Unscramble mysql password

Dolemite50

I'm having trouble here. I am storing passwords from a form. They are scrambled. I am unable to come up with the right text to check the password when somebody attempts to login. Through this forum I found PASSWORD() function, but while trying to run it I got an error that it was an undefined function. Another post told me to change that to md5(), which gives an error that the passwords do not match.
Here is my latest version of an attempt.

First page (Signup, data goes into mysql):
$sql = "insert into member set loginName ='$loginName', createDate = '$createDate', password = "md5($password)", email='$email'";

Second page (verification):
$checkp = "SELECT password from $table WHERE loginName = \"$loginName\"" ;
$result2 = mysql_query($checkp, $conn) or die(mysql_error());
$pass = mysql_result($result2, 0);

										if ($pass != md5('$password')){

Thanks for helping a rookie.,

eoghain

Typically what I do is use the MySQL function password() to store excrypted passwords in my database and then do the same thing to check them.

//First page (Signup, data goes into mysql): 
$sql = "insert into member set loginName ='$loginName', createDate = '$createDate', password = password($password), email='$email'";

//Second page (verification): 
$checkp = "SELECT count(*) from $table WHERE loginName = \"$loginName\" AND password = password($password)" ; 
$result2 = mysql_query($checkp, $conn) or die(mysql_error()); 
$pass = mysql_result($result2, 0); 

if ($pass != 0){

Dolemite50

Thank you so much for the help.
I got an error when trying to sign in. Using my name and "testpass" as the password, it created the entry in the mysql table with a scrambled password. When I then tried to login I got:

Unknown column 'testpass' in 'where clause'

Heres the line:

else{
$checkp = "SELECT count(*) from $table WHERE loginName = \"$loginName\" AND password = password($password)" ;

$result2 = mysql_query($checkp, $conn) or die(mysql_error());

$pass = mysql_result($result2, 0);

								if ($pass != 0){

Another prob that I had, on the first line it wouldnt work with password($password) I got a syntax error. When I used single quotes it was able to enter the record.
password('$password') Is that the way it should be or is that the cause of my new problem.

Thanks again.

Dolemite50

btw, when I say it wouldnt work on the first line I was referring to this line:

//$sql = "insert into member loginName ='$loginName', createDate = '$createDate', password = password($password), email='$email'";

eoghain

That's why my signature says none of my code is tested. Yes you need to add single quotes to show that the value of $password is supposed to be taken as a string, and not as a column name. You'll have to quote it in the second query as well.

Dolemite50

You are the man.
Thanks alot for helpin a newb.