supplied argument is not a valid MySQL result. Ehh?

[itandme]

This is a page for the guilds to edit their info. But it keeps giving me "not a valid resource". Am I not using fetch_array on the correct way?

<?
#CONNECT
$conn = mysql_connect ("localhost","","**") or die
("Could not create a connection");
$db = mysql_select_db("lorens_sfs",$conn);

#DISPLAY
$sql = "SELECT name,abbrev,charter,website,location,uosite,password,rp,pk,pvp,faction,align,waring FROM guilds WHERE id = $id";
$result = mysql_query($sql);

$info = mysql_fetch_array($result, MYSQL_ASSOC);
$name = $info['name'];
$abbrev = $info['abbrev'];
$charter = $info['charter'];
$location = $info['location'];
$website = $info['website'];
$uosite = $info['uosite'];
$rp = $info['rp'];
$pk = $info['pk'];
$pvp = $info['pvp'];
$faction = $info['faction'];
$alignment = $info['alignment'];
$waring = $info['waring'];

#IF SAVE
if (($REQUEST_METHOD=='POST')) {
$selection = "SELECT password FROM guilds WHERE id = $id" or die ("cant do it");

$execute = mysql_query($selection);
$row = mysql_fetch_array($execute, MYSQL_ASSOC);
$password = $row['password'];
if ($pass == $password) {

$write = "UPDATE guilds SET name='$name', charter='$charter', abbrev='$abbrev', location='$location', website='$site', uosite='$uosite', rp='$rp', pk='$pk', pvp='$pvp', faction='$faction', alignment='$alignment', waring='$waring' WHERE id=$id";
$insert = mysql_query($write);
header("Location: saved.php?name=$name");
}
else {
header("Location: saved.php?id=$id&name=$name&charter=$charter&abbrev=$abbrev&location=$location&site=$site&uosite=$uosite");
}
}

?>

[trooper]

What that means is that your initial sql query failed due to the syntax of the query (perhaps you are calling for a non existant field)

where you have $result = mysql_query($sql) change it to

$result = mysql_query($sql) or die(mysql_error());

then if the sql is incorrect it will tell you the problem with the query

ALWAYS ALWAYS check that calls to mysql functions are successfull by putting the "or die(mysql_error())" after the call to the function. A few extra letters after a call to a function will save a LOT OF HEARTACHE later.

HTH

GM

[itandme]

ok, i did so thanks.

now it gives me "You have an error in your SQL syntax near '' at line 1"

Eh? Is that IN the mysql table? LINE 1?

[nnichols]

The ref to LINE 1 means the first line of the query - as mysql only sees the query, not the rest of your php code.

Run the query from the mysql client on the command line or through a GUI so you can get the syntax right. The most likely thing is a misspelled column_name.

In your code above you have "password" with a line break in it - that would be enough to cause the syntax error.

Hope this helps 😉

[itandme]

Well, here comes whats really odd.

The code DOES DISPLAY the guild info.. You can go to http://lorenscorn/guilds/editguild.php?id=28 and check it out. So code that connects to the database, get the info, and display it works just fine. I also ran the first SELECT query and the UPDATE query in the phpmyadmin, and they both worked just fine!

Its when I try to UPDATE it, when the syntax error comes! I have checked the spellings a hundred times, and I actually found two misstakes, altough correcting them gave me no result.

<?
$conn = mysql_connect("localhost","gggggg","ggggggg") or die
(mysql_error());
mysql_select_db(lorens_sfs) or die(mysql_error());

$query = "SELECT name,abbrev,charter,website,location,uosite,password,rp,pk,pvp,faction,align,waring,email FROM guilds WHERE id = $id";
$result = mysql_query($query) or die(mysql_error());

$info = mysql_fetch_array($result);
$name =$info['name'];
$abbrev =$info['abbrev'];
$charter = $info['charter'];
$location = $info['location'];
$website = $info['website'];
$uosite = $info['uosite'];
$rp = $info['rp'];
$pk = $info['pk'];
$pvp = $info['pvp'];
$faction = $info['faction'];
$alignment = $info['align'];
$waring = $info['waring'];
$email = $info['email'];
$password = $info['password'];

THE ABOVE CODE SEEMS TO WORK!!!

if (($REQUEST_METHOD=='POST')) {
if ($pass == $password) {
$write = "UPDATE guilds SET name='$name', charter='$charter', abbrev='$abbrev', location='$location', website='$site', uosite='$uosite', rp='$rp', pk='$pk', pvp='$pvp', faction='$faction', align='$alignment', waring='$waring' WHERE id=$id";
$insert = mysql_query($write);
header("Location: savedornot.php?page=saved&name=$name");
}
else {
header("Location: savedornot.php?page=passerror&id=$id&name=$name&charter=$charter&abbrev=$abbrev&location=$location&site=$site&uosite=$uosite");
}
}

?>

Thanks!

[nnichols]

Try this and see what errors it prints -

if (($REQUEST_METHOD=='POST')) {

if ($pass == $password) {
	$write = "UPDATE guilds SET name='$name', charter='$charter', abbrev='$abbrev', location='$location', website='$site', uosite='$uosite', rp='$rp', pk='$pk', pvp='$pvp', faction='$faction', align='$alignment', waring='$waring' WHERE id=$id";
	$insert = mysql_query($write)or die('Update query failed.<BR>Error: ' . mysql_error() . '<BR>Query: ' . $write);
	header("Location: savedornot.php?page=saved&name=$name");
	}
else {
	header("Location: savedornot.php?page=passerror&id=$id&name=$name&charter=$charter&abbrev=$abbrev&location=$location&site=$site&uosite=$uosite");
	}
}

If it fails and outputs the errors as I added them - check that the syntax of the printed query is correct.

Hope this helps 😉

EDIT: there was a "." missing

[itandme]

hmm, that gives me a parse error as the die( ) are not closed properly with the " before the last ). And I cant close it as it will be $insert = mysql_query($write)or die('Update query failed.<BR>Error: ' mysql_error() . '<BR>Query: ' . $write');, which gives me a parse error aswell.

However i named all errors differently so I could see what one it was who erro..red..

It erro..red on: $query = "SELECT name,abbrev,charter,website,location,uosite,password,rp,pk,pvp,faction,align,waring,email FROM guilds WHERE id = $id";
$result = mysql_query($query) or die("RESULT"); <-- THIS GAVE ERROR

The little boring this is, that its totally impossible. That query was run when displaying the info(sucessfully), and i can not see how it can error at the time I dont even use it.

This makes no sense at all.. Right? uh?

[nnichols]

The code I gave you can't cause a parse error (OOPS).

I suggest you post your full code, and make sure you put it between [ PHP][/code] tags so it is easy to read.

[itandme]

Ahh sorry it doesnt give me a parse error i out it in the wrong query, sorry. However it still keeps giving me the You have an error in your SQL syntax near '' at line 1 thingy.

<?
$conn = mysql_connect("localhost","lorens_lennart","postbil") or die
("DIE");
mysql_select_db(lorens_sfs) or die(mysql_error());

$query = "SELECT name,abbrev,charter,website,location,uosite,password,rp,pk,pvp,faction,align,waring,email FROM guilds WHERE id = $id";
$result = mysql_query($query) or die(mysql_error());

$info = mysql_fetch_array($result);
$name =$info['name'];
$abbrev =$info['abbrev'];
$charter = $info['charter'];
$location = $info['location'];
$website = $info['website'];
$uosite = $info['uosite'];
$rp = $info['rp'];
$pk = $info['pk'];
$pvp = $info['pvp'];
$faction = $info['faction'];
$alignment = $info['align'];
$waring = $info['waring'];
$email = $info['email'];
$password = $info['password'];


if (($REQUEST_METHOD=='POST')) {
if ($pass == $password) {
$write = "UPDATE guilds SET name='$name', charter='$charter', abbrev='$abbrev', location='$location', website='$site', uosite='$uosite', rp='$rp', pk='$pk', pvp='$pvp', faction='$faction', align='$alignment', waring='$waring' WHERE id=$id";
$insert = mysql_query($write)or die('Update query failed.<BR>Error: ' . mysql_error() . '<BR>Query: ' . $write);
header("Location: savedornot.php?page=saved&name=$name");
}
else {
header("Location: savedornot.php?page=passerror&id=$id&name=$name&charter=$charter&abbrev=$abbrev&location=$location&site=$site&uosite=$uosite");
}
}

?>

[nnichols]

Try this -

<?
$conn = mysql_connect("localhost","lorens_lennart","postbil") or die("DIE");
mysql_select_db("lorens_sfs") or die('Failed to select db: ' .mysql_error());

$query = "SELECT  name,abbrev,charter,website,location,uosite,password,rp,pk,pvp,faction,align,waring,email FROM guilds WHERE id = $id";
$result = mysql_query($query) or die('SELECT query failed.<BR>Error: ' . mysql_error() . '<BR>Query: ' . $query);

$info = mysql_fetch_array($result);
$name =$info['name'];
$abbrev =$info['abbrev'];
$charter = $info['charter'];
$location = $info['location'];
$website = $info['website'];
$uosite = $info['uosite'];
$rp = $info['rp'];
$pk = $info['pk'];
$pvp = $info['pvp'];
$faction = $info['faction'];
$alignment = $info['align'];
$waring = $info['waring'];
$email = $info['email'];
$password = $info['password'];


if (($REQUEST_METHOD=='POST')) {

if ($pass == $password) {
	$write = "UPDATE guilds SET name='$name', charter='$charter', abbrev='$abbrev', location='$location', website='$site', uosite='$uosite', rp='$rp', pk='$pk', pvp='$pvp', faction='$faction', align='$alignment', waring='$waring' WHERE id=$id";
	$insert = mysql_query($write)or die('UPDATE query failed.<BR>Error: ' . mysql_error() . '<BR>Query: ' . $write);
	header("Location: savedornot.php?page=saved&name=$name");
	}
else {
	header("Location: savedornot.php?page=passerror&id=$id&name=$name&charter=$charter&abbrev=$abbrev&location=$location&site=$site&uosite=$uosite");
	}
}

?>

[itandme]

it gives me this :

SELECT query failed.
Error: You have an error in your SQL syntax near '' at line 1
Query: SELECT name,abbrev,charter,website,location,uosite,password,rp,pk,pvp,faction,align,waring,email FROM guilds WHERE id = 

That cant be right! I have a guy on my ICQ thats saying that the code is loosing the $id variable.... Can that be right?

[nnichols]

YES. That is exactly what it means. The script isn't receiving the $id var. Where does it come from? Do you have register_globals on? Should it be $_POST['id']?

[itandme]

I put <form method="post" action="editguild.php?id=29"> in the form action, and that DID work, altough the update query gave me error..

HOW can that be? Should not the code keep the id variable?

[nnichols]

What was the error from the update?

[itandme]

Its not my server...

For some reason <form action="editguild.php?id=$id"> did NOT work.. the heck..

[itandme]

the problem might be solved puting

<form method="post" action="editguild.php?id=<? echo $id; ?>">

and picking up the id from the query aswell..

must walk my dog now.. argh.. bleeding

[itandme]

But how should I do this?
I need to send the id and password aswell. But I cant use hidden fields as they are shown in the source, and the password needs to be secret... Sooo.. Whats the options?uh.. thanks!