Re: [PHP] php Login script issue
On 9/16/07, Chris Carter [EMAIL PROTECTED] 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']==$password)){
include ../include/newsession.php;
echo p class=data centerSuccessfully,Logged inbrbr
logout.php Log OUT brbr welcome.php Click here if your browser is not
redirecting automatically or you don't want to wait. br/center;
print script;
print self.location='submit-store-details.php';; // Comment this
line if you don't want to redirect
print /script;
}
}
else {
session_unset();
echo Wrong Login. Use your correct Userid and Password and Try
brcenterinput type='button' value='Retry'
onClick='history.go(-1)'/center;
}
?
I am getting this error when I am using this code:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
resource in thispage.php on line 37
Wrong Login. Use your correct Userid and Password and Try
Why does it show up everytime and whats wrong with mysql_fetch_array().
Please advice also if there is some other way available please help me try
that.
Thanks,
Chris
I advice you to split the code up in 2 seperate actions, and check for errors.
if($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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RE: [PHP] php Login script issue
Hi,
$result = mysql_query(SELECT * FROM tablename WHERE
userName='$userName' AND password = '$password');
if($rec = mysql_fetch_array($result)){
//your code
}
Try like this it may solve. It may solve your problem
Don't try to fetch the result from one single line code.
Warm Regards,
Sanjeev
http://www.sanchanworld.com/
http://webdirectory.sanchanworld.com - Submit your website URL
http://webhosting.sanchanworld.com - Choose your best web hosting plan
-Original Message-
From: Chris Carter [mailto:[EMAIL PROTECTED]
Sent: Sunday, September 16, 2007 3:10 PM
To: php-general@lists.php.net
Subject: [PHP] php Login script issue
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);
$password=mysql_real_escape_string($password);
if($rec=mysql_fetch_array(mysql_query(SELECT * FROM tablename WHERE
userName='$userName' AND password = '$password'))){
if(($rec['userName']==$userName)($rec['password']==$password)){
include ../include/newsession.php;
echo p class=data centerSuccessfully,Logged inbrbr
logout.php Log OUT brbr 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
brcenterinput 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
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RE: [PHP] php Login script issue
argh! hotmail sucks
I don't see in the script where you are using $_POST / $_GET / $_REQUEST to
access tha data from the form. Its likely that the example you are following is
old and uses 'register_globals'. Since register_globals is a huge security hole
and is not active in any new installations of PHP you need to change your
script to use the above methods to get the form data. The error you are getting
is due to the fact that you are not passing in the values to the sql and not
getting a valid result
Note that the below example fixes your issue but DOES NOT do any validation,
which you really should do before passing your data to the sql...
$userid=mysql_real_escape_string($_POST['userid']);
$password=mysql_real_escape_string($_POST['password']);
if($rec=mysql_fetch_array(mysql_query(SELECT * FROM tablename WHERE
userName='$userName' AND password = '$password'))){
if(($rec['userName']==$userName)($rec['password']==$password))
bastien
Date: Sun, 16 Sep 2007 02:39:57
-0700 From: [EMAIL PROTECTED] To: php-general@lists.php.net Subject: [PHP]
php Login script issue 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);
$password=mysql_real_escape_string($password);
if($rec=mysql_fetch_array(mysql_query(SELECT * FROM tablename WHERE
userName='$userName' AND password = '$password'))){
if(($rec['userName']==$userName)($rec['password']==$password)){ include
../include/newsession.php; echo Successfully,Logged in logout.php Log OUT
welcome.php Click here if your browser is not redirecting automatically or
you don't want to wait. ; print ; } } else { session_unset(); echo
Wrong Login. Use your correct Userid and Password and Try
onClick='history.go(-1)'; } ? 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 -- View this message in
context: http://www.nabble.com/php-Login-script-issue-tf4450691.html#a12698139
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