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torrentz

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  2. Palīdziet salikt kopā skriptu :)

    in PHP un datubāzes

    Posted · Edited by 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...)?

  3. Palīdziet salikt kopā skriptu :)

    in PHP un datubāzes

    Posted · Edited by 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];

  5. Palīdziet salikt kopā skriptu :)

    in PHP un datubāzes

    Posted · Edited by 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];

  6. Palīdziet salikt kopā skriptu :)

    in PHP un datubāzes

    Posted

    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)?

  7. Palīdziet salikt kopā skriptu :)

    in PHP un datubāzes

    Posted

    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

  9. kļūda skriptā

    in PHP skripti

    Posted · Edited by torrentz

    man

    <?php

    if (strstr($_SERVER['USER_AGENT'], 'Windows NT 5')){$os='Windows NT 5';}

    else {$os='Other';}

    print"$os<br/>";

    $ua = $_SERVER['HTTP_USER_AGENT'];

    print"$ua";

    ?>

    izvada:

    Other

    Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.6) Gecko/20070725 Firefox/2.0.0.6

  10. kļūda skriptā

    in PHP skripti

    Posted

    bet probleema ir taa ka jebkuraa gadiijumaa vienmeer sanaak user agent ir "Other". Li gan mans user agnet ir Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.8.1.6) Gecko/20070725 Firefox/2.0.0.6, un shajaa user agentaa ir Windows NT 5, tomeer skripts pievieno datubaazee +1 vieniibu pie other..

  13. kļūda skriptā

    in PHP skripti

    Posted · Edited by torrentz

    $string = $_SERVER['USER_AGENT'];

     

    if (preg_match('@SymbianOS/9.1@', $string)) { $os = "SymbianOS/9.1";}

    else if { (preg_match('@SymbianOS/9.2@', $string)){$os = "SymbianOS/9.2";}

    else if { (preg_match('@Series80@', $string)){$os = "Series80";}

    utt....

    else {$os = "Other";}

    Vai tā ir pareizi? Es īsti nezinu kādus simbolus jāizmanto @ vietā :) Vai kāds nevarētu pateikt priekšā?

  14. kļūda skriptā

    in PHP skripti

    Posted · Edited by torrentz

    I taada lieta, ka man visi user agenti sanaak kaa "Other", lai gan vismaz 40% user agentu vajadzeetu buut ar kaadu noshiim OS. Doma ir taada ka pilnajaa user agentaa, piemeeram: Nokia6680/1.0 (4.04.07) SymbianOS/8.0 Series60/2.6 Profile/MIDP-2.0 Configuration/CLDC-1.1, Mozilla/4.0 (compatible; MSIE 6.0; Windows 95; PalmSource; Blazer 3.0) 16;160x160 ir jaatrod kaada no shiim opereetaajsisteemaam..

  25. paste.php.lv

    in Atsauksmes, ierosinājumi, lamu vārdi

    Posted · Edited by torrentz

    es itkaa jau tiku galaa, kad serverim buus lielaaka noslodze tad arii kjershos optimizaacijai klaat (taa kaa neesmu dizhs programmettajs, naaksies luugt paliidziibu visdriizaak kaadam citam). skripts jau ir salabots un tas skripts ir tikai kaa piemeers :) http://wapcount.info te var apskatiities kaa tas izskataas dziivajaa :)

    starp citu man ir doma ielikt papildus statistiku (mp3, video utt.), bet skriptam jaabuut tad baigi optimizeetam...

    man ir telefona datubaaze un ja skripts neatrod user agentu datubaazee, tad vinsh tiek automaatiski pielikts ar default uzstaadiijumiem (noscreen,nocamera utt.) un peec tam es rokos netā un mekleeju informaaciju par konkreeto user agent ...

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