php-general Digest 16 Sep 2007 21:24:47 -0000 Issue 5022
Topics (messages 262141 through 262149):
php Login script issue
262141 by: Chris Carter
262142 by: Tijnema
262146 by: Sanjeev N
262147 by: Bastien Koert
Re: Configure mail to use Gmail smtp
262143 by: debussy007
262144 by: Thomas Bachmann
262148 by: mike
Re: split in to multiple pages (I think)
262145 by: Thomas Bachmann
Finding next recored in a array
262149 by: Richard Kurth
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--- Begin Message ---
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> <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
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--- End Message ---
--- Begin Message ---
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']==$password)){
> include "../include/newsession.php";
> echo "<p class=data> <center>Successfully,Logged in<br><br>
> logout.php Log OUT <br><br> welcome.php Click here if your browser is not
> redirecting automatically or you don't want to wait. <br></center>";
> print "<script>";
> print " self.location='submit-store-details.php';"; // Comment this
> line if you don't want to redirect
> print "</script>";
>
> }
> }
> else {
>
> session_unset();
> echo "Wrong Login. Use your correct Userid and Password and Try
> <br><center><input type='button' value='Retry'
> onClick='history.go(-1)'></center>";
>
> }
> ?>
>
> I am getting this error when I am using this code:
>
> Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
> resource in thispage.php on line 37
> Wrong Login. Use your correct Userid and Password and Try
>
> Why does it show up everytime and whats wrong with mysql_fetch_array().
>
> Please advice also if there is some other way available please help me try
> that.
>
> Thanks,
>
> Chris
I advice you to split the code up in 2 seperate actions, and check for errors.
> if($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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--- Begin Message ---
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: [EMAIL PROTECTED]
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> <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
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--- Begin Message ---
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: [EMAIL PROTECTED]> 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>
Sent from the PHP - General mailing list archive at Nabble.com.>> --> PHP
General Mailing List (http://www.php.net/)> To unsubscribe, visit:
http://www.php.net/unsub.php>
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--- End Message ---
--- Begin Message ---
I succeeded using the Zend Framework.
debussy007 wrote:
>
> Hello,
>
> I have read here : http://www.geekzone.co.nz/tonyhughes/599
> that I can use Gmail as a free SMTP server.
>
> Is it possible to change the php.ini in order to have this running ?
>
> I need to specify in some way the following :
>
> Outgoing Mail (SMTP) Server - requires TLS: smtp.gmail.com (use
> authentication)
> Use Authentication: Yes
> Use STARTTLS: Yes (some clients call this SSL)
> Port: 465 or 587
> Account Name: your Gmail username (including '@gmail.com')
> Email Address: your original isp address ([EMAIL PROTECTED])
> Password: your Gmail password
>
>
>
> I can see in the php.ini that I can specify the port, smtp server and the
> from.
>
> But how do I do for he other parameters ?
>
> Thank you !!
>
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--- Begin Message ---
ZF rocks ;)
debussy007 schrieb:
I succeeded using the Zend Framework.
debussy007 wrote:
Hello,
I have read here : http://www.geekzone.co.nz/tonyhughes/599
that I can use Gmail as a free SMTP server.
Is it possible to change the php.ini in order to have this running ?
I need to specify in some way the following :
Outgoing Mail (SMTP) Server - requires TLS: smtp.gmail.com (use
authentication)
Use Authentication: Yes
Use STARTTLS: Yes (some clients call this SSL)
Port: 465 or 587
Account Name: your Gmail username (including '@gmail.com')
Email Address: your original isp address ([EMAIL PROTECTED])
Password: your Gmail password
I can see in the php.ini that I can specify the port, smtp server and the
from.
But how do I do for he other parameters ?
Thank you !!
--- End Message ---
--- Begin Message ---
i'm quite sure you can't with php's built-in mail() function.
however more advanced modules should support it. look in PEAR, google
for phpmailer, etc. i'm sure there's got to be some. worst case i
think you have all the tools in PHP to make your own anyway.
On 9/16/07, debussy007 <[EMAIL PROTECTED]> wrote:
> debussy007 wrote:
> >
> > Hello,
> >
> > I have read here : http://www.geekzone.co.nz/tonyhughes/599
> > that I can use Gmail as a free SMTP server.
> >
> > Is it possible to change the php.ini in order to have this running ?
> >
> > I need to specify in some way the following :
> >
> > Outgoing Mail (SMTP) Server - requires TLS: smtp.gmail.com (use
> > authentication)
> > Use Authentication: Yes
> > Use STARTTLS: Yes (some clients call this SSL)
> > Port: 465 or 587
> > Account Name: your Gmail username (including '@gmail.com')
> > Email Address: your original isp address ([EMAIL PROTECTED])
> > Password: your Gmail password
> >
> >
> >
> > I can see in the php.ini that I can specify the port, smtp server and the
> > from.
> >
> > But how do I do for he other parameters ?
> >
> > Thank you !!
--- End Message ---
--- Begin Message ---
Why don't post an example of your news file too?
Joker7 schrieb:
Hi,
I'm using the code below to display news articles-which works great apart
from. I can control the number of articles,but I would like to add a link to
the bottom of the page to the un-displayed articles ( nexted 5 articles and
so on) any pointers would be most welcome as most know my PHP skills are not
the best :).
Cheers
Chris
<?
$max_latest = 5;
$filename = "articlesum.php";
#- open article
if(file_exists($filename)){
$fh = fopen($filename, "r");
$old_news = fread($fh, filesize($filename));
fclose($fh);
}
#- get article
$articles = explode("<!--ARTICLE-->", $old_news);
$i=0;
foreach ( $articles as $article ){
if(count($articles)>$i){
if($max_latest >= $i++){
print $article;
}
}
}
?>
--- End Message ---
--- Begin Message ---
$Campaign_array| = array('0','1','3','5','8','15','25');|
I know that I can find the next recored in a array using next. What I do
not understand is if I know the last number was say 5 how do I tell the
script that that is the current number so I can select the next record
||
--- End Message ---
