$check = mysql_query("SELECT * FROM `accounts` WHERE username='". strtolower($username) . "'"); if (mysql_num_rows($check) == 1){ $erros = 1; if ($lang == "pt") { $errmsg .= "A Conta selecionada já existe!<br>"; } else { $errmsg .= "The Selected Account already exists!<br>"; } }
Ele retorna: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/doug/public_html/index.php on line 133
a estrutura desta tabela:
CREATE TABLE IF NOT EXISTS `accounts` ( `username` VARCHAR(16) NOT NULL, `password` CHAR(64) NOT NULL, `fullname` VARCHAR(64) NOT NULL, `location` VARCHAR(64) NOT NULL, `email` VARCHAR(64) NOT NULL, `computer` VARCHAR(64) NOT NULL, `hdid` INTEGER NOT NULL, `regip` VARCHAR(15) NOT NULL, `lastip` VARCHAR(15) DEFAULT NULL, `created` INTEGER NOT NULL, `lastused` INTEGER DEFAULT NULL, PRIMARY KEY (`username`) );