User input script fails. each user belongs to one our more groups.

LordGarfield · Dec 16, 2003

Hi all,

I post my source code here

This one works.
********user_i.php*********************
<html>
<body>
<table>
<form method="post" action="user_o.php">
<tr>
<td>GEBRUIKERSNAAM:</td><td><input type="text" name="unaam"></td>
</tr>
<tr>
<td>NAAM:</td><td><input type="text" name="anaam"></td>
</tr>
<tr>
<td>VOORNAAM:</td><td><input type="text" name="vnaam"></td>
</tr>
<tr>
<td>PASWOORD:</td><td><input type="password" name="paswoord"></td>
</tr>
<tr>
<td>PASWOORD NOGMAALS:</td><td><input type="password" name="paswoordc"></td>
</tr>

<tr>
<td colspan=2>IS LID VAN</td>
</tr>

<?php
$con = mysql_connect("localhost", "root", "&quot

;
mysql_select_db("rcomdb&quot

;
$rec = mysql_query("select * from groups&quot

;
$i=1;
while (list($id, $gnaam) = mysql_fetch_row($rec))
{
echo "<tr><td>" . $gnaam . "</td><td><input type='checkbox' name='groep[]' value='" . $id . "'></td></tr>";
$i=$i + 1;
}
?>
<tr>
<td>
<input type="submit" name="Submit" value="Toevoegen">
</td>
</tr>
</form>
</table>
</body>
</html>
****************end user_i.php***********************
This one has the error's
***************user_o.php**********************

<?php
$unaam = $_POST['unaam'];
$paswoord = $_POST['paswoord'];
$paswoordc = $_POST['paswoordc'];
$anaam = $_POST['anaam'];
$vnaam = $_POST['vnaam'];
echo $unaam . $paswoord . $anaam . $vnaam;
if ($unaam <> '' && $paswoord <> '' && $anaam <> '' && $vnaam <> '')
{

if ($paswoord == $paswoordc)
{

$db = mysql_connect("localhost","root","&quot

;
mysql_select_db ("rcomdb&quot

;
$sql = "select usern from users where usern = ". $unaam;
$result1 = mysql_query($sql);
if (mysql_num_rows($result1)>0)
{
echo 'Gebruiker bestaat reeds';
}
else
{
$sql = "insert into users (usern, userp, naam, vnaam) values (" .$unaam. "," .$paswoord. "," .$anaam. "," .$vnaam. &quot

";
$result2 = mysql_query($sql);
echo 'Gebruiker Toegevoegd';
$result3 = mysql_query("select userid from users where usern = $unaam&quot

;
$userid = mysql_fetch_row($result3);
$result4 = mysql_query("select groupid from groups&quot

;
$tot_row = mysql_num_rows($result4);
for ($i=0; $i < $tot_row; $i++)
{
$groupid = $_POST['groep'][$i];
$result5 = mysql_query("insert into usergroup (userid, groupid) values ( $userid, $groupid)&quot

;
}
}
mysql_close();
}
else
{
echo 'Paswoorden komen niet overeen.';
}
}
else
{
echo 'U moet alle velden invullen.';
}
?>

*****************end user_o.php*********************

And I shall post the php myadmin exports of the tables that are used.

CREATE TABLE `users` (
`userid` int(11) NOT NULL auto_increment,
`usern` varchar(100) NOT NULL default '',
`userp` varchar(100) NOT NULL default '',
`naam` varchar(100) NOT NULL default '',
`vnaam` varchar(100) NOT NULL default '',
PRIMARY KEY (`userid`)
) TYPE=MyISAM CHARSET=latin1 AUTO_INCREMENT=2;

CREATE TABLE `groups` (
`groupid` int(11) NOT NULL auto_increment,
`naam` varchar(100) NOT NULL default '',
PRIMARY KEY (`groupid`)
) TYPE=MyISAM CHARSET=latin1 AUTO_INCREMENT=4 ;

INSERT INTO `groups` VALUES (1, 'Administratie');
INSERT INTO `groups` VALUES (2, 'Opvoeders');
INSERT INTO `groups` VALUES (3, 'Onderhoud');

CREATE TABLE `usergroup` (
`ugid` int(11) NOT NULL auto_increment,
`userid` int(11) NOT NULL default '0',
`groupid` int(11) NOT NULL default '0',
PRIMARY KEY (`ugid`)
) TYPE=MyISAM CHARSET=latin1 AUTO_INCREMENT=1 ;

The database is called rcomdb

When I execute this script I get this output

testtesttesttest
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in c:\inetpub\

http://wwwroot\website1\user_o.php

on line 18
Gebruiker Toegevoegd
Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in c:\inetpub\

http://wwwroot\website1\user_o.php

on line 28

Notice: Undefined offset: 2 in c:\inetpub\

http://wwwroot\website1\user_o.php

on line 33

Notice: Undefined offset: 3 in c:\inetpub\

http://wwwroot\website1\user_o.php

on line 33

Also there is no user created in the database so it seems that even the insert into is failed but it gave not an error.

can anyone helping me??

I installed PHP 4.3.3.3 and MySQL 4.1 on IIS (Standard version on winXP pro I think it is 5)

Tank you so much.

hos2 · Dec 17, 2003

I would write the queries like this

Code:

$sql = &quot;select usern from users where usern = '$unaam'&quot;;

$sql = &quot;insert into users (usern, userp, naam, vnaam) values ('$unaam','$paswoord','$anaam','$vnaam')&quot;;
            $result2 = mysql_query($sql);

            $result3 = mysql_query(&quot;select userid from users where usern = '$unaam'&quot;);

don't know if that will solve all your problems

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User input script fails. each user belongs to one our more groups.

LordGarfield

IS-IT--Management

hos2

Programmer

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