[RESOLVED] Help with mysql query

[2005]

$result = mysql_query("SELECT u.userid, f.favoriteid FROM session u, favorites f WHERE u.userid=f.favoriteid and f.favoriteowner='$id'"); 

This query selects and returns to the user the number of favorites who are online.

"userid" in table "session" may contain ID numbers of users or IP addresses of guests who are not logged in.

I have a favorite with userid 194.
A quest with IP address 194.XX.XX.XX comes to the website. This query for some strange reason shows that I have 1 favorite online.

I don't use "LIKE" statement, so how is it possible that 194 = 194.XX.XX.XX ?

Thanks.

[NogDog]

I think that what is happening is that you are comparing an integer to a string, so MySQL is casting the string to an integer. So '194.XX.XX.XX' is converted to integer 194, since the way strings are cast to an integer is to read from left to right until some character is encountered that cannot be interpreted as part of a numeric value. You may therefore need to cast the first integer value into a string to do the comparison:

. . . AND CAST(f.favoriteowner AS CHAR) = '$id'

[2005]

NogDog, thank you very much.