Simple Member script help?

ViciOuS · 04-12-2005

Hello once again. I am trying to make a simple member system for my clan site. If you would please take a moment to view my site (http://www.clanadvanced.com) and see how simple indeed the member system is.

Now, the whole script works just fine and dandy except for one part, please take a look at this page to see what i mean. I need this to be like the members list on the site if you know what i mean, lol.

Here are my 2 pages.

AddMember.php

Code:

<?php
include('session_vars.php');
// Check to see if they are logged in..
if ($enpw != $pw){
header("Location: ../ad_login.php");}

else if(!$action){
?>

<BODY background="../images/image_22.gif" OnLoad="self.focus();document.addmember.name.focus()">

<div align="center">
<br><a href="#" target="iframe" title="Remove Member(s)">Remove Member(s)</a>
<br><br>

 <FORM ACTION="?action=submit" method="POST" name="addmember">
  <TABLE width="80%" valign="center">

    <tr bgcolor="#CCCCCC">
     <td><font face="Verdana, Arial, Helvetica, sans-serif" size="2">
&nbsp;Member Name: </td>
     <td><INPUT type="text" name="member" tabindex="1" size="30"></td>
    </tr>

<br><br>

    <tr bgcolor="#EEEEEE">
     <td><font face="Verdana, Arial, Helvetica, sans-serif" size="2">
&nbsp;Rank: </td>
     <td>
        <select name="rank">
        <option value="Member">Member</option>
        <option value="Senior Member">Senior Member</option>
        <option value="Council">Council</option>
        </select>
     </td>
    </tr>

  </TABLE>
  <TABLE width="80%" valign="center">
    <tr bgcolor="#CCCCCC">
     <td><center>
        <INPUT type="submit" tabindex="5" name="submit" value="Add Member">
       </center></td>
    </tr>
  </TABLE>
 </FORM>
<br><br>
[<a href="Loggedin.php" target="iframe"> Back to Admin CP </a>]
<br>
<a href="../members.php" target="iframe">Members List</a>
</div>

<?php

 }else if($action == "submit") {

include('mysql_connect.php');

 if(empty($member)) exit("Please Select a Rank and provide a Member to Add");

// add slashes
$member = addslashes($member);

// add data into database
mysql_query("INSERT INTO members (member,rank) VALUES ('$member','$rank')") or die(mysql_error());


echo "Successfully Added $member as $rank<br><br><a href=\"Loggedin.php\" target=\"iframe\">Return to the Admin Panel</a>";
}
?>

members.php

Code:

<html>
<head>
<title>Members List</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<style>
<!--
a{text-decoration:none}
//-->
</style>
<style>
a:link{color:orange; text-decoration:underline; }
a:hover{color:orange; text-decoration:underline; }
</style>
</head>

<BODY BGCOLOR=#332815 background="images/image_22.gif" link="#000000" vlink="#F00000" bgproperties="fixed" LEFTMARGIN=2 TOPMARGIN=0 MARGINWIDTH=0 MARGINHEIGHT=0>
<?php
include('admincp/mysql_connect.php');

$query = mysql_query('SELECT * FROM members ORDER BY id DESC');

while($row = mysql_fetch_object($query)) {

$rank = $row->rank;
$name = $row->member;

 if($rank == "Member") {
$member_list = $name;}

 else if($rank == "Senior Member") {
$senior_list = $name;}

 else if($rank == "Council") {
$council_list = $name;}
?>

<table width="100%" border="0" cellspacing="0" cellpadding="3">
  <tr> 
    <td><b><font face="Verdana, Arial, Helvetica, sans-serif" size="2">Members List</font></b></td>

  </tr>
  <tr>
    <td bgcolor="#F5F5F5">
<table align="center" width="100%" border="0">
  <tr>
    <td align="center"><font face="Verdana, Arial, Helvetica, sans-serif" size="2"><br>

<u><b>Leader and Founder</b></u><br>Dahnkuana[aD]

     <br><br>

<u><b>Council</b></u><br>
<?php echo $council_list; ?><br>

     <br><br>

<u><b>SeniorMembers</b></u><br>
<?php echo $senior_list; ?><br>

     <br><br>

<u><b>Members</b></u><br>
<?php echo $member_list; ?><br>

     <br><br>

</font>
    </td>
  </tr>
</table>
    </td>
  </tr>
</table>

</body>
</html>

<?php
}
?>

Can someone please tell me what i'm doing wrong?

scrowler · 04-12-2005

the second page doesn't load :O

ViciOuS · 04-12-2005

Yeah, currently theres something wrong with my server. I'm going to have to talk to my host and see what he can do about it. My whole site isn't working right and I can't get into my cpanel :-\

Anyways, I will bump this topic once my server is fixed unless you can figure out whats wrong with my code without seeing that page.

