how do you make user id to username

[riskmod]

I have a db that when i get some results i get the user_id like this:

<?php echo $report_row['user_id']; ?>

How can i change this to username - i have this in the form already:

<?php
$user_result = mysql_query("SELECT * FROM users ORDER BY username", $db);
while($user_row = mysql_fetch_array($user_result))
{
?>
<option value="<?php echo $user_row['user_id']; ?>"><?php echo $user_row['username']; ?></option>

I tried username on the report_row but looks like im missing something..

[NogDog]

What is the column name in the database for the user name in that table? That will be your array key (case-sensitive).

[riskmod]

What is the column name in the database for the user name in that table? That will be your array key (case-sensitive).

well its 'users' on the 'main' db

and in users theres:
Field:

user_id
username [SELECT COUNT( * ) AS `Rows` , `username` ]
first_name
last_name

[NogDog]

Unless there is some syntax error I'm not seeing, the code you posted originally looks OK to me. The only problem I could see was if "username" was not the correct column name. If it is, then you code ought to work OK, I think. You could do a quick check by changing it to this just for debugging:

<?php
$user_result = mysql_query("SELECT * FROM users ORDER BY username", $db);
while($user_row = mysql_fetch_array($user_result))
{
   // delete or comment out after done debuggin:
   echo "<pre>".print_r($user_row, true)."</pre>";
   exit;
   // end debug code
?>
<!-- rest of code -->

That will show you exactly what data was retrieved by your query and what the array keys are.

PS: Just a little "style" note: if you are only going to reference the row results by column name, I'd suggest using mysql_fetch_assoc() instead of mysql_fetch_array(), as the latter stores each column twice: once with the column name key and once with a numeric index in the array.

[riskmod]

Unless there is some syntax error I'm not seeing, the code you posted originally looks OK to me.

NogDog thanks for the mysql_fetch_assoc() tip.

Ok checked good but I still cant get it as a result like this:

				while($report_row = mysql_fetch_array($report_result))
				{
					?>
<tr>
<td <?php if(($count &#37; 2) == 0) echo 'class="alt_row"'; ?>><?php echo getStatus($report_row['status'], $db); ?></td>
<td <?php if(($count % 2) == 0) echo 'class="alt_row"'; ?>><?php echo date("n/d/y g:i a", strtotime($report_row['created'])); ?></td>
<td <?php if(($count % 2) == 0) echo 'class="alt_row"'; ?><?php echo $report_row['user_id']; ?></td>

Again it works with user_id but I tried username from users in the row but still haven't had luck

Here's another way i can get it but not in the results:

<?php
if(!empty($user))
{
?>								
<b>User:</b>
<?php				
array_walk($user, "arrayChangeUser", $db);
echo implode(", ", $user);?><br />
<?php
}
?>

[riskmod]

<?php
$user_result = mysql_query("SELECT * FROM users ORDER BY username", $db);
while($user_row = mysql_fetch_array($user_result))
{
   // delete or comment out after done debuggin:
   echo "<pre>".print_r($user_row, true)."</pre>";
   exit;
   // end debug code
?>
<!-- rest of code -->

That will show you exactly what data was retrieved by your query and what the array keys are.

Oh forgot when I did that it hid/erased the usernames in the list fyi.

I tried the user_rows just still not implementing it correctly I guess

[riskmod]

Anyone else have any ideas?

I just want to know what to input here:

<?php echo $report_row['user_id']; ?>

how can I change this to username ('username' will define nothing)

Am I just missing an edit on the result/report row?:

while($report_row = mysql_fetch_array($report_result))

[Dasher]

It seems that you must not know the correct names of the columns in your table. Apparently `username` is not one of the column names, maybe it is user_name, or usernames or last_name, first_name which could be combined in mysql with something like;

$query = "SELECT CONCAT(first_name, " ", last_name) AS username FROM users ORDER BY username ASC"; 

If I am having a query problem I often sort out the problem using phpMyAdmin and doing queries there before committing them to php code.

[riskmod]

Thanks Dasher yeah I'll look into that but how was this define then?

<select>
<?php
$user_result = mysql_query("SELECT * FROM users ORDER BY username", $db);
while($user_row = mysql_fetch_array($user_result))
{
?>
<option value="<?php echo $user_row['user_id']; ?>"><?php echo $user_row['username']; ?></option>.......

It shows the username there as a select box...I can select the user and then get the results....RESULTS is where i want a column with the USERNAME there...so I added another column in the result where it showed from what user it was made from which it does work:

<td <?php if(($count &#37; 2) == 0) echo 'class="alt_row"'; ?><?php echo $report_row['user_id']; ?></td>

I just need this to be username. Obviously it doesn't make sense if i looked it under the username, but it will make sense if i do a report base on the 'date' and so on....

I need to define the $user_row for the report_row..

Thanks

[Dasher]

Does this work? I don't like to constantly switch back and forth between html