postion:absolute;
left: 100px;
top: 500px;}
<div class="div1">(1st row of data)</div>
<div class="div2">(2nd row of data)</div>
Heres my PHP code:
<?php
$hostname_connPHP = "localhost";
$database_connPHP = "scores";
$username_connPHP = "root";
$password_connPHP = "*****";
$Connection1 = mysql_pconnect($hostname_connPHP, $username_connPHP, $password_connPHP) or die(mysql_error());
mysql_select_db("scores") or die(mysql_error());
$result = mysql_query("SELECT * FROM scores")
or die(mysql_error());
$rowsPerPage = 12;
// by default we show first page
$pageNum = 1;
// if $_GET['page'] defined, use it as page number
if(isset($_GET['page']))
{
$pageNum = $_GET['page'];
}
// counting the offset
$offset = ($pageNum - 1) * $rowsPerPage;
$query = "SELECT name, school, score
FROM scores
order by score desc
LIMIT $offset, $rowsPerPage";
$result = mysql_query($query) or die('Error, query failed');
// print the student info in table
echo '<table cellpading="0" cellspacing="0">
<tr>
<td>Name</td>
<td>School</td>
<td>Score</td>
</tr>';
while(list($name, $school, $score) = mysql_fetch_array($result))
{
echo "<tr><td>$name</td><td>$school</td><td>$score</td></tr>";
}
echo '</table>';
echo '<br>';
$query = "SELECT *
FROM scores ";
$pagingQuery = "LIMIT $offset, $rowsPerPage";
$result = mysql_query($query . $pagingQuery) or die('Error, query failed');
// ... the code that prints the result in a table
// how many rows we have in database
$result = mysql_query($query) or die('Error, query failed');
$numrows = mysql_num_rows($result);
?>
Any help would be great, thanks :thumbsup: