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  #1  
Old July 3rd, 2003, 07:23 PM
Tiger in a Tube Tiger in a Tube is offline
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passing variables around several php pages

Hi there,

I have a user authentication set up that then (is supposed) to use the username to provide specific content (in this case a welcome message) per user.

The mechanic is you log in, are validated using a mySQL database. You then go to the main page where a message stored as part of your mySQL record is displayed.

My problem is that I can't get the record to display. $username is the name of the variable through my pages.

I'm using frames. The query code is;
PHP Code:
function print_message($username)
{
    
$conn=db_connect();

    
$query "select * from content_admin where username='$username'";
    
$message mysql_query($query$conn);
    
$use mysql_fetch_row($message);
    return 
$use[4];


The print_message($username) function is called from this piece of code.


PHP Code:
 print_r($username); //for checking what's coming out of the query
$use print_message($username);

session_start();
check_valid_user();

?>
<HTML>
//irrelevant HTML here

<TABLE WIDTH=136 BORDER=0 CELLPADDING=0 CELLSPACING=0>
    <TR>

    <TD WIDTH=136 HEIGHT=152 align="left" valign="top" background="images/home_right01.gif" class="content" ALT="">
      <font size="1"></font>
      <table width="88%" border="0" align="center" cellpadding="1" cellspacing="0">
        <tr>
          <td height="129"><font size="1" face="Trebuchet MS"> <br>
            <?=$use ?></font></td>
        </tr>
      </table> 

The result of this is that I can't get the message to display. I'm assuming that the problem is to do with passing the $username because when I put a hard coded user name, I get the correct message.

thanks in advance for your help,
TiaT

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  #2  
Old July 3rd, 2003, 08:12 PM
DudeThatWasRude DudeThatWasRude is offline
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Quote:
$query = "select * from content_admin where username='$username'";


This is not good coding style.
Select * from.....
will never guarantee you that some field will appear in the specific posion.

An example of better code for your context:

<?php
mysql_connect($host, $user, $password);
mysql_select_db("database");
$result = mysql_query("select user_id, fullname from table");
while ($row = mysql_fetch_array($result)) {
echo "user_id: ".$row["user_id"]."<br>\n";
echo "user_id: ".$row[0]."<br>\n";
.....
.....

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  #3  
Old July 3rd, 2003, 09:24 PM
Tiger in a Tube Tiger in a Tube is offline
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thanks Dude, I'll change that query accordingly.

It still doesn't solve my problem though. .

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  #4  
Old July 4th, 2003, 03:44 AM
digitallysmooth digitallysmooth is offline
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Looks like you are trying to pass an array to a function that is looking for a string:

PHP Code:
 print_r($username); //for checking what's coming out of the query
$use print_message($username); 
I'm not sure of how you structured your $username array (you should have posted the info from 'print_r'), however, I assume you could call your function like this:
PHP Code:
 $use print_message($username[0]); 
__________________
__________________________________________________ _
Wil Moore III, MCP | Integrations Specialist | Senior Consultant
Are You Listed...? | DigitallySmooth Inc.

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  #5  
Old July 4th, 2003, 08:35 PM
Tiger in a Tube Tiger in a Tube is offline
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yes, I've fixed that so it gives the right record out of the array.

I don't actually get any value or result from the print_r($username[0]) funtion, so I can't post it.

But the problem still remains that I can't get the appropriate value to come up when I use the variable. If I hard code the the username in to the query, I get a result.

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