sql

[pinehead18]

$db = mysql_select_db("pinehead", $con) or die(mysql_error());
$sql = "INSERT INTO users (uname) VALUES ('$uname')";
$sth = mysql_query($sql) or die(mysql_error());

Ok, the var $uname will echo the correct information associated with it. The code will add another row to the db however it does not add anycontent so lets say $uname = "me";

The script will ad another line without anything under uname so "me" is not getting into the db.

Any ideas?
Thanks
Anthony

[drew010]

$sql = "INSERT INTO users (uname) VALUES ('$uname')";

change that to:

$sql = "INSERT INTO users (uname) VALUES ($uname)";

try pulling the single quotes off, its looking at that as a literal string i think.

[HalfaBee]

It looks OK.
Try echoing the $sql and see what it shows.

I prefer "INSERT INTO users SET uname='$uname' "
but that is just me 🙂

HalfaBee

[HalfaBee]

If you don't put the quotes it looks for a column called $uname ( me ) in this case.

HalfaBee

[pinehead18]

 
$sql = "INSERT INTO users SET uname='$uname'";

that doesn't seem to work either.. Just posts a blank row.

[HalfaBee]

Did you echo the $sql?

HalfaBee

[pinehead18]

yes, it echo's the right name. So it seems to be working.. but still doens't post to the db

[HalfaBee]

The SQL is correct.
How are you checking the contents of the db?

HalfaBee

[pinehead18]

I have webmin.. it is a compelete db controler.

[HalfaBee]

Well you have me stumped!

HalfaBee

[pinehead18]

lol i'm stumped as well.. and quite pissed

[HalfaBee]

Could it be how your table is created?

Maybe uname is an INT and defaulting to NULL.

HalfaBee

[pinehead18]

its a var char

[HalfaBee]

Well, unless you want to give me ftp & mysql access I don't think I can fix your problem.

HalfaBee