PHP Server Side Scripting Forum

echo "the db error is: ".mysql_error($query);

it gave me this:

Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in /home/por/public_html/petition/submit_petition.php on line 72

the db error is:

I notice earlier you have this:

$dbtable = $petition;

So, now does $petition hold any value to pass to the new variable? And my third questions is, why assign something to $dbtable? Why not just the $petition variable in your query?

And welcome to WebmasterWorld GlobalFusion :)

Not a valid MySQL link resource.

You are not sending the query to the database, as the connection is not valid.

Try adding the connection handle in the query:

mysql_query($query, $link);

You could also var_dump($link) to check if it is a MySQL resource.

$dbtable DOES hav a value, and the reason it isnt HARDCODED, is because it is part of a dynamic form that uses a different table for different forms.

I added:
//insert data into DB
mysql_query($query, $link);
var_dump($link);

------------------------------------------
FORM PAGE GOES TO 'SUBMIT_PETITION.PHP'
------------------------------------------

-------------------------------------
Heres The Code (SUBMIT_PETITION.PHP)
-------------------------------------

connecttodb($servername,$dbname,$dbusername,$dbpassword);
function connecttodb($servername,$dbname,$dbuser,$dbpassword)
{
global $link;
$link=mysql_connect ("$servername","$dbuser","$dbpassword");
if(!$link){die("Could not connect to MySQL");}
mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error());
}

//set query
$query = "insert INTO $dbtable (ID, firstname, lastname, email, area, comments, date, time, ip) VALUES ('','$firstname','$lastname','$email','$area','$comments','$date','$time','$ip')";

//insert data into DB
$result = mysql_query($query);

echo "the db error is: ".mysql_error($query);
echo '<br />--------------<br /> '.$query.'<br />--------------------<br /><br />';

mysql_close();

//redirect to confirmation page
include('submit_confirm.php');

------------------------------------------
OUTPUT SHOWN ON 'SUBMIT_CONFIRM.PHP'
------------------------------------------

OutPut Shows as:

Warning: mysql_error(): supplied argument is not a valid MySQL-Link resource in /home/por/public_html/petition/submit_petition.php on line 41
the db error is:
--------------
insert INTO 001 (ID, firstname, lastname, email, area, comments, date, time, ip) VALUES ('','John','Doe','SomeWhere','johndoe@nonames.com','In the end.. this will work..','Sunday 20th of March 2005','02:57:25 PM','202.#*$!.xxx.xxx')
--------------------

---------------------
connecttodb($servername,$dbname,$dbusername,$dbpassword);
function connecttodb($servername,$dbname,$dbuser,$dbpassword)
{
global $link;
$link=mysql_connect ("$servername","$dbuser","$dbpassword");
if(!$link){die("Could not connect to MySQL");}
mysql_select_db("$dbname",$link) or die ("could not open db".mysql_error());
}
return $link;
---------------------

And i tried in the closing bracket too. but still nothing.

The (out of bracket attempt) prevented my 'output' page from showing the data.

*currently running with the 'return $link' inside the brackets.

if you are enclosing a variable name in a single quote, '$var' then try this instead: (enclose in double quotes)

$query = "INSERT INTO.... VALUES ('". $var1 ."', '". $var2 ."');";

Your code should work without any problems, but I find that sometimes this method is better (even if it's more difficult to look at).

No harm in trying... I've never had problems doing it this way.

Tried this:
-----------
//set query
$query = "INSERT INTO '$dbtable' ('ID', 'firstname', 'lastname', 'email', 'area', 'comments', 'date', 'time', 'ip') VALUES ('', '".$firstname."', '".$lastname."', '".$email."', '".$area."', '".$comments."', '".$date."', '".$time."', '".$ip."')";
//insert data into DB
mysql_query($query);

To No Avail either.. :(
------------------------

And This:
---------

//set query
$query = "INSERT INTO '$dbtable' ('firstname', 'lastname', 'email', 'area', 'comments', 'date', 'time', 'ip') VALUES ('".$firstname."', '".$lastname."', '".$email."', '".$area."', '".$comments."', '".$date."', '".$time."', '".$ip."')";
//insert data into DB
mysql_query($query);

Neither Worked..
----------------

:(

The only time it succeded.. was when i made an insert via PhpMyAdmin.

This is the code it produced:

PHPMYADMIN
__________________________________________________
SQL-query:
INSERT INTO `001` ( `ID` , `firstname` , `lastname` , `email` , `area` , `comments` , `date` , `time` , `ip` )
VALUES (
'', 'test', 'user', 'test@user.com', 'this area', 'some comments', 'some date', 'some time', '#*$!.#*$!.xxx.xxx'
);
__________________________________________________

But if i use that EXACT same code in my php file.. it doesnt INSERT. :( could it be something else?

$query = "INSERT INTO `$dbtable` (`firstname`, `lastname`, `email`, `area`, `comments`, `date`, `time`, `ip`) VALUES ('".$firstname."', '".$lastname."', '".$email."', '".$area."', '".$comments."', '".$date."', '".$time."', '".$ip."')";
//insert data into DB
mysql_query($query);

Also, with regards to an earlier post, the return $link; should have been INSIDE the function.

well, i think it was a combination of several things..

But it works now!

Thankyou soooo much for all your help guys!

I have only one thing to add (ask), is the code you posted EXACTLY as it is in your program? If it is then this is probably your problem: