[SOLVED] how to open page and make sure users sees the right information

garydt · March 27, 2007

I'm making a website where registered users can upload a photo as well as information. The main page will display all of the users' photos. When user1 clicks on, say, user5's photo i want a page to open up and user5's information to be displayed. Is there a way to do this?

Every user has a username stored in the database along with their information and the url to their photo.

Code of the main page-

<?php require_once('Connections/elvisdb.php'); ?>
<?php
if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? "'" . doubleval($theValue) . "'" : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}


?><!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
<style type="text/css">
<!--
.style1 {color: #FFFFFF}
.style3 {color: #000099}
-->
</style>
</head>

<body>
<table width="1100"  border="0" cellpadding="5">
  <tr>
    <td width="151"><table width="150" bgcolor="#F48D02" border="0" cellpadding="0">
      <tr>
        <td class="style3"> </td>
      </tr>
    </table></td>
    
<?php

mysql_select_db($database_elvisdb, $elvisdb);
$query_Recordset1 = "SELECT * FROM images ORDER BY `count` DESC";
$Recordset1 = mysql_query($query_Recordset1, $elvisdb) or die(mysql_error());

while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)) {
for ($i=1;$i<=3;$i++) { 
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
echo $i;

?>	
<td>
<p><img src="<?php print $row_Recordset1['imageName']?>"></p>
<p><?php print $row_Recordset1['usnm']?></p> </td>

<?php
if ($i=3) {
?>
  <br />
<?php
}
}
}
?>
  </tr>
</table>
</body>
</html>
<?php
mysql_free_result($Recordset1);
?>

only one · March 27, 2007

very simple, just select from all the different tables where the information is stored, and where their username is....

garydt · March 27, 2007

Thanks.

How do i pass an username from the main page to the next page when the logged-in user clicks on another user's photo to see that user's info? I'm guessing i can't use the session variable mm_username as thats taken up with the current logged-in user?

As you can see, i use a while loop to display all the photos. Everytime it loops do i need to assign each photo a variable/session variable? If so how do i pass it to the next page to retrieve that user's info, (not the logged-in user's info)?

I hope i explained it clearly.

joshi_v · March 27, 2007

while displaying the photos , select user name for that photo from DB and pass it as a hidden variable with that photo. so when u click on that photo pass this hidden variable too. so that u can use it in query.

I hope i could understans what u r trying!

Regards,

Joshi.

JasonLewis · March 27, 2007

here is my step by step guide, with no code.

1. make a page which lists out all the users and have there username linked like this: view.php?user=$username

2. on the view to php page grab the username from the url using $_GET

3. have a query to select all information regarding that user.

4. display the information.

quite simple really.

garydt · March 27, 2007

Thanks very much. I will try writing the code for that

jitesh · March 27, 2007

Set links like this

<a href='user_info.php?userid='.$userid[$i]>Username</a>

in 'user_info.php' get userid like this

$userid = $_GET['userid'];

Then you can fetch data like this

"SELECT * FROM user WHERE UserID =".$userid;

garydt · March 27, 2007

Thanks alot.

Heres what i've got-

main page

    
<?php

mysql_select_db($database_elvisdb, $elvisdb);
$query_Recordset1 = "SELECT * FROM images ORDER BY `count` DESC";
$Recordset1 = mysql_query($query_Recordset1, $elvisdb) or die(mysql_error());

while ($row_Recordset1 = mysql_fetch_assoc($Recordset1)) {
$uname = $row_Recordset1['usnm'];
for ($i=1;$i<=3;$i++) { 
$totalRows_Recordset1 = mysql_num_rows($Recordset1);
echo $i;
echo $usernm;
?>	

<td><p><a href="userpage.php?$userid='.$uname'"><img src="<?php print $row_Recordset1['imageName'] ?>"></a></p>

userpage

<?php
session_start();
?>
<?php require_once('Connections/elvisdb.php'); ?>
<?php

if (!function_exists("GetSQLValueString")) {
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  $theValue = get_magic_quotes_gpc() ? stripslashes($theValue) : $theValue;

  $theValue = function_exists("mysql_real_escape_string") ? mysql_real_escape_string($theValue) : mysql_escape_string($theValue);

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? "'" . doubleval($theValue) . "'" : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}
}

$user = $_GET['userid'];
$colname_Recordset1 = "-1";
if (isset($_SESSION['MM_Username'])) {
  $colname_Recordset1 = (get_magic_quotes_gpc()) ? $_SESSION['MM_Username'] : addslashes($_SESSION['MM_Username']);
}
mysql_select_db($database_elvisdb, $elvisdb);
$query_Recordset1 = sprintf("SELECT * FROM userinformation WHERE usernm = $uname", GetSQLValueString($colname_Recordset1, "text"));
$Recordset1 = mysql_query($query_Recordset1, $elvisdb) or die(mysql_error());
$row_Recordset1 = mysql_fetch_assoc($Recordset1);
$totalRows_Recordset1 = mysql_num_rows($Recordset1);

$colname_Recordset2 = "-1";
if (isset($_SESSION['MM_Username'])) {
  $colname_Recordset2 = (get_magic_quotes_gpc()) ? $_SESSION['MM_Username'] : addslashes($_SESSION['MM_Username']);
}

mysql_select_db($database_elvisdb, $elvisdb);
$query_Recordset2 = sprintf("SELECT * FROM images WHERE usnm = $uname", GetSQLValueString($colname_Recordset2, "text"));
$Recordset2 = mysql_query($query_Recordset2, $elvisdb) or die(mysql_error());
$row_Recordset2 = mysql_fetch_assoc($Recordset2);
$totalRows_Recordset2 = mysql_num_rows($Recordset2);
$img = $row_Recordset2['imageName'];
$big = $row_Recordset2['bigimagename'];
echo $img;

When i click on a photo and the userpage opens i get-

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

garydt · March 27, 2007

I've put this-

<td><p><a href="userpage.php?username=$uname"><img src="<?php print $row_Recordset1['imageName'] ?>"></a></p>

But it doesn't pass on as a variable, only as $uname

I got this on userpage

$user = $_GET['username];

What have i done wrong?

garydt · March 27, 2007

any suggestions?

Edward · March 27, 2007

Hey,

I'm not too advanced but I think I might be able to give you some tips.

Your Code:

<td><p><a href="userpage.php?$userid='.$uname'"><img src="<?php print $row_Recordset1['imageName'] ?>"></a></p>

New Code:

<td><p><a href="userpage.php?userid=<?php echo $uname; ?>"><img src="<?php print $row_Recordset1['imageName'] ?>"></a></p>

Firstly, see that I removed the dollar sign for the variable in the URL (but importantly, not for the value). Secondly, assuming you were showing me your code in its full context, the quotes seemed over complex. I have reduced these and only moved from HTML to PHP and back to display the relevent username in the URL, which was necessary so that $username will display the username, as opposed to displaying '$username'.

I think it would also help you to be more consistent with your naming of data. For example, you have given the 'username' four different names throughout your site.

The database uses 'usnm'

The main page uses 'uname'

The URL uses 'userid'

The User page 'user'

Perhaps change them all to 'username', or 'database_username' and 'website_username' if you really want to distinguish them.

Hopefully that might help, if not, post back.

garydt · March 27, 2007

Thanks alot, that did it!

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[SOLVED] how to open page and make sure users sees the right information

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