[SOLVED] Need big help (URGENT!)

[studgate]

Hi Guys, I need some help, I have written a report page

but I am having problem loading the data from the database

here is the code, please help:

 

<?
$num = 0;
$dbconn = @mysql_connect("localhost","username","password");
@mysql_select_db("database");

if (empty($bywhat))
{
?>
	<font size="+2"><strong>Report Page</strong></font><br><br>
   Links:<br>
	 <br>
	     <a href="reports.php?bywhat=date">Report by date</a><br>
	     <a href="reports.php?bywhat=state">Report by state</a><br>
	     <a href="reports.php?bywhat=city">Report by city</a><br>
	     <a href="reports.php?bywhat=situation">Report by activity</a><br>
	<br>

	<a href="form.php">Form Page</a>		
<?
}
else
{
	$strsql = "SELECT ".$bywhat." , COUNT(*) as num FROM database ORDER BY ".$bywhat." DESC";
	$ressql = @mysql_query($strsql);
?>
<table class="sort-table" id="table-1" cellspacing="0">
<thead> 
	<tr>
		<td align="left" width="300"> <? if ($bywhat=='date') echo("Date"); elseif ($bywhat=='city') echo("City"); elseif ($bywhat=='situation') echo("Situation"); elseif ($bywhat=='state') echo("State"); ?></td>
		<td align="right" width="100"> Number. </td>
	</tr>
  </thead>
	<tbody>
<?
	while(list($subj, $num)=@mysql_fetch_row($ressql))
	{
	 if(strlen($subj)<1){
		continue;
	}
	$count++;
	if($count%2!=0){
		$class="<tr class=\"odd\">";
	}else{
		$class="<tr class=\"even\">";
	}

	echo "$class";
?>

  <td align="left"> <a href="results.php?subj=<?=$subj?>&bywhat=<?=$bywhat?>"><?=$subj?></a></td>
			<td align="right"> <?=$num?> </td>
		</tr>

<?
	}
	@mysql_free_result($ressql);
?>
</tbody>
</table>

<?
}
@mysql_close($dbconn);
?>

 

I can't find the problem, any help is welcome.

Thanks!

[btherl]

The first thing to do is replace

 

	$dbconn = @mysql_connect("localhost","username","password");
@mysql_select_db("database");

 

with

 

	$dbconn = mysql_connect("localhost","username","password") or die("Couldn't connect to database\n");
mysql_select_db("database") or die("Couldn't select database\n");

 

Similarly with every call to mysql_query().  You should add 'or die("Query failed: " . mysql_error());' to each one.  But DON'T add it to mysql_num_rows(), or mysql_fetch_row() or any of the result fetching functions.

 

@ will ignore errors.  die() will tell you when there is an error.

[btherl]

Example:

 

	$strsql = "SELECT ".$bywhat." , COUNT(*) as num FROM database ORDER BY ".$bywhat." DESC";
	$ressql = mysql_query($strsql) or die("Error in $strsql\n" . mysql_error());

[studgate]

btherl, thank you very much, I found the problem.

that's why I love this site.

I completely forgot to put the

	
$ressql = mysql_query($strsql) or die("Error in $strsql\n" . mysql_error());

and I was able to find the problem and solved the issue.

Thank you again.