Pagination

[raptor30506090]

Hi every one

 

im getting an 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 '' at line 1

SELECT * FROM

 

if(isset($_GET['subid'])){
                      $id = $_GET['subid'];
					}						
                     $query = mysql_query("SELECT * FROM sub_nav WHERE subNav_ID=" . mysql_real_escape_string((int)$id));
					 $row = mysql_fetch_assoc($query);
					 $table =  strtolower($row['subNavName']);
					 // ******************************* This if statment controls the content ************************
					 if($row['show_hide'] == 1){
						                      $per_page = 2;

$pages_query = mysql_query("SELECT COUNT('id') FROM textiles");

$pages = ceil(mysql_result($pages_query, 0) / $per_page);

$page = (isset($_GET['p']) AND (int)$_GET['p'] > 0) ? (int)$_GET['p'] : 1;
  
$start = ($page-1) * $per_page;

$query = mysql_query("SELECT imageName FROM textiles LIMIT $start, $per_page");





while ($query_row = mysql_fetch_assoc($query)){
echo '<p>',$query_row['imageName'] , '</p>';
}


if ($pages >= 1){

	for($x=1; $x<=$pages; $x++){
		echo '<a href="?p='.$x.'">'.$x.'</a> '; 
		}
	}

 

Can any one help please

[MarPlo]

Hi

I think the error is because the column 'id' is between simple quotes, try this:

"SELECT COUNT(id) FROM textiles"

[raptor30506090]

sorry tryed that no good

 

it shows 2 items from the database and 2 links as it should you only get the error when i hit the link 1 or 2

[Jessica]

Post the entire error, and you say you get an error when you click on a link, is that a link to this page or a different page?

[Psycho]

When running queries - especially dynamic ones (i.e. they have variables) I would highly encourage you to create the queries as string variables first. That way if there is an error you can echo the query out for debugging purposes.

 

Also, there is absolutely no reason to use this:

mysql_real_escape_string((int)$id)

If you have forced the value to be an integer, then mysql_real_escape_string() is not doing anything of value. using (int) is already making the value safe for a DB query

 

Based upon the partial error you posted the only query which that would come from is this one:

$query = mysql_query("SELECT * FROM sub_nav WHERE subNav_ID=" . mysql_real_escape_string((int)$id));

 

BUt, you don't have any error handling that would have displayed that error! Are you sure that is the correct code that is producing the error? Try changing to the following

$id = (int) $id;
$sql = "SELECT * FROM sub_nav WHERE subNav_ID = {$id}";
$query = mysql_query($sql);
if(!$query)
{
    die("Query: {$sql}<br>Error: " . mysql_error());
}

[xyph]

You should echo out this query:

 

"SELECT imageName FROM textiles LIMIT $start, $per_page"

 

And post what it outputs. I'm assuming that's where the error is occurring.

 

 

 

All of Psychos advice is great, and escaping the variable after using (int) IS redundant, though I don't think it should cause an error.

[raptor30506090]

Hi all thank you for your help yes jesirose that the paganation link iv tryed the above with no luck

 

still getting the same 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 '' at line 1

SELECT * FROM

[Jessica]

As was said above, there's no way that code can produce that error.

Post the code for the page with any changes you've made since you said you tried suggestions.

[raptor30506090]

Hi here is the new code

 

if(isset($_GET['subid'])){
                      $id = $_GET['subid'];
					}



$id = (int) $id;
$sql = "SELECT * FROM sub_nav WHERE subNav_ID = {$id}";
$query = mysql_query($sql);
if(!$query)
{
    die("Query: {$sql}<br>Error: " . mysql_error());
}



					 $row = mysql_fetch_assoc($query);
					 $table =  strtolower($row['subNavName']);
					 // ******************************* This if statment controls the content ************************
					 if($row['show_hide'] == 1){
$per_page = 2;

$pages_query = mysql_query("SELECT COUNT('id') FROM textiles");

$pages = ceil(mysql_result($pages_query, 0) / $per_page);

$page = (isset($_GET['p']) AND (int)$_GET['p'] > 0) ? (int)$_GET['p'] : 1;
  
$start = ($page-1) * $per_page;

$query = mysql_query("SELECT imageName FROM textiles LIMIT $start, $per_page");

while ($query_row = mysql_fetch_assoc($query)){
echo '<p>',$query_row['imageName'] , '</p>';
}


if ($pages >= 1){

	for($x=1; $x<=$pages; $x++){
		echo '<a href="?p='.$x.'">'.$x.'</a> '; 
		}
	}


											   }

 

Thank you

[raptor30506090]

still nedd help guys thanks

[xyph]

Can you copy and paste the error directly? With the 'Query: ' and 'Error: ' strings included?

It's odd your query is being truncated to SELECT * FROM, if that's indeed what's happening.

 

Also, don't bump your threads until they've dropped off the first page, it's against the rules and will discourage help. We find it rude and selfish.

[raptor30506090]

ok this incudes the else

 

<?php
if(isset($_GET['subid'])){
                      $id = $_GET['subid'];
					}



$id = (int) $id;
$sql = "SELECT * FROM sub_nav WHERE subNav_ID = {$id}";
$query = mysql_query($sql);
if(!$query)
{
    die("Query: {$sql}<br>Error: " . mysql_error());
}



					 $row = mysql_fetch_assoc($query);
					 $table =  strtolower($row['subNavName']);
					 // ******************************* This if statment controls the content ************************
					 if($row['show_hide'] == 1){
$per_page = 2;

$pages_query = mysql_query("SELECT COUNT('id') FROM textiles");

$pages = ceil(mysql_result($pages_query, 0) / $per_page);

$page = (isset($_GET['p']) AND (int)$_GET['p'] > 0) ? (int)$_GET['p'] : 1;
  
$start = ($page-1) * $per_page;

$query = mysql_query("SELECT imageName FROM textiles LIMIT $start, $per_page");

while ($query_row = mysql_fetch_assoc($query)){
echo '<p>',$query_row['imageName'] , '</p>';
}


if ($pages >= 1){

	for($x=1; $x<=$pages; $x++){
		echo '<a href="?p='.$x.'">'.$x.'</a> '; 
		}
	}


											   }else{  
												      if(isset($_GET['subid'])){
                                                                                $id = $_GET['subid'];
					                                                        }						
                                                                            $query = mysql_query("SELECT * FROM sub_nav WHERE subNav_ID=" . (int)$id);
					                                                        $row = mysql_fetch_assoc($query);
					                                                        $table =  strtolower($row['subNavName']);
						                                                    $sql = "SELECT * FROM $table";

																			$query = mysql_query($sql) or die(mysql_error() . '<br />' . $sql);
																			$row = mysql_fetch_assoc($query);
																			echo '<p>'.$row['content'].'</p>';
												                            }

?>

hope this helps if i remove the else then it returns nothing

Thanks guys and girls

[xyph]

$table is undefined.

 

You need to make sure your data contains what you expect (or even exists) before using it in a query.

 

You should code with error reporting set to -1. See my signature. Your code would have generated at the very least an undefined index notice.

[raptor30506090]

Hi

 

this is what im getting

 

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 '' at line 1

SELECT * FROM

 

 

[Jessica]

$table is undefined.

 

You need to make sure your data contains what you expect (or even exists) before using it in a query.

 

You should code with error reporting set to -1. See my signature. Your code would have generated at the very least an undefined index notice.

[raptor30506090]

Hi all thanks for your help its not the Pagination part thats going wrong im going to have to have a re think on how to do this thanks again

[raptor30506090]

Hi jesirose

 

Yes i can see what the problem is now thanks for your help im going to have a re think on how to do this