NEWBIE question

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arrowhead
Forum Newbie
Posts: 4
Joined: Wed Mar 25, 2009 5:02 pm

NEWBIE question

Post by arrowhead »

I am sorry for asking this but i need help. What i write is giving me correct result but i am very sure there is an easier and better way to write this code.

Code: Select all

 
$query_Recordset1 = "SELECT `name`,`pid` FROM list WHERE  `sanid` = 14 ORDER BY name";
$query_Recordset2 = "SELECT `name`,`pid` FROM list WHERE  `sanid` = 15 ORDER BY name";
$query_Recordset3 = "SELECT `name`,`pid` FROM list WHERE  `sanid` = 16 ORDER BY name";
$query_Recordset4 = "SELECT `name`,`pid` FROM list WHERE  `sanid` = 18 ORDER BY name";
$query_Recordset5 = "SELECT `name`,`pid` FROM list WHERE  `sanid` = 19 ORDER BY name";
 
$Recordset1 = mysql_query($query_Recordset1, $French) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
 
$Recordset2 = mysql_query($query_Recordset2, $French) or die(mysql_error());
$row_Recordset2 = mysql_fetch_assoc($Recordset2);
$totalRows_Recordset2 = mysql_num_rows($Recordset2);
 
$Recordset3 = mysql_query($query_Recordset3, $French) or die(mysql_error());
$row_Recordset3 = mysql_fetch_assoc($Recordset3);
$totalRows_Recordset3 = mysql_num_rows($Recordset3);
 
$Recordset4 = mysql_query($query_Recordset4, $French) or die(mysql_error());
$row_Recordset4 = mysql_fetch_assoc($Recordset4);
$totalRows_Recordset4 = mysql_num_rows($Recordset4);
 
$Recordset5 = mysql_query($query_Recordset5, $French) or die(mysql_error());
$row_Recordset5 = mysql_fetch_assoc($Recordset5);
$totalRows_Recordset5 = mysql_num_rows($Recordset5);
 
 
Charles256
DevNet Resident
Posts: 1375
Joined: Fri Sep 16, 2005 9:06 pm

Re: NEWBIE question

Post by Charles256 »

Code: Select all

 
$queries=array();
$results=array();
$num_records=array();
for($i=14;$i<20;$i++)
{
  $queries[] = "SELECT `name`,`pid` FROM list WHERE  `sanid` = $i ORDER BY name";
}
foreach($queries as $query)
{
  if(isset($temp)){  unset($temp);}
  $temp=mysql_query($query,$French) or die(mysql_error());
  $results[]=mysql_fetch_assoc($temp);
  $num_records[]=mysql_num_rows($temp);
}
 
that's not debugged at all and it's the same thing you did, just a bit more condensed. Hope that helps some.
arrowhead
Forum Newbie
Posts: 4
Joined: Wed Mar 25, 2009 5:02 pm

Re: NEWBIE question

Post by arrowhead »

then how i am going to insert the code below to my page. I was using this:

Code: Select all

<li style="list-style:none"><a href="sanatci.php?sid=<?php echo $row_Recordset5['sanid']; ?><?php echo $row_Recordset5['name']; ?></a>
   <ul> 
     <?php
  do {?>
<li style="list-style:none"><a href="sarki.php?sarki=<?php echo $row_Recordset5['sarkid']; ?>"><?php echo $row_Recordset5['sarkname']; ?></a></li>
   <?php } while ($row_Recordset5 = mysql_fetch_assoc($Recordset5)); ?>
      </ul>
</li>
Result looks like:

MADONNA
* Song Name
* Song Name
* Song Name

METALLICA
* Song Name
* Song Name
User avatar
califdon
Jack of Zircons
Posts: 4484
Joined: Thu Nov 09, 2006 8:30 pm
Location: California, USA

Re: NEWBIE question

Post by califdon »

You need to explain what you're trying to do. I'm guessing that you only need one query, but it's not at all clear what data you have in your database or what part of it you're trying to extract.
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