MySQL

aenirtower

hi, I'm a newbie to mysql, and I need help with this error:

Warning: Supplied argument is not a valid MySQL result resource in c:\instantmessenger\window.php on line 47

Warning: Supplied argument is not a valid MySQL result resource in c:\instantmessenger\window.php on line 48

can you tell me how to fix this error?

siwis

Look at your code on the lines quoted.

Generally, you get that error when you list() after a query and yet the query has failed for one reason or another, or if you attempt to query using a non existent or invalid variable.

Post the code here and I'm sure some genius will spot it for you :-)

aenirtower

<script lnaguage="javascript">

var limit="0:5"
if (document.images){
var parselimit=limit.split(":")
parselimit=parselimit[0]60+parselimit[1]1
}
function beginrefresh(){
if (!document.images)
return
if (parselimit==1)
window.location.reload() && window.scroll(0,20000)
else{
parselimit-=1
curmin=Math.floor(parselimit/60)
cursec=parselimit%60
if (curmin!=0)
curtime=curmin+" minutes and "+cursec+" seconds left until page refresh!"
else
curtime=cursec+" seconds left until page refresh!"

setTimeout("beginrefresh()",1000)
}
}

window.onload=beginrefresh

</script>
<body >
<?
include ("mysql.php");
$query="select from users where username='$username' and status='yes'";
$result=mysql_query($query, $link);
if($result!=0)
{
print"<center><body bgcolor=\"000066\"> $username you have logged out ";
exit;
}
else
{
$query="delete from chatdata where message =' ' or buddy='to '";
mysql_query($query,$link);
$query="select sender, buddy, message from chatdata where sender = '$username' or sender like '%$buddy%' and buddy ='$buddy' or buddy like '%$username%' order by id desc";
$result=mysql_query($query, $link);
$num_rows=mysql_num_rows($result);
while($fresult=mysql_fetch_array($result))
{
print"$fresult[0] $fresult[1]$fresult[2]";
print" ";
}
}
?>

ben_g

in one of your lines you have a single quote arround a varriable name. You don't need it.

$query="select from users where username='$username' and status='yes'

aenirtower

Nope .. I still get it.

siwis

Assuming the code lines tally, the sql spitting out is:
$query="select sender, buddy, message from chatdata where sender = '$username' or sender like '%$buddy%' and buddy ='$buddy' or buddy like '%$username%' order by id desc";

Try formatting it with parentheses:
$query="SELECT sender, buddy, message FROM chatdata WHERE ((sender = '$username' or sender like '%$buddy%') and (buddy ='$buddy' or buddy like '%$username%')) ORDER BY id DESC";

No guarantees ;-)

siwis

Actually, this might sound daft, but how are the values for the variables $buddy, $username or are they held in cookies?

Try echoing their values to screen prior to the sql that is causing the probkem, just to check whether they have values correctly set.

aenirtower

okay .. This maybe the leading error:

$query="insert into users set username='$username', password='$password', status='yes'";
$result=mysql_query($query);

now .. this causes the same error as above (scroll up to see the window.php error)

siwis

Originally posted by aenirtower
okay .. This maybe the leading error:

$query="insert into users set username='$username', password='$password', status='yes'";
$result=mysql_query($query);

now .. this causes the same error as above (scroll up to see the window.php error)

$query="INSERT INTO users VALUES ('$username', '$password', 'yes')";
$result=mysql_query($query);

This assumes that your table has 3 columns :-)

aenirtower

but like .. it'll look for "username", "password" and "status" and it'll return undefined.