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ziedinjsh
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Sveiki!

 

ir tāda lieta ka dtubāzē tiek saglabātās šādas lietas: ip, url, domain.

 

pie izvadīšanas ar while protams viss ir pēc kārtās. Jautājums ir sekojoš: Kā var izvadīt tā lai parādas piem tikai domain ar count cik ir saistīts ar šo domēnu zem kura ir atkal dažādi url kuriem ir šis domēns un pie šī url ir atkal count ja ir identiski url.

 

Piemēram:

 

www.google.lv (3):

www.facebook.com?page=4 (x1);

www.facebook.com?page=7 (x5);

www.kkas.lv?id=3 (x1)

 

 

www.facebook.com (2):

www.google.com?search=lapa (x1)

www.google.com?search=savaadaak (x4)

Edited by ziedinjsh
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Paldies par ideju, bet uzmocīju šādi:

$hvz = mysql_query("select domain, count(1) as rpt_count from referer group by domain");
while($data = mysql_fetch_array($hvz)){
echo '<p>'.$data['domain'].'('.$data['rpt_count'].')</p>';

 $d = mysql_query("select url, count(1) as rp_count from referer where domain='".$data['domain']."' group by url") or die(mysql_error());
  while($data = mysql_fetch_array($d)){
   echo '<p style="font-size:12px;color:#ccc;">'.$data['url'].' ('.$data['rp_count'].')</p>';
  }
}

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$hvz = mysql_query("select domain, count(1) as rpt_count from referer group by domain");

 

Gadījumā vēlams nebija rakstīt

$hvz = mysql_query("SELECT domain, count(1) AS rpt_count FROM referer GROUP by domain");

?

 

Kapēc vispār ir jātaisa "echo" ar "HTML" saturu?

Noderētu palasīt - "MVC"

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