[SOLVED] Pagination

[adam291086]

I have just done a tutorial on pagination. As i am new to this i am not sure what this error means. I am using the include syntax to call on the pagination document to act upon my view an album page. Can anyone see whats wrong with this code?

 

Error

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'table LIMIT 0, 25' at line 1

 

<?php 

WRONG USERNAMES AND PASSWORDS ECT ECT 
$db_server ="db11sdfds25.oneandone.co.uk";
$db_user = "dbo21dsfdsf8351273";
$db_pass = "adawmasdfdsn291086";
$db_name = "db21835sdfds1273";
$con = mysql_connect("$db_server","$db_user","$db_pass");

if (!$con)
{
die('Could not connect: ' . mysql_error());
}

$limit = 25; 
$query_count = "SELECT count(*) FROM table"; 
$result_count = mysql_query($query_count); 
$totalrows = mysql_num_rows($result_count); 
if(empty($page)){ 
$page = 1; 
} 

$limitvalue = $page * $limit - ($limit); 
$query = "SELECT * FROM table LIMIT $limitvalue, $limit"; 
$result = mysql_query($query) or die("Error: " . mysql_error()); 
if(mysql_num_rows($result) == 0){ 
echo("Nothing to Display!"); 
} 
$bgcolor = "#E0E0E0"; // light gray 
echo("<table>"); 

while($row = mysql_fetch_array($result)){ 
if ($bgcolor == "#E0E0E0"){ 
$bgcolor = "#FFFFFF"; 
}else{ 
$bgcolor = "#E0E0E0"; 
} 
echo("<tr bgcolor=".$bgcolor.">n<td>"); 
echo($row["users"]); 
echo("</td>n<td>"); 
echo($row["usersID"]); 
echo("</td>n</tr>"); 
} 
echo("</table>"); 
if($page != 1){ 
$pageprev = $page--; 

echo("<a href=\"$PHP_SELF&page=$pageprev\">PREV".$limit."</a> "); 
}else{ 
echo("PREV".$limit." "); 
} 
$numofpages = $totalrows / $limit; 

for($i = 1; $i <= $numofpages; $i++){ 
if($i == $page){ 
echo($i." "); 
}else{ 
echo("<a href=\"$PHP_SELF?page=$i\">$i</a> "); 
} 
} 


if(($totalrows % $limit) != 0){ 
if($i == $page){ 
echo($i." "); 
}else{ 
echo("<a href=\"$PHP_SELF?page=$i\">$i</a> "); 
} 
} 
if(($totalrows - ($limit * $page)) > 0){ 
$pagenext = $page++; 

echo("<a href=\"$PHP_SELF?page=$pagenext\">NEXT".$limit."</a>"); 
}else{ 
echo("NEXT".$limit); 
} 

mysql_free_result($result); 
?> 

[MadTechie]

is the mySQL table called "Table" ?

$query = "SELECT * FROM table LIMIT $limitvalue, $limit";

$query_count = "SELECT count(*) FROM table";

[adam291086]

sorry forgot to change that. Have changed it now and i still get the same problem.

[MadTechie]

the "table" probably doesn't exist,

is it the exact same error

[haaglin]

add quotes:

 

$query = "SELECT * FROM 'table' LIMIT $limitvalue, $limit"; 

[adam291086]

Right the code seems to process. All i get is a list on 'N' down the page and nothing happens to the number of pictures. I am using the syntax "include" to use this pagination code on the php code that gets everything from the database to display the pictures. You can see the results at http://adamplowman.co.uk/display_album.php?album_id=30.

[adam291086]

Sorry im being an idiot.It's all sorted now. Cheers