On 9/16/07, Chris Carter <chandan9sharma@xxxxxxxxx> 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
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