Display Results

spy0012

Hi, i am trying to display results from this query

<?php
$res = mysql_query("SELECT * FROM authuser WHERE team='Clan Leader'");
?>

But cannot get it to work any help, does not have to be in a table but has to be seprate lines per result

E.g It pulls the team='clan leader' from authuser and will display uname from authuser from the query above

I can run this query in phpmyadmin and pulls 2 results but cannot make it display in a php page

YappyDog

What code are you using to try to display the results?

spy0012

I don't actually have code to display i tried many ways but cannot get it to display.

I know how to display a count of members but not to display a quered result.

This is what i use to display a whole list of members

Calling code: $res = mysql_query("SELECT uname,team,status FROM authuser ORDER BY team='Clan Leader' DESC LIMIT 100");

Display code:

<? echo("<table width=407 border=0 align='center' cellpadding=0 cellspacing=0>
<tr>
<td width=75% align=center><table width=498 border=1 cellpadding=0 cellspacing=0 bgcolor='#CCCCCC'>
<tr><td width='362' align=left ><div align='center' class='style1'><strong>Clan Name</strong></div></td>
<td width='130' align=left><div align='center' class='style1'><strong>Member Type</strong></div></td>
<td width='130' align=left><div align='center'><strong>Status</strong></div></td>
</tr>");

      for ($i = 0; $i < mysql_num_rows($res); $i++)
     {
	       $tref = round($tref, 2);
           $name = mysql_result($res, $i, "uname");
           $id = mysql_result($res, $i, "team");
		   $stat = mysql_result($res, $i, "status");
           $nn = $i + 1;
           echo("<tr style=\"background-color: #f7f7f7\">
             <td><div align='center'><font face=$fontface size=2>$name</font></div>                 <font face=$fontface size=2><center></td><td><div align='center'>$id</div></td>
             <td><div align='center'>$stat</div></td>
           </tr>");
        }

echo("
</table></td>

</table><br>");

mysql_close;
?

That code uses a table to display results in split form,

The first post i want it to query the team field and display the uname of them results

so team='clan member' i want it to display the users in that team
Just uname is all i want, not all the members just the clan members uname

dnast

Edit: The following is incorrect!

while ($row = mysql_fetch_array($res)) {
  echo $row['authuser']."<br />";
}

spy0012

I have no idea where to put that as where ever i put it, it never works

dnast

It can be placed anywhere after the query in your first post as long as $res haven't been changed. Try putting it right after

$res = mysql_query("SELECT * FROM authuser WHERE team='Clan Leader'");

By the way, what happens when you use that code?

spy0012

I get the following:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/kilopa3/public_html/temp.php on line 3

Full code now:
<?php
$res = mysql_query("SELECT * FROM authuser WHERE team='Clan Leader'");
while ($row = mysql_fetch_array($res)) {
echo $row['authuser']."<br />";
}

I need the results in a certain area of the page
www.kiloparsec.com/temp.php

dnast

Looks like your query isn't producing a result. Either there are no matches or there's an error.

Change the query to this:

$res = mysql_query("SELECT * FROM authuser WHERE team='Clan Leader'")
  or die("Database Error: ".mysql_error());

to display any errors. You can also try typing the query string into the MySQL console and see what happens.

And you can output the results anywhere on the page -- I was just telling you to put it right after the query to narrow down the problem.

spy0012

The help sofar is good, i can get it to display how many clan leaders we have but not there names:

$res = mysql_query("SELECT * FROM authuser WHERE team='Clan Leader'")
or die("Database Error: ".mysql_error());
----> added this $leader = mysql_result($res, 0); <------

I want it to display there names not count them, how can i do that, if i remove the $leader = mysql_result($res,0); the code dosen't display errors

But i want the names to display, the database name for the username is uname

So i want to only display uname from selected team='clan leader'

Any ideas

Thanks for the sofar

dnast

My apologies. I was reading too fast and was using your table name as a field name and never went back to see if I did anything wrong until now.

To keep with the way you're doing it, use

$leader = mysql_result($res,0,"uname");

to retrieve the name. It'll only retrieve the first name in your result set, though. You'll need a loop to get all of them.

I have to go, but I'll be back later. Once again, sorry. I really messed up on that one 🙂

spy0012

I fixed the code using the following:

include($DOCUMENT_ROOT . '/vars1.php');
mysql_connect($db_host, $db_user, $db_pwd);
mysql_select_db($db_name);

$res = mysql_query("SELECT uname FROM authuser WHERE team='Clan Leader'");
$leader = mysql_result($res,0,'uname');
$leader1 = mysql_result($res,1,'uname');
?>

And called results by this:

<? echo("$leader"); ?><br><? echo("$leader1"); ?>

If there is an easier way to call the results then i am open to suggestions

Thanks for the help on this coding

spy0012

Based on the above how would i say display 5 newest members that have joined, e.g displays new members username until it reaches 5 members then updates by newest member thant joins and drops one to display new username at top of list

Hope you understand what i mean

dnast

Glad you got it working. If you only need 2 names like you wrote previously, then it's easier that way. If you need a larger number it would be best to use a loop.

(Note: This is what I meant to put the first time, heh. And you can change what's in the echo to how you need it to fit in your page. This is just a simple example.)

while ($row = mysql_fetch_array($res)) {
  echo $row['uname']."<br />";
}

To show the 5 newest you will need a date field or a unique identifier (a.k.a. ID) that auto-increments. You can then add "ORDER BY date_joined_or_id DESC LIMIT 5" to your query.