Why Why Why?

Anon

Ok, I've trawled through all the relevent posts on this forum and I'm still stuck.

I keep getting the following error:
Warning: Supplied argument is not a valid MySQL result resource in /var/www/html/dws-support/index.php on line 99

I can't figure out why. Try it 193.122.28.177/index.php
login:bill password:gates

The code is attached below. and line 99 reads:
if (mysql_num_rows($verify_user) > 0) {

If anyone could help I would be grateful.

I've put * in the code where the db user and password go, but they are correct.

Thanks,

Nick Gushlow.

---oOo---

<head>
<title>[Dynamic Web Support] User Login</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">

 echo $verify_user."<br>";
 echo $verify_hash."<br>";
 echo $access."<br>";

// process form
// Check to see if the user exists in the database
if (mysql_num_rows($verify_user) > 0) {
// check to see if the password is correct
if (mysql_num_rows($verify_hash) > 0) {
// check the access level of the user
if ($access == 1) {
// start the session
session_start();
/
// log session
$time_in = time();
mysql_db_query("INSERT INTO sessions
(sess_id,username,time_in)
VALUES
('$PHPSESSID','username','$time_in')
");
/
/
echo " just performed insert (allegedly)<br>";
echo $PHPSESSID;
echo $data[0];
echo $time_in;
/
echo "<script language=\"JavaScript\"> ";
echo "location=\'debug.php\';";
echo "</script>";
} else {
$error_result = 3;
}
} else {
// error result = wrong password
$error_result = 2;
}
} else {
// error result = user does not exist
$error_result = 1;
}
$error_number = mysql_errno();
$error_msg = mysql_error();
echo "MySQL error ".mysql_errno().": ".mysql_error()."<br>";
echo $username." <br>";

switch ($error_result) {
case 0:
echo "<br> There has been an internal error. <br>";
echo " Please contact the";
echo " <a href=mailto:technical@dynamicweb.co.uk>technical team</a>";
echo " at Dynamic Web Solutions. <br>";
exit;
case 1:
echo "<br> No such user (".$username.") exists. <br>";
exit;
case 2:
echo "<br> Wrong password. <br>";
exit;
case 3:
echo "<br> You do not have the correct Access Level. <br>";
echo " Please contact the";
echo " <a href=mailto:technical@dynamicweb.co.uk>technical team</a>";
echo " at Dynamic Web Solutions if you feel this is incorrect.";
echo "<br>";
exit;
}
$error_number = mysql_errno();
$error_msg = mysql_error();
echo "MySQL error $error_number: $error_msg";

}
?>

<form method="post" action="<?php echo $PHP_SELF?>" name="login">
<table width="100%" border="0" cellspacing="0" cellpadding="0" vspace="0" hspace="0" height="100%">
<tr>
<td height="50%"> </td>
</tr>
<tr align="center">
<td height="95">
<table width="400" border="0" cellspacing="0" cellpadding="0" height="94" vspace="0" hspace="0" align="center">
<tr>
<td width="301" height="45" valign="bottom"><img src="im/splash_top.gif" width="301" height="45" border="0" vspace="0" hspace="0"></td>
<td rowspan="3" width="99">
<table width="99" border="0" cellspacing="0" cellpadding="0" height="94">
<tr>
<td><a href="signup.html" onMouseOut="MM_swapImgRestore()" onMouseOver="MM_swapImage('signup','','im/splash_signup_over.gif',1)"><img src="im/splash_signup.gif" width="99" height="15" border="0" vspace="0" hspace="0" alt="signup for caller support" name="signup"></a></td>
</tr>
<tr>
<td><img src="im/splash_right.gif" width="99" height="79" border="0" vspace="0" hspace="0"></td>
</tr>
</table>
</td>
</tr>
<tr>
<td background="im/splash_mspacer.gif" height="25">
<table width="301" border="0" cellspacing="0" cellpadding="0" height="25">
<tr>
<td width="12"> </td>
<td valign="middle">
<FONT size=-1 point-size="7">
<input type="text" name="username" size="15" class="formInputLogin" value="username">
</font>
</td>
<td valign="middle">
<FONT size=-1 point-size="7">
<input type="text" name="passwd" size="15" class="formInputLogin" value="password">
</font>
</td>
<td valign="middle" width="49">
<input type="image" border="0" name="submit" src="im/splash_login.gif" alt="login" width="49" height="10">
</td>
</tr>
</table>
</td>
</tr>
<tr>
<td width="301" height="24" valign="bottom" background="im/splash_bspacer.gif" align="right"><a href="forgotpass.html" onMouseOut="MM_swapImgRestore()" onMouseOver="MM_swapImage('forgotpwd','','im/splash_forgotpwd_over.gif',1)"><img name="forgotpwd" border="0" src="im/splash_forgotpwd.gif" width="194" height="18"></a></td>
</tr>
</table>
</td>
</tr>
<tr>
<td height="50%"> </td>
</tr>
</table>
</form>
<?php
?>
</body>

Anon

Shouldn't you establish the db connection before running the query?

