Re: php Login script issue

On 9/16/07, Chris Carter <chandan9sharma@yahoo.com> wrote: > > Hi, > > Its just a login and password validation that I am trying to achieve. If the > username is correct then the person is able to view certain page, if > incorrect then he is directed elsewhere. > > <? > $userid=mysql_real_escape_string($userid); Here you call it $userid > $password=mysql_real_escape_string($password); > > if($rec=mysql_fetch_array(mysql_query("SELECT * FROM tablename WHERE > userName='$userName' AND password = '$password'"))){ and here you call it $userName. If this is the full code, $userName is not set here, and it would result in query userName='' and mysql_query will return FALSE, which isn't a valid mysql resource for mysql_fetch_array. > > if(($rec['userName']==$userName)&&($rec['password']==$password)){ > include "../include/newsession.php"; > echo "<p class=data> <center>Successfully,Logged > in<br><br> > logout.php Log OUT <br><br> welcome.php Click here if your browser is not > redirecting automatically or you don't want to wait. <br></center>"; > print "<script>"; > print " self.location='submit-store-details.php';"; // Comment this > line if you don't want to redirect > print "</script>"; > > } > } > else { > > session_unset(); > echo "Wrong Login. Use your correct Userid and Password and Try > <br><center><input type='button' value='Retry' > onClick='history.go(-1)'></center>"; > > } > ?> > > I am getting this error when I am using this code: > > Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result > resource in thispage.php on line 37 > Wrong Login. Use your correct Userid and Password and Try > > Why does it show up everytime and whats wrong with mysql_fetch_array(). > > Please advice also if there is some other way available please help me try > that. > > Thanks, > > Chris I advice you to split the code up in 2 seperate actions, and check for errors. > if($rec=mysql_fetch_array(mysql_query("SELECT * FROM tablename WHERE > userName='$userName' AND password = '$password'"))){ would become: $result = mysql_query("SELECT * FROM tablename WHERE userName='$userName' AND password = '$password'") or die (mysql_error()); // You could also add some checks here with mysql_num_rows for example... if($rec=mysql_fetch_array($result)){ Tijnema -- If this is a mailing list: DO NOT TOP POST! why?: http://www.caliburn.nl/topposting.html Vote for PHP Color Coding (aka Syntax Highlighting) in Gmail! -> http://gpcc.tijnema.info