i can't send my userid to the next page..can some1 check for

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fnleong
Forum Newbie
Posts: 18
Joined: Wed May 12, 2004 5:49 am

i can't send my userid to the next page..can some1 check for

Post by fnleong »

thanks~~
//this is the 1st page allow user to click and then link to another profile page..

Code: Select all

<?php
<html>
	
<HEAD><TITLE>Member Section</TITLE> 
</HEAD> 
<body>
<a href="userview.php?userID=<?php echo $row["userID"];?>">View Profile </a></div></td> 


</body> 
</html>
?>
Last edited by fnleong on Thu May 13, 2004 3:43 am, edited 2 times in total.
fnleong
Forum Newbie
Posts: 18
Joined: Wed May 12, 2004 5:49 am

Post by fnleong »

this is the page link to

Code: Select all

<?php
<html>
<head>
  <title>List MY Records</title>
</head>


<h1>My Profile</h1></center>
<?php

  @ $db = mysql_pconnect('localhost','root');

  if (!$db)
  {
     echo 'Error: no record to database.  Please try again later.';
     exit;
  }

  mysql_select_db('mobile');
  $userID = $_GET['userID'];
  $query = "select * from users where userID =$userID";
  
  $result = mysql_query($query);
 
 
	printf("<TABLE BORDER WIDTH="100%%">\n");
	printf("<TR>
   
         <TD><B>User Name</B></TD>
         <TD><B>Password</B></TD>
	<TD><B>first Name</B></TD>
	<TD><B>Last Name</B></TD>
	<TD><B>Email</B></TD>
         <TD><B>Address</B></TD>
 	<TD><B>PhoneNumber</B></TD>
         <TD><B>Gender</B></TD>
 	<TD><B>userID</B></TD>
	     
        </TR>\n");


	while (($row = mysql_fetch_object($result))){

	printf("<TR>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
	 <TD>%s</TD>
	 <TD>%s</TD>
         <TD>%s</TD>
         <TD><A HREF="modifyuser.php?userID=%s"><I>Modify</I></A></TD>
        </TR>\n",
	$row->username, $row->password, $row->first_name, $row->last_name,$row->email,$row->address,$row->PhoneNumber,$row->gender, $row->userID,$row->userID,$row->userID);}
	printf("</TABLE>\n");
	
	

?>
Last edited by fnleong on Thu May 13, 2004 3:38 am, edited 1 time in total.
User avatar
JayBird
Admin
Posts: 4524
Joined: Wed Aug 13, 2003 7:02 am
Location: York, UK
Contact:

Post by JayBird »

PLEASE PLEASE PLEASE READ THIS BEFORE I ANSWER ANY OF YOURS POSTS!

viewtopic.php?t=21171

Once you have edited your posts into the correct format, i will try to help.

Mark
malcolmboston
DevNet Resident
Posts: 1826
Joined: Tue Nov 18, 2003 1:09 pm
Location: Middlesbrough, UK

Post by malcolmboston »

replace your query with this, see what happens

Code: Select all

$query = "SELECT * from USERS where userID ='$userID'";
// added this print statement to view query
print $query;
$result = mysql_query($query) or die (mysql_error());
?>
fnleong
Forum Newbie
Posts: 18
Joined: Wed May 12, 2004 5:49 am

Post by fnleong »

this is the error msg

Code: Select all

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in C:\Program Files\Apache Group\Apache2\htdocs\viewprofile.php on line 47
it display a table but is without contents
Last edited by fnleong on Thu May 13, 2004 3:39 am, edited 1 time in total.
malcolmboston
DevNet Resident
Posts: 1826
Joined: Tue Nov 18, 2003 1:09 pm
Location: Middlesbrough, UK

Post by malcolmboston »

Code: Select all

while (($row = mysql_fetch_object($result)))
to

Code: Select all

while (($row = mysql_fetch_array($result))){
fnleong
Forum Newbie
Posts: 18
Joined: Wed May 12, 2004 5:49 am

stil cant

Post by fnleong »

i hav change as what u said, but still cant

Code: Select all

<?php
<?php 
   session_start(); 
  
   if (isset($HTTP_SESSION_VARS['valid_user'])) 
   { echo '<p>You are logged in as '.$HTTP_SESSION_VARS['valid_user'].'</p>';
?> 
<html>
<head>
  <title>List MY Records</title>
</head>


<h1>My Profile</h1></center>
<?php

  @ $db = mysql_pconnect('localhost','root');

  if (!$db)
  {
     echo 'Error: no record to database.  Please try again later.';
     exit;
  }

  mysql_select_db('mobile');
  $userID = $_GET['userID'];
  $query = "SELECT * from users where userID ='$userID'"; 
  
  print $query; 
  $result = mysql_query($query) or die (mysql_error()); 

 
	printf("<TABLE BORDER WIDTH="100%%">\n");
	printf("<TR>
   
         <TD><B>User Name</B></TD>
         <TD><B>Password</B></TD>
	<TD><B>first Name</B></TD>
	<TD><B>Last Name</B></TD>
	<TD><B>Email</B></TD>
         <TD><B>Address</B></TD>
 	<TD><B>PhoneNumber</B></TD>
         <TD><B>Gender</B></TD>
 	<TD><B>userID</B></TD>
	     
        </TR>\n");


	while (($row = mysql_fetch_array($result))){ 


	printf("<TR>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
         <TD>%s</TD>
	 <TD>%s</TD>
	 <TD>%s</TD>
         <TD>%s</TD>
         <TD><A HREF="modifyuser.php?userID=%s"><I>Modify</I></A></TD>
        </TR>\n",
	$row->username, $row->password, $row->first_name, $row->last_name,$row->email,$row->address,$row->PhoneNumber,$row->gender, $row->userID,$row->userID,$row->userID);}
	printf("</TABLE>\n");
	
	

?>
<br><br>
<a href="adminusermenu.php">BACK</br></a>

</body> 
</html><? 
exit; 
   }else 
   { 
     echo '<p>You are not logged in..</p>'; 
     echo '<p>Only logged in admins may see this page.</p>'; 

   } 

?> 
?>
?>
fnleong
Forum Newbie
Posts: 18
Joined: Wed May 12, 2004 5:49 am

Post by fnleong »

it still can't get the user id from the previous page i think..
undefined variable row in line 16
teckyan888
Forum Commoner
Posts: 40
Joined: Tue May 11, 2004 10:46 am

Post by teckyan888 »

Code: Select all

<?php
mysql_select_db('mobile'); 
  $userID = $_GET['userID']; 
  $query = "select * from users where userID =$userID"; 

?>
u try like this:

Code: Select all

<?php
mysql_select_db('mobile'); 
$query = 'select * from Users '
."where userID = '{$_REQUEST['userID']}' ";

?>
fnleong
Forum Newbie
Posts: 18
Joined: Wed May 12, 2004 5:49 am

Post by fnleong »

still cant teck yan.is not this proiblm, is it cant get the userid from the previous page, how to pass the id to this page?
hav u do the view profile part?for user to view their profile
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