Palīdziet salikt kopā skriptu :)

[torrentz]

Pirmaa tabula:

CREATE TABLE `category` (
`uid` INT(16),
`category` enum('0','1','2','3','4','5','6','7','8','9','10') NOT NULL default '0',
PRIMARY KEY ( `uid` )
) TYPE=MyISAM COMMENT='kategorijas';

 

Otraa tabula:

CREATE TABLE `weeks` (
 `id` int(10) unsigned NOT NULL auto_increment,
 `uid` int(10) unsigned NOT NULL default '0',
 `next_mon` int(10) unsigned NOT NULL default '0',
 `date` int(10) unsigned NOT NULL default '0',
 `day_week` smallint(5) unsigned NOT NULL default '0',
 `count` int(10) unsigned NOT NULL default '0',
 `host` int(10) unsigned NOT NULL default '0',
 `in` int(10) unsigned NOT NULL default '0',
 `out` int(10) unsigned NOT NULL default '0',
 `Siemens` int(10) unsigned NOT NULL default '0',
 `Nokia` int(10) unsigned NOT NULL default '0',
 `Samsung` int(10) unsigned NOT NULL default '0',
 `Motorola` int(10) unsigned NOT NULL default '0',
 `LG` int(10) unsigned NOT NULL default '0',
 `Sagem` int(10) unsigned NOT NULL default '0',
 `SonyEricsson` int(10) unsigned NOT NULL default '0',
 `Alcatel` int(10) unsigned NOT NULL default '0',
 `Opera` int(10) unsigned NOT NULL default '0',
 `Mozilla` int(10) unsigned NOT NULL default '0',
 `Panasonic` int(10) unsigned NOT NULL default '0',
 `Other` int(10) unsigned NOT NULL default '0',
 PRIMARY KEY  (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=cp1251 AUTO_INCREMENT=16336 ;

 

Koda gabals:

 $result = mysql_query("SELECT count(`uid`) FROM `weeks` WHERE `host` > 0 and `date`=$today");
$cntData = mysql_fetch_row($result);
$count_users = $cntData[0];
$result=mysql_query("SELECT DISTINCT uid,count,host FROM weeks WHERE host > 0 AND date=$today ORDER BY host DESC LIMIT $start,$end");
$count_users_on_page = mysql_num_rows($result);

 

Kaa man paarveidot vaicaajumus, lai php skripts neizveeleetos visus UID, bet gan tikai tos UID, kas ir, piemeeram, kategorijā 5?

Pilns skipta kods: http://paste.php.lv/6232

[andrisp]

Es nezinu, vai tas būs optimālākais kverijs, bet:

 

SELECT count(`uid`) 
FROM `weeks` LEFT JOIN kat ON (
weeks.uid = kat.uid
) 
WHERE `host` > 0 and `date`=$today AND kat.category = 5

[torrentz]

Ir tāda lieta, ka

$result = mysql_query("SELECT count(`uid`)
FROM `weeks` LEFT JOIN kat ON (
weeks.uid = kat.uid
) WHERE `host` > 0 and `date`=$today AND kat.category = 0");
$cntData = mysql_fetch_row($result);
$count_users = $cntData[0];

tas count_users paraada neko un pie tam uzmet taadu erroru:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/wapcount/public_html/portal.php on line 132

 

visiem UID ieksh category tabulas, kategorija pagaidaam ir 0, taatad buutu jaaizvada cipars kaut kur ap ~140

 

Kaa ir japārveido ši vaicājums, lai vinjsh izveeleetos tikai noteiktu kategorijas UID?

 

$result=mysql_query("SELECT DISTINCT uid,count,host FROM weeks WHERE host > 0 AND date=$today ORDER BY host DESC LIMIT $start,$end");
$count_users_on_page = mysql_num_rows($result);

 

Un vai šis vaicajums vispār ir jāpārveido (tas izvada tikai pirmos piecs UID vienaa lapaa)?

[andrisp]

Klau, tas kods, ko es iemetu bija ar neīstu tabulas nosaukumu ? Tu pirms kopē kaut kur kodu, nepārbaudi, ko viņš dara ?

[xPtv45z]

Vai tad lai COUNT strādātu nav nepieciešams arī GROUP BY?

[andrisp]

Nē.

[torrentz]

Tur jau tā lieta, ka tiešām es neko nejeedzu vairaak par update table set a=$b where uid=$uid :)

kat bija jaanomaina uz category?

taa ir pareizi? veeljoprojaam nestraadaa un count_USERS NEKO NERAADA

$result = mysql_query("SELECT count(`uid`)
FROM `weeks` LEFT JOIN category ON (
weeks.uid = category.uid ) WHERE `host` > 0 and `date`=$today AND category.category = 0");
$cntData = mysql_fetch_row($result);
$count_users = $cntData[0];

Edited October 12, 2007 by torrentz

[andrisp]

protams.

[andrisp]

AND category.category = 0

 

to nullīti liec iekš pēdiņām. Kāpēc ? http://laacz.lv/2006/05/08/mysql-gotcha/

[torrentz]

neesmu ielicis peedinjas vnk tas viss ir ieksh $result = mysql_query("");

cik izlasiiju no tava bloga, tad man taas peedinjas tomeer ir jaaliek?

[andrisp]

No mana bloga ? Un kāds tam sakars ar mysql_query() ?

 

Pēdiņās jāliek tā nullīte pie category.category = 0.

 

Tas nozīmē, kas es domāju šādi:

 

category.category = '0'

 

Un kāpēc tā jādara, vari izlasīt iekš laacz.lv bloga. Un ja tu domāji, ka tas ir mans blogs, tad tu maldies. :)

[torrentz]

ups, sajaucu apblog ar laacz :D

shitaa ir pareizi?

$result = mysql_query("SELECT count(`uid`)
FROM `weeks` LEFT JOIN category ON (
weeks.uid = category.uid
)
WHERE `host` > 0 and `date`=$today AND category.category = '0'");
	$cntData = mysql_fetch_row($result);
	$count_users = $cntData[0];

Edited October 12, 2007 by torrentz

[andrisp]

A tu pamēģini.

 

Bet nu manām acīm izskatās, ka viss ok.

[torrentz]

Nestraadaa, veeljoprojaam

$result = mysql_query("SELECT count(`uid`)

FROM `weeks` LEFT JOIN category ON (

weeks.uid = category.uid

)

WHERE `host` > 0 and `date`=$today AND category.category = '0'");

$cntData = mysql_fetch_row($result);

$count_users = $cntData[0]

izvada shito:

Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/wapcount/public_html/portal.php on line 133

 

lai gan category tabulaa visiem UID no 1 liidz 3000 kategorija ir 0...

 

Varbuut probleema ir tajaa, ka weeks tabulaa ir mazaak UID nekaa categoy tabulaa (tur es saliku vairaak UID priekshdienaam...)?

Edited October 12, 2007 by torrentz

[Vebers]

Vienigā problēma var būt $today mainīgajā, neizskatījās, ka viņš k-kur tiktu definēts. Ja pareizi sapratu ko vēlies panākt vari iemēģināt šito :

 

SELECT count(*) as cnt, w.uid, w.count, w.host FROM weeks w, category c WHERE w.host > 0 AND w.date=$today and c.category = '5'  GROUP BY uid ORDER BY w.host DESC