Could anyone please help me on this??
I have a php script page, which is basically quiz. Visitors (after login in
with their email address) are supposed to answer each question, and when
they click the button at the bottom, the next page will show which problems
they got right or wrong.
Now my visitors, after seeing some problems they got wrong, click Back
button at the Explorer menu, and if they do, they get the below message:
=============== =============== =============== =============== =============== =
=============== =========
Warning: Page has Expired The page you requested was created using
information you submitted in a form. This page is no longer available. As a
security precaution, Internet Explorer does not automatically resubmit your
information for you.
To resubmit your information and view this Web page, click the Refresh
button.
=============== =============== =============== =============== =============== =
=============== =========
Below are the script that I am using. This scripts are after initial login
page, which asks for visitor's email address. I have authorized visitors
already in my MySQL table, and only the visitors whose email address is in
my MySQL table are allowed to login.
=============== =============== =============== =============== =============
<?php
session_start() ;
define('HOST', 'mysql');
define('USER', 'byung');
define('PASS', 'xiclcvjq');
define('DB', 'Learning');
if (empty($HTTP_SE SSION_VARS['emailadd']))
{
mysql_connect(H OST, USER, PASS);
mysql_select_db (DB);
$result = mysql_query("SE LECT COUNT(*) AS numfound FROM emailAddress WHERE
emailaddress='{ $HTTP_POST_VARS['emailadd']}'") or die(mysql_error ());
$result_ar=mysq l_fetch_array($ result);
if ($result_ar['numfound']<1)
{
header('Locatio n: jinuform.php?er ror=1');
exit;
}
$emailadd=$HTTP _POST_VARS['emailadd'];
session_registe r('emailadd');
$insert = mysql_query("IN SERT INTO StuAccess (emailaddress, IssueNo)
VALUES ('{$HTTP_POST_V ARS['emailadd']}', 1)") or die(mysql_error ());
}
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>Jinu Math Study Guide</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body>
<div align="center">
<table width="745" border="1" cellspacing="0" cellpadding="0" >
<tr>
<td width = "200">  ;</td>
<td width = "545">  ;</td>
</tr>
<tr>
<td> </td>
<td>> </td>
</tr>
<tr>
<td height="459">&n bsp;</td>
<td><form name="form1" method="post" action="grade.p hp">
<table width="545" border="1" cellspacing="0" cellpadding="0" >
<tr>
<td><div align="center"> Today's Exercise</div></td>
</tr>
<tr>
<td bgcolor="#E0E0E 0"><strong><fon t color="#0000A0" >Q<font
size="1">UESTIO N
</font>1</font></strong></td>
</tr>
<tr>
<td><p>What is the length of l times w?</p>
<input type="radio" name="Q1" value="1">
<input type="radio" name="Q1" value="2">
<input type="radio" name="Q1" value="3">
<input type="radio" name="Q1" value="4"><br>
</td>
</tr>
<tr>
=============== =============== =============== =============== =============
What above codes do is getting the email address from the login page (from
the previous page), and then allow login only if the visitors' email address
they typed in is in my MySQL database.
Below are the codes, which process the quiz form:
<?php
define('HOST', 'mysql');
define('USER', 'byung');
define('PASS', 'xiclcvjq');
define('DB', 'JinuAcademy');
mysql_connect(H OST, USER, PASS);
mysql_select_db (DB);
$q1 = $HTTP_POST_VARS['Q1'];
$q2 = $HTTP_POST_VARS['Q2'];
$q3 = $HTTP_POST_VARS['Q3'];
$q4 = $HTTP_POST_VARS['Q4'];
$q5 = $HTTP_POST_VARS['Q5'];
$q6 = $HTTP_POST_VARS['Q6'];
$q7 = $HTTP_POST_VARS['Q7'];
if ($q1 == '' || $q2 == '' || $q3 == '' || $q4 =='' || $q5 == '' || $q6 ==
'' || $q7 == '')
{
echo '<h1>Sorry!! You need to fill in your answer for all
questions</h1>';
}
else
{
$score = 0;
if ($q1 == 1)
{
$score++;
$q1Grade = "Correct";
echo '<B>1. Correct!!</B><p>';
}
else
{
$q1Grade = "Incorrect" ;
echo '<B>1. Incorrect..</B><p>';
}
if ($q2 == 1)
{
$score++;
.....
Now above codes present how the visitor did in the Quiz. But from this
page, if the visitor click the Back button, then they get the error message
above. How would I be able to make my visitor to go back to the previous
page with answers they have clicked in showing up again rather than getting
page expired message??
Any help or comment will be deeply appreciated.
Thanks.
