[SOLVED] Image gallery repeating same image?

[selliott]

I'm putting together a simple image gallery, but I can't figure out how to get it to display ALL the images with a specific CatID.  This is what I've pieced together, but it just keeps repeating the same image.

 

<?php 
$CatID = $_REQUEST['CatID'];
$Photos = mysql_query("SELECT ID, Title, ImageURL, CatID, Details FROM Photo WHERE CatID='$CatID'", $connection) or die("error querying database");
$rows_nb = mysql_num_rows($Photos);

$pic_num = 0;
$pic_code = mysql_fetch_array($Photos);


					echo '<table width="75%" border="0" align="center">'; 

					while($row = mysql_fetch_assoc($Photos))


					{ 
						echo '<tr>'; 

						for ($tb_rows=0;$tb_rows<3;$tb_rows++) 
						{

							if ($pic_num < $rows_nb)
							{ 
							echo '<td><div align="center"><img src="'.$pic_code['ImageURL'] .'" border="1" /></div></td>';
							$pic_num++;
							}else{
							echo '<td></td>';
							}

						} 

					echo '</tr>'; 
					} 

					echo '</table>';
?>

[landavia]

bwawhahahaha...

let me laugh to myself..

 

simple.. try add random after catid

example:

mypic.php?catid=3

change into

mypic.php?catid=3&time=3648364

fyi: 3648364 is random number

 

it happen 2 me.. and i don't know why my image always repeated..

 

fyi.. my script for image like this

 

mypic.php?id=134

 

and i try 2 upload new image to id 134..

when i refresh.. WAKZZZ the image don't shown right T_T

 

and i discover about cache .. cache that always browser do in soo you can save a lot space and time with it.

 

and that's the reason i give the answer

[selliott]

bwawhahahaha...

let me laugh to myself..

 

simple.. try add random after catid

example:

mypic.php?catid=3

change into

mypic.php?catid=3&time=3648364

fyi: 3648364 is random number

 

it happen 2 me.. and i don't know why my image always repeated..

 

fyi.. my script for image like this

 

mypic.php?id=134

 

and i try 2 upload new image to id 134..

when i refresh.. WAKZZZ the image don't shown right T_T

 

and i discover about cache .. cache that always browser do in soo you can save a lot space and time with it.

 

and that's the reason i give the answer

 

Maybe I'm misunderstanding what you're saying (sorry if I am), but changing gallery.php?CatID=1  to gallery.php?CatID=1&time=23423 doesn't change anything.  I click on a gallery link on one page (photos.php), which inserts the Category ID in the URL (CatID=1 in this example), then when I get to that gallery page I have the script on that page pulling out the images from the photos table in the db with a category ID (CatID) that matches that category's ID.  I'm guessing it just ignores everything in the url string except the CatID?

[landavia]

i realy need see the source..

i think the problem not in galery but in photo or the logic

[printf]

This is still not how I would personally do it, it's just the easiest way to fix your code so you can do what you want...

 

<?php

// mysql connection and select db here

// you should validate this variable...

$CatID = $_REQUEST['CatID'];

$Photos = mysql_query ( "SELECT ID, Title, ImageURL, CatID, Details FROM Photo WHERE CatID= '" . $CatID. "';", $connection ) or die ( "error querying database" );

if ( mysql_num_rows ( $Photos ) > 0 )
{
$coulmns = 3;
$start   = 0;
$data    = '';

echo '<table width="75%" border="0" align="center">'; 

while ( $row = mysql_fetch_assoc ( $Photos ) )
{
	$data .= '<td><div align="center"><img src="' . $row['ImageURL'] . '" border="1" /></div></td>';

	$start++;

	// table row is done, build it and dump it

	if ( $start == $columns )
	{
		$start = 0;

		echo '<tr>' . $data . '</tr>';

		$data = '';
	}
}

// fix any offset columns to keep the table columns in line

if ( $start != 0 )
{
	for ( $i = $start; $i <= $coulmns; $i++ )
	{
		$data .= '<td></td>';
	}

	echo '<tr>' . $data . '</tr>';

	unset ( $i );
}

echo '</table>';

unset ( $columns, $start, $data, $row );
}

mysql_free_result ( $Photos );

?>

[landavia]

wakakakaka.. i'm a fool.. in here..

if you laugh to me.. i will accept it

 

>>$pic_code = mysql_fetch_array($Photos);

this goes line before while($row=mysql_fetch_array($Photos))

and below is always repeating again and again..

TS should type this

<?php 
$CatID = $_REQUEST['CatID'];
$Photos = mysql_query("SELECT ID, Title, ImageURL, CatID, Details FROM Photo WHERE CatID='$CatID'", $connection) or die("error querying database");
$rows_nb = mysql_num_rows($Photos);


$pic_num = 0;
$pic_code = mysql_fetch_array($Photos);

echo '<table width="75%" border="0" align="center">'; 

while($row = mysql_fetch_assoc($Photos))
{ 
$pic_code =$row; //wakakaka... this is the problem
echo '<tr>'; 

for ($tb_rows=0;$tb_rows<3;$tb_rows++) 

{

if ($pic_num < $rows_nb)

{ 

echo '<td><div align="center"><img src="'.$pic_code['ImageURL'] .'" border="1" /></div></td>';

$pic_num++;


}else{

echo '<td></td>';

}

} 

echo '</tr>'; 

} 

echo '</table>';
?>

i think i know what TS want.. but the above replies is the correct ^^

 

[selliott]

Thanks for the help guys!  I still plan on adding more to the script, but I was stuck here. 

 

Note: small typo in printf's post.  $coulmns = 3;    should be  $columns = 3;