$verify_user = mysql_query("SELECT * FROM user
WHERE username = '$username'");

You should place these two lines ahead of the query above:

mysql_connect($sys_dbhost,$sys_dbuser,$sys_dbpasswd);
mysql_select_db ($sys_dbname);

Hope this help...

Anon

Ok I just tried the code below and it works.
That says to me the the variable $username from the form is wrong, but when I output the variable using echo it shows correctly!

The field username is type varchar(8). Which should be ok.

Help.

Nick.

---oOo---
<?
// Debug / Testing file
//
// 30.10.2000

include ($DOCUMENT_ROOT.'/core/header_user.inc');
$sys_dbhost='localhost'; // MySQK Host Name
$sys_dbuser='**'; // MySQL Username
$sys_dbpasswd='**'; // MySQL User Password
$sys_dbname='dws_support'; // MySQL Database

mysql_connect($sys_dbhost,$sys_dbuser,$sys_dbpasswd);
mysql_select_db ($sys_dbname);

$result = mysql_query("SELECT * from user where username = 'bill'");

if (mysql_num_rows($result) > 0) {
echo "loads of users!! <br>";
} else {
echo "What users?! <br>";
}

Anon

It's a good point. I assumed that as the line
$verify_user = mysql_query("SELECT * FROM user
WHERE username = '$username'");

is only defined before I connect and not called until the line

if (mysql_num_rows($verify_user) > 0) {

which is after I connect, that it wouldn't matter.

Am I looking at this the wrong way? I shall try connecting earlier and see what happens.

Nick.

Anon

Just tried it and it didn't help.

🙁

Anon

Reading manuals can be a big help there :-)

mysql_query() runs the query, it does not define anything.
Thus you must connect before you can mysql_query()

Anon

But I am connecting before I run the query.

mysql_connect($sys_dbhost,$sys_dbuser,$sys_dbpasswd);
mysql_select_db ($sys_dbname);

if ($submit_x) {

echo $verify_user."<br>";
echo $verify_hash."<br>";
echo $access."<br>";
// process form
// Check to see if the user exists in the database
if (mysql_num_rows($verify_user) > 0) {
.....

Where $verify_user is the query.

Anon

Well max was right (Thanks max).

In a debug script I just wrote that just takes a username and connects and compares; if I connect after I define the query (whether I've called it or not) I get the same error. If I connect before I don't.

BUT I've moved the connect in my script to the front of the code (just after I define the connection details) and I STILL get the error.

?? Much wierdness ??

Anon

HMmm you might want to read the manual again.

$database_link=mysql_connect($sys_dbhost,$sys_dbuser,$sys_dbpasswd);

mysql_select_db ($sys_dbname);

$sql_query="select * from table";

$results=mysql_query($sql_query,$database_link);

print "total results found:".mysql_num_rows($results);

while ($row=mysql_fetch_array($results,MYSQL_ASSOC))
{
print $row["field1"];
print $row["field2"];
};

Anon

Vincent, both you and max were right about connecting first. (see my post in this thread entitled 'Ok making some progress)

Trouble is even when connecting is the first thing my script does I still get the error.

I'm well confused. 🙁

Anon

did you print the mysql errors:

echo mysql_errno().": ".mysql_error()."<BR>";

this will tell you exactly what the errors are.

Anon

Yeah.

Error number is 0 and it doesn't give me a message.

Any help?

Anon

Ok so mysql error number 0 means there is no error.

I trawled through my code and found that during my attempts to 'fix' the problem I removed some qoutes ('). I put them in and that combined with moving the connection sequence solved the problem.

Cheers Guys.

If anyone else has this problem remember:
Always connect to the database BEFORE doing anything else.
🙂
-- Nick.

Anon

Nick, I had the same problem and was told it had to be extra quotes or an extra braket or to do it another way. Well, as I think we both know, it should work and it is the best and easiest way so I kept looking. I don't know if this will help you but in my case I found a column name that was incorrect in my select. eg....

$numresults = mysql_query("SELECT * FROM table1 WHERE(cuisine='$cuisine')");

In my case I added some columns and therefor I actually needed to say (cuisine1='cuban') in the select.

I would double check all your selects and the spelling used in them. This is especially true when variables are used in a select
eg....
(cuisine='$cuisine') or even

($dynamic_column_variable='$something')

which is the case with my problems.

The select may look and be right at the select but if the column name specified is variable then you can easily assign an incorrect column name to it and the selct will still be correct and still not the problem.

Hope this helps!
Mark

Anon

Thanx for the tip Mark, I have noted and logged it.

I'm rather glad I'm not the only one who has had problems.

I believe the solution to my problem was just the position of my mysql connect statements (they weren't at the front of the script) and some missing quotes. Can't quite remember now, but I'm pretty sure.

Nick.