Displaying a portion of a result set

bcarl314 · 07-03-2002, 02:45 AM

How can I nodify the code below to, by default display the 1st 10 resutls from the query, then at the bottom, display a list Page 1 of 2 | 3 | 4 | 5 and so on.?

Code:

<?php
	include('dbConn.php');
	$i="";
	$linkID = mysql_connect("localhost", "$dbUser", "$dbPass");
	if ($linkID != FALSE) {
		mysql_select_db("$dbName", $linkID);
		$resultID = mysql_query("SELECT * FROM images WHERE catID='$catID'", $linkID);
		$tot=mysql_num_rows($resultID);
		while($row = mysql_fetch_assoc($resultID)) {
			$iName= $row['imageName'];
			$iID= $row['imageID'];
			$iWords = $row['keywords'];
			$desc = $row['descTxt'];
			print "<img src='images/small/".$iID.".jpg' align='left' alt='".$iWords."' border='2' bordercolor='black' height='120'>";
			print "<b>".$iName."</b><br>";
			print "<font class='main'>";
			if($desc!=null) {
				include("$desc");
			}
			//print "This is where I need to add a nice little description for this image.<br>Now I've just got to get off my lazy but and do it.";
			print "</font>";
			print "<a href='photos.php?imageID=".$iID."'>Click here to see more about this photograph!</a>";
			print "<br clear='all'><hr>";
		}
	}
	else {
		print "Bad Connection";
	}
	mysql_close($linkID);
?>

I've tried using the limit for my sql, but I get errors. Any help is appreciated.
Thanks.

Cloudski · 07-03-2002, 07:55 AM

Well, I uh... don't know much about PHP, but couldn't you set the displaying part of the code(the whole thing) In a for-loop, and st a variable so that whenever it reaches a number, set next page?

Sorry if I am not much help... I am still a newbie at this

Jeewhizz · 07-03-2002, 09:02 AM

Use something like this

PHP Code:


			
$query = "SELECT * FROM table LIMIT $limit,25";

$result = mysql_db_query($db,$query,$connect);

while($row = mysql_fetch_assoc($result)) {

$iName= $row['imageName'];

$iID= $row['imageID'];

$iWords = $row['keywords'];

$desc = $row['descTxt'];

print "<img src='images/small/".$iID.".jpg' align='left' alt='".$iWords."' border='2' bordercolor='black' height='120'>";

print "<b>".$iName."</b><br>";

print "<font class='main'>";

if($desc!=null) {

include("$desc");

    }

//print "This is where I need to add a nice little description for this image.<br>Now I've just got to get off my lazy but and do it.";

print "</font>";

print "<a href='photos.php?imageID=".$iID."'>Click here to see more about this photograph!</a>";

print "<br clear='all'><hr>";

}

}

else {

print "Bad Connection";

}



$total = mysql_num_rows($result);



$foo = $total / 25;

$foo=$foo+1;

$limit=0;

echo "<p>&nbsp;</p>";

echo "<center>Go to page: | ";

for($i=1;$i<$foo;$i++)

{

  echo "<a href=\"$PHP_SELF?limit=$limit\">$i</a> | ";

  $limit = $limit + 25;

}

Hope that helps

Jee

bcarl314 · 07-03-2002, 11:18 AM

Jee,

Thanks for the code, that almost exactly what I was trying to do, then I get a undefined variable limit error the first time through.

When someone enters this page the string is...
http://localhost/photos.php?catID=1

Do I need to change all those links to
http://localhost/photos.php?catID=1&limit=0
for the initial viewing?

I trind to use something like
if(!isset($limit)) {
$limit=0;
}
but that seems to throw an error???undef var limit grr

I keep thinking it should work, but it don't.