ViciOuS · 04-13-2005

Okay, my site is working and you can view that page now.

Thanks for your patience in advance.

scrowler · 04-13-2005

Code:

<?php

mysql_connect("localhost", "user", "pass");
mysql_select_db("db_name");

$main = "SELECT * FROM members WHERE rank = ";

# edit the ends of the following to mimic the MySQL stored data that determines ranking.
# edit the keys of the array to mimic the desired output ranking.
$sql["Leader"] = $main . "Leader";
$sql["Council"] = $main . "Council";
$sql["Senior Members"] = $main . "SeniorMembers";
$sql["Members"] = $main . "Members";

echo '<div style="text-align: center">';

foreach($sql as $key => $current){

echo "<p><u>$key</u></p><p>";

$do = mysql_query($current);
while($row = mysql_fetch_assoc($do)){
echo $row['name'] . '<br />';
}

echo '</p><p>&nbsp;</p>';

}

?>

try that

ViciOuS · 04-13-2005

Okay, here is my code for members.php:

Code:

<?php
include('admincp/mysql_connect.php');

$main = "SELECT * FROM members WHERE rank = ";

# edit the ends of the following to mimic the MySQL stored data that determines
# the ranking.
# edit the keys of the array to mimic the desired output ranking.
$sql["Council"] = $main . "Council";
$sql["Senior Members"] = $main . "Senior Member";
$sql["Members"] = $main . "Member";

echo '<div style="text-align: center">';

foreach($sql as $key => $current){

echo "<p><u>$key</u></p><p>";

$do = mysql_query($current);
while($row = mysql_fetch_assoc($do)){
echo $row['member'] . '<br />';
}

echo '</p><p>&nbsp;</p>';

}

?>

Even though i'm not exactly sure how $current works in foreach($sql as $key => $current){

I still changed everything else that needed to be changed to fit my MySQL db and etc.

Check the page to see the error.

scrowler · 04-13-2005

sorry, i missed a bit, change this:

Code:

# after the last }
echo '</div>';

also, foreach is a constructor built for arrays, $sql is the array representor, $key is the dynamic variable that represents the name of the key in each iteration of the array, and $current will contain the current value (SQL code) as it iterates through the array.

this appears to be an SQL error, try modifying the following:

Code:

$do = mysql_query($current);
# change to
$do = mysql_query($current) or die(mysql_error());

and report if there is an error and if so, what is it?

ViciOuS · 04-13-2005

Now i get: Unknown column 'Council' in 'where clause'

I don't see how thats an unknown column though. Btw, thanks for explaining that foreach() constructor, it helps a lot.

Order · 04-13-2005

try 'council' as opposed to 'Council'.. PHP can be touchy with formatting, caps are evil :P

ViciOuS · 04-13-2005

nah, now it just says: Unknown column 'council' in 'where clause'

lol..

Order · 04-14-2005

I see the problem, $main is a sql return value, not a string. You will need to use one of the mysql functions to turn it into usable values.

ViciOuS · 04-14-2005

Is there somewhere I can learn about these mysql functions that turn sql return values into string? or can you show me?

Either way, It would be of much help.

Order · 04-14-2005

Something like this may help:

PHP Code:


			
$main = "SELECT * FROM members ORDER BY id DESC";



if($ret = mysql_query($main))

{

  while($row = mysql_fetch_array($ret))

  {

    echo '<pre>';

    print_r($row);

    echo '</pre>';

  }

} else {

  die('MySQL Error: ' . mysql_error());

}

Take a look at the results, it will show you what the $row arrays look like for your members, to adapt it to your site, remove the code within the while loop. Boom, instant member list.

Also, I am not one for doing all the hard work, I will only get you where you need to get, my help requires your intelligence, otherwise there is no learning.

ViciOuS · 04-14-2005

Well, first off, I want it to display the right names under the correct rank. Not just a list of all the members. Although I did try and use your code in with mine and here is what I came up with:

Code:

<?php
include('admincp/mysql_connect.php');

$main = "SELECT * FROM members WHERE rank=";

$sql["Council"] = $main . "Council";
$sql["Senior Members"] = $main . "Senior Members";
$sql["Members"] = $main . "Members";

echo '<div style="text-align: center">';

foreach($sql as $key => $current){

echo "<p><u>$key</u></p><p>";


 if($ret = mysql_query($main))
 {
   while($row = mysql_fetch_array($ret))
   {
    echo $row['member'] . '<br />';;
   }
 } else {
   die('MySQL Error: ' . mysql_error());
 } 

 echo '</p><p>&nbsp;</p>';

}
?>

I get an error under the council rank saying:
"You have an error in your SQL syntax. Check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1"

I tryed taking out (WHERE rank=") but it just displays all the members under all the ranks.

scrowler · 04-14-2005

Code:

if($ret = mysql_query($main))
# change to
if($ret = mysql_query($current))