php/mysql help

-Ray- · 06-20-2008, 09:23 AM

-----SOLVED---------

I am in the middle of creating a search by letter script, except, i have ran into some trouble..

i want the user to be able to click a letter and it show all the results with that letter. no problem.. already done using..

PHP Code:

   
$alp=$_GET[a];

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

 
$word= $row['title'];

$id=$row['id'];

 
 
$ab = substr($word, 0, 1); 

 
$str = strtolower($ab);

 
if($alp != $str){

 
}

else{

 
echo ("<br>" . $id  ."-<a href=\"$id/\">" . $data[0] . "</a>");

 
}

But, now, i want to make a paging script to page the data, but i cannot due that because i cannot count the mysql data in this way. I do not know how to and i am soo confused. every time i make a paging function i used a mysql count

this is what i believe it would have to look like but, we no its not going to work.. lol

PHP Code:

  select title where title="substr($word, 0, 1)"

-NC- · 06-20-2008, 09:33 AM

select title where LEFT(title,1) = $yourletter

http://dev.mysql.com/doc/refman/5.0/...#function_left

-Ray- · 06-20-2008, 09:38 AM

thanks.. ill try that...

-NC- · 06-20-2008, 09:41 AM

no problem, glad it worked, you could also use

where title LIKE '$yourletter%'
http://dev.mysql.com/doc/refman/5.0/...functions.html

not sure which one is more efficient...

edit: hey! you edited your post, did it not work after all?

-Ray- · 06-20-2008, 09:48 AM

i trying it on function one... it worked... im trying it on function 2 now

final code... works... thanks

PHP Code:

  <?php

 
$alp=$_GET[a];

require("pagerclass.php");

$limit="20";

 
mysql_connect("localhost", "------------") or die(mysql_error());

mysql_select_db("---------") or die(mysql_error());

 
$p = new Pager;

$start = $p->findStart($limit);

 
 
$count = mysql_num_rows(mysql_query("SELECT * FROM drink WHERE LEFT(title,1) = '$alp'"));

 
$pages = $p->findPages($count, $limit);

 
$pagelist = $p->pageList($_GET['page'], $pages, $urlqry);

echo $pagelist;

 
 
$result = mysql_query("SELECT title,id FROM drink where LEFT(title,1) = '$alp' ORDER BY title ASC LIMIT $start, $limit")

or die(mysql_error());  

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

 
$word= $row['title'];

$id=$row['id'];

 
$data = explode(' recipe',$word); 

 
echo ("<br>" . $id  ."-<a href=\"$id/\">" . $data[0] . "</a>");

 
} 

?>

06-20-2008, 09:23 AM	#1 (permalink)
-Ray- An American Soldier Join Date: Jun 2005 Location: Pennsylvania Posts: 1,630 2.53 NP$ (Donate)	php/mysql help -----SOLVED--------- I am in the middle of creating a search by letter script, except, i have ran into some trouble.. i want the user to be able to click a letter and it show all the results with that letter. no problem.. already done using.. PHP Code: `$alp=$_GET[a]; while($row = mysql_fetch_array( $result )) { $word= $row['title']; $id=$row['id']; $ab = substr($word, 0, 1); $str = strtolower($ab); if($alp != $str){ } else{ echo ("<br>" . $id ."-<a href=\"$id/\">" . $data[0] . "</a>"); }` But, now, i want to make a paging script to page the data, but i cannot due that because i cannot count the mysql data in this way. I do not know how to and i am soo confused. every time i make a paging function i used a mysql count this is what i believe it would have to look like but, we no its not going to work.. lol PHP Code: `select title where title="substr($word, 0, 1)"` __________________ Completely Free cPanel Hosting -- Deployed to Afghanistan -- "There is no such thing as rich over-night unless you owned Google stock in 2006" Last edited by -Ray-; 06-20-2008 at 10:07 AM.

06-20-2008, 09:33 AM	#2 (permalink)
-NC- Traveller Join Date: Mar 2007 Location: Yet another city Posts: 1,392 614.57 NP$ (Donate)	select title where LEFT(title,1) = $yourletter http://dev.mysql.com/doc/refman/5.0/...#function_left __________________ Internet.geek.nz NameCooler.com Unlimited Domain Name Web Hosting Travel Money Rates

06-20-2008, 09:38 AM	#3 (permalink)
-Ray- An American Soldier Join Date: Jun 2005 Location: Pennsylvania Posts: 1,630 2.53 NP$ (Donate)	thanks.. ill try that... __________________ Completely Free cPanel Hosting -- Deployed to Afghanistan -- "There is no such thing as rich over-night unless you owned Google stock in 2006"

06-20-2008, 09:41 AM	#4 (permalink)
-NC- Traveller Join Date: Mar 2007 Location: Yet another city Posts: 1,392 614.57 NP$ (Donate)	no problem, glad it worked, you could also use where title LIKE '$yourletter%' http://dev.mysql.com/doc/refman/5.0/...functions.html not sure which one is more efficient... edit: hey! you edited your post, did it not work after all? __________________ Internet.geek.nz NameCooler.com Unlimited Domain Name Web Hosting Travel Money Rates

06-20-2008, 09:48 AM	#5 (permalink)
-Ray- An American Soldier Join Date: Jun 2005 Location: Pennsylvania Posts: 1,630 2.53 NP$ (Donate)	i trying it on function one... it worked... im trying it on function 2 now final code... works... thanks PHP Code: <?php $alp=$_GET[a]; require("pagerclass.php"); $limit="20"; mysql_connect("localhost", "------------") or die(mysql_error()); mysql_select_db("---------") or die(mysql_error()); $p = new Pager; $start = $p->findStart($limit); $count = mysql_num_rows(mysql_query("SELECT * FROM drink WHERE LEFT(title,1) = '$alp'")); $pages = $p->findPages($count, $limit); $pagelist = $p->pageList($_GET['page'], $pages, $urlqry); echo $pagelist; $result = mysql_query("SELECT title,id FROM drink where LEFT(title,1) = '$alp' ORDER BY title ASC LIMIT $start, $limit") or die(mysql_error()); while($row = mysql_fetch_array( $result )) { $word= $row['title']; $id=$row['id']; $data = explode(' recipe',$word); echo ("<br>" . $id ."-<a href=\"$id/\">" . $data[0] . "</a>"); } ?> __________________ Completely Free cPanel Hosting -- Deployed to Afghanistan -- "There is no such thing as rich over-night unless you owned Google stock in 2006" Last edited by -Ray-; 06-20-2008 at 10:06 AM.

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