Re:[SOLVED] PHP Newbie Database Records?

André, Thank you for your patience, solution one works fine. Regards Col "André cerqueira" <accerqueira@superig.com.br> wrote in message news:20040216051618.7697.qmail@pb1.pair.com... > -----BEGIN PGP SIGNED MESSAGE----- > Hash: SHA1 > > i think the query: "SELECT count(uid) FROM users WHERE > username='$username'", will give a one row result, that row will have > one field, and that field is the number of rows that match the where > clause (you are using count(uid)) > if there is no match, the result will be a row, with a field, with 0 as > value > > mysql_num_rows return the number of rows of the given result > > you can do this: > > $result = mysql_query("SELECT count(uid) FROM users WHERE username = '$username'") > > or die("Invalid query: " . mysql_error()); > > $num_rows = mysql_result($result, 0, 0); > > or this: > > $result = mysql_query("SELECT uid FROM users WHERE username = '$username'") > > or die("Invalid query: " . mysql_error()); > > $num_rows = mysql_num_rows($result); > > i think the first one is better > > > Petcol wrote: > > Andre, > > > > Thank you, I had read the manual, understanding it fully when you've > > been coding in another language is another matter. > > > > I had relied on DreamweaverMX to create the Database connection which > > > > > > > meant > > I didn't really understand how it was returning the results. > > > > I was able to output the results by altering the code DreamweaverMX > > had created. > > > > Thanks, now using the database query method you showed, I've been > > able to get the number of rows returned. > > > > However, whenever I change the username I still get a row count of 1 > > ? <?php if (isset($_POST["USERNAME"])) { $username = > > $_POST["USERNAME"]; > > > > $result = mysql_query("SELECT count(uid) FROM users WHERE username = > > '$username'") or die("Invalid query: " . mysql_error()); > > $num_rows = > > mysql_num_rows($result); echo $username ."&nbsp;&nbsp;\n". $num_rows; > > } ?> > > > > echo $username returns the username entered into the field before the > > submit button > is hit. > > $num_rows; always returns one numeric 1 even though there is no > > username like > $username > > in the database ?????? > > > > Have I missed something? > > > > Col > > > > "André cerqueira" <accerqueira@superig.com.br> wrote in message > > news:20040216023035.29662.qmail@pb1.pair.com... > > > > I believe you forgot to read the manual. > > > > > >> $query_qUsername = "SELECT * FROM users WHERE username = > >> '$username'"; $num_rows = mysql_num_rows($query_qUsername); > > > > > > $query_qUsername is a String > > > > mysql_num_rows want a result resource as argument You have no result > > till you do mysql_query > > > > > > Read these please: http://www.php.net/mysql > > http://www.php.net/mysql_query > > > > > > > > Petcol wrote: > > > >> PHP Newbie, > > > >> I'm trying to find out if a username already appears in my > >> database. > > > >> I have two users in my table, one is colin so when I enter a > >> username of colin I would normally under ColdFusion anyway just to > >> something like: <cfif $query_qUsername.RecordCount GT 0> User > >> already exists do something else, like throw the user back an > > > >> error. > > > >> </cfif> > > > >> My Query works $query_qUsername = "SELECT * FROM users WHERE > >> username = '$username'"; > > > >> I've tried a number of different things here and everything either > >> comes back with nothing, or throws an error. $num_rows = > >> mysql_num_rows($query_qUsername); echo $num_rows; // this comes > >> back with nothing, I was expecting a > > > >> numeric > > > >> of 1 exit; > > > >> Regards Col > > > -----BEGIN PGP SIGNATURE----- > Version: GnuPG v1.2.4 (MingW32) > Comment: Using GnuPG with Mozilla - http://enigmail.mozdev.org > > iD8DBQFAMFIhaxdA/5C8vH8RAocmAKCiJsq+ZjfwKT8+Nh/AbNQ9vB5ZOwCeKCRd > hKO0x1rckabl8/++P995sCE= > =2epc > -----END PGP SIGNATURE-----