[SOLVED] help showing search results

[cs1h]

Hi,

 

I have a search script that checks two databases and then returns the results, but its not showing any results or coming up with any errors.

 

I asked for help a few weeks ago but had no success and I have had none on my own ether.

 

The script is,

 

Code:

<?php 

$username = $_COOKIE['loggedin'];

$db=mysql_connect("localhost","xxx","xxx");
mysql_select_db("xxx") or die("Unable to select database");
$result=mysql_query("SELECT Code_1,Code_2 FROM xxx WHERE username='$username'") or die(mysql_error());
if(!$result){$sw=0;$error="No results";}
else
{
$sw=1;
$row=mysql_fetch_array($result);
$code1=$row["code1"];
$code2=$row["code2"];
mysql_close($db);
$db=mysql_connect("localhost","yyy","yyy");
mysql_select_db("yyy") or die("Unable to select database");
$result=mysql_query("SELECT name,code1,code2 FROM `yyy` WHERE (codeb1='$code1' AND codeb2='code2' AND accepted='no') ORDER BY id") or die(mysql_error());
if(!$result){$sw=0;$error="You have no friend requests at the moment";}
else
{
  $sw=1;
  $asker=array();
  $asker_code1=array();
  $asker_code2=array();
  while($row=mysql_fetch_array($result))
  {
   $asker[]=$row["name"];
   $asker_code1[]=$row["code1"];
   $asker_code2[]=$row["code2"];
   }
  mysql_close($db);
  $db=mysql_connect("localhost","xxx","xxx");
  mysql_select_db("xxx") or die("Unable to select database");
  for($i=0;$i<count($title);$i++)
  {
   $result=mysql_query("SELECT avatar,name1 FROM xxx WHERE username='".$asker[$i]."'")or die(mysql_error());
   if(!$result){$sw=0;$error="No results 2</a>.";}
   else
   {
    $sw=1;
    $row=mysql_fetch_array($result);
     print "<div id='comment'>
             <img src='/avatar/".$row["avatar"]."' width='56' height='56' class='comment_photo' />
             <div id='comment_header'><span class='comment_author'> ".$row["name1"]." </span><span class='comment_wrote'>wants to be friends</span><span class='comment_stars'></span></div>
             <div class='comment_text'></div>
             <div id='comment_footer'><span class='comment_delete'><a href='ssi_css_my_profile.php?c1={$asker[$i]}&c2={$asker[$i]}'>View Profile</a> | <a href='friend_accept.php?name=".$row["name1"]."&c1={$asker_code1[$i]}&c2={$asker_code2[$i]}'>Accept</a> | <a href='friend_deny.php?name=".$row["name1"]."&c1={$asker_code1[$i]}&c2={$asker_code2[$i]}'>Deny</a></span></div>
            </div>";
		}
		}
}
}
If($sw==0){echo $error;}
if($db){mysql_close($db);}
?>

 

Can anyone help?

 

Thanks,

Colin

[jonsjava]

added some error checking on your results.

<?php 

$username = $_COOKIE['loggedin'];

$db=mysql_connect("localhost","xxx","xxx");
mysql_select_db("xxx") or die("Unable to select database");
$result=mysql_query("SELECT Code_1,Code_2 FROM xxx WHERE username='$username'") or "ERROR: ".mysql_error();

if(!$result || strstr($result, "ERROR: "))
{
$sw=0;$error="No results";
}
else
{
$sw=1;
$row=mysql_fetch_array($result);
$code1=$row["code1"];
$code2=$row["code2"];
mysql_close($db);
$db=mysql_connect("localhost","yyy","yyy");
mysql_select_db("yyy") or die("Unable to select database");
$result=mysql_query("SELECT name,code1,code2 FROM `yyy` WHERE (codeb1='$code1' AND codeb2='code2' AND accepted='no') ORDER BY id") or "ERROR: ".mysql_error();
if(!$result || strstr($result, "ERROR: ")){
	$sw=0;
	$error="You have no friend requests at the moment";
}
else
	{
	$sw=1;
	$asker=array();
	$asker_code1=array();
	$asker_code2=array();
	while($row=mysql_fetch_array($result))
	{
		$asker[]=$row["name"];
		$asker_code1[]=$row["code1"];
		$asker_code2[]=$row["code2"];
	}
	mysql_close($db);
	$db=mysql_connect("localhost","xxx","xxx");
	mysql_select_db("xxx") or die("Unable to select database");
	for($i=0;$i<count($title);$i++)
	{
		$result=mysql_query("SELECT avatar,name1 FROM xxx WHERE username='".$asker[$i]."'")or "ERROR: ".mysql_error();
		if(!$result || strstr($result, "ERROR: ")){
			$sw=0;
			$error="No results 2</a>.";
		}
		else
		{
		$sw=1;
		$row=mysql_fetch_array($result);
		print "<div id='comment'>
		<img src='/avatar/".$row["avatar"]."' width='56' height='56' class='comment_photo' />
		<div id='comment_header'><span class='comment_author'> ".$row["name1"]." </span><span class='comment_wrote'>wants to be friends</span><span class='comment_stars'></span></div>
		<div class='comment_text'></div>
		<div id='comment_footer'><span class='comment_delete'><a href='ssi_css_my_profile.php?c1={$asker[$i]}&c2={$asker[$i]}'>View Profile</a> | <a href='friend_accept.php?name=".$row["name1"]."&c1={$asker_code1[$i]}&c2={$asker_code2[$i]}'>Accept</a> | <a href='friend_deny.php?name=".$row["name1"]."&c1={$asker_code1[$i]}&c2={$asker_code2[$i]}'>Deny</a></span></div>
		</div>";
		}
	}
}
}
if($sw==0){
echo $error;
}
if($db){
mysql_close($db);
}
?>

[cs1h]

Hi thanks for the help,

 

I'm now getting the error message

 

You have no friend requests at the moment

 

So I put in an echo statement for the two search codes that are retrieved from the first database search and used to search the second one. It shows both statements and the data is in the database.

 

Any idea what could be wrong?

 

Thanks,

Colin