I have a php script page, which is basically quiz. Visitors (after login in
with their email address) are supposed to answer each question, and when
they click the button at the bottom, the next page will show which problems
they got right or wrong.
Now my visitors, after seeing some problems they got wrong, click Back
button at the Explorer menu, and if they do, they get the below message:
=============== =============== =============== =============== =============== =
=============== =========
Warning: Page has Expired The page you requested was created using
information you submitted in a form. This page is no longer available. As a
security precaution, Internet Explorer does not automatically resubmit your
information for you.
To resubmit your information and view this Web page, click the Refresh
button.
=============== =============== =============== =============== =============== =
=============== =========
Below are the script that I am using. This scripts are after initial login
page, which asks for visitor's email address. I have authorized visitors
already in my MySQL table, and only the visitors whose email address is in
my MySQL table are allowed to login.
=============== =============== =============== =============== =============
<?php
session_start() ;
define('HOST', 'mysql');
define('USER', 'byung');
define('PASS', 'xiclcvjq');
define('DB', 'Learning');
if (empty($HTTP_SE SSION_VARS['emailadd']))
{
mysql_connect(H OST, USER, PASS);
mysql_select_db (DB);
$result = mysql_query("SE LECT COUNT(*) AS numfound FROM emailAddress WHERE
emailaddress='{ $HTTP_POST_VARS['emailadd']}'") or die(mysql_error ());
$result_ar=mysq l_fetch_array($ result);
if ($result_ar['numfound']<1)
{
header('Locatio n: jinuform.php?er ror=1');
exit;
}
$emailadd=$HTTP _POST_VARS['emailadd'];
session_registe r('emailadd');
$insert = mysql_query("IN SERT INTO StuAccess (emailaddress, IssueNo)
VALUES ('{$HTTP_POST_V ARS['emailadd']}', 1)") or die(mysql_error ());
}
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">
<html>
<head>
<title>Jinu Math Study Guide</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
</head>
<body>
<div align="center">
<table width="745" border="1" cellspacing="0" cellpadding="0" >
<tr>
<td width = "200">  ;</td>
<td width = "545">  ;</td>
</tr>
<tr>
<td> </td>
<td>> </td>
</tr>
<tr>
<td height="459">&n bsp;</td>
<td><form name="form1" method="post" action="grade.p hp">
<table width="545" border="1" cellspacing="0" cellpadding="0" >
<tr>
<td><div align="center"> Today's Exercise</div></td>
</tr>
<tr>
<td bgcolor="#E0E0E 0"><strong><fon t color="#0000A0" >Q<font
size="1">UESTIO N
</font>1</font></strong></td>
</tr>
<tr>
<td><p>What is the length of l times w?</p>
<input type="radio" name="Q1" value="1">
<input type="radio" name="Q1" value="2">
<input type="radio" name="Q1" value="3">
<input type="radio" name="Q1" value="4"><br>
</td>
</tr>
<tr>
=============== =============== =============== =============== =============
What above codes do is getting the email address from the login page (from
the previous page), and then allow login only if the visitors' email address
they typed in is in my MySQL database.
Below are the codes, which process the quiz form:
<?php
define('HOST', 'mysql');
define('USER', 'byung');
define('PASS', 'xiclcvjq');
define('DB', 'JinuAcademy');
mysql_connect(H OST, USER, PASS);
mysql_select_db (DB);
$q1 = $HTTP_POST_VARS['Q1'];
$q2 = $HTTP_POST_VARS['Q2'];
$q3 = $HTTP_POST_VARS['Q3'];
$q4 = $HTTP_POST_VARS['Q4'];
$q5 = $HTTP_POST_VARS['Q5'];
$q6 = $HTTP_POST_VARS['Q6'];
$q7 = $HTTP_POST_VARS['Q7'];
if ($q1 == '' || $q2 == '' || $q3 == '' || $q4 =='' || $q5 == '' || $q6 ==
'' || $q7 == '')
{
echo '<h1>Sorry!! You need to fill in your answer for all
questions</h1>';
}
else
{
$score = 0;
if ($q1 == 1)
{
$score++;
$q1Grade = "Correct";
echo '<B>1. Correct!!</B><p>';
}
else
{
$q1Grade = "Incorrect" ;
echo '<B>1. Incorrect..</B><p>';
}
if ($q2 == 1)
{
$score++;
.....
Now above codes present how the visitor did in the Quiz. But from this
page, if the visitor click the Back button, then they get the error message
above. How would I be able to make my visitor to go back to the previous
page with answers they have clicked in showing up again rather than getting
page expired message??
Any help or comment will be deeply appreciated.
Thanks.
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