Php and sql question...long shot tho.

oops - my mistake. Change this line:

$userid = (int)GET['userid'];

PHP:

to this:

$userid = (int)$_GET['userid'];

PHP:

Brew

 

ok ta, still get this - Parse error: parse error, unexpected '['

 

What line do you get the error on ?

Brew

 

Line 7 - $userid = (int)GET_['userid'];

 

Change that line to this:

$userid = (int)$_GET['userid'];

PHP:

Brew

 

Again I get this:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource

 

Then I guess you have no gifts for that user. It's a bit difficult to debug without you showing exactly what the value of $userid is and what data you have in your database. Can you provide this ?

Brew

 

Here is some data from the tables. Maybe a pic would have been better?!


giftlogtable

Logid Giftid Recipientid Fromid   
7 1 nat12 Helen Edit Delete 
10 1 nat12 helen Edit Delete 
11 1 helen Nat12 Edit Delete 
12 1 helen helen Edit Delete 

HTML:

gifts table

Giftid Name Image   
1 Kiss <img src="/gifts/Kiss.png"> Edit Delete 
2 Hot <img src="/gifts/hot.png"> Edit Delete 
3 Leaf <img src="/gifts/leaf.png"> Edit Delete 

HTML:

 

Ok, so far-

*user send info from the form into the giftlog table, info goes in fine.

*When user tries to see what gifts have been sent to them they is an error.(many infact!)

 

And how are we supposed to help if we don't know what these errors are?

Please post all error messages. And the exactly code you're using.

 

Ok here is every thing I have-

Basic test form to submit data to table -

<form action="gift.php" method="post">
<input type="text" name="giftid" value="$name">
<input type="text" name="recipientid" value="$recipientid">
<input type="text" name="username" value="$username">
<input type="submit" name="Submit" value="Submit" class="mybutton">
</form>

HTML:

<form action="gift.php">from above !

<?
require ("config.php");
include("functions.php");



// Insert a row of information into the table "giftlog"
mysql_query("INSERT INTO giftlog
(Giftid, Recipientid,Fromid) VALUES('$giftid', '$recipientid', '$username' ) ")
or die(mysql_error());



echo "Data Inserted!";

?>

PHP:

To view gifts in member area -

<?
$username = (int)$_GET['username'];
$result = mysql_query("SELECT giftlog.name AS name, giftlog.image
AS image FROM giftlog
LEFT JOIN gifts
ON giftlog.giftid = gifts.giftid
WHERE giftlog.recipientid=" . $username);

while ($row = mysql_fetch_assoc($result))
{
echo $row['name'];
echo $row['image'];
}
?>

PHP:

error message- Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource



Also please not I have change userid to username, I am playing around with this so either change it back or ignore it.

 

OK, the problem seems to be that $userid is not a numeric value. Try replacing this:

$userid = (int)$_GET['userid'];

PHP:

with this:

$userid = mysql_real_escape_string( $_GET['userid'] );

PHP:

and call your page like this:

{page name}.php?userid=helen

Brew

 

hhhmmm still getting the same error message -

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource.


Would it change anything if I changed userid to username as thats is whats used on the site ?!

 

Yes, however if you are going to change it to this:

$username = mysql_real_escape_string( $_GET['username'] );

PHP:

The you will have to call the page like this:

{page name}.php?username=helen

Brew

 

Ok, but really that shouldnt effect it either way. ?


Anyhow im still get the same message. - Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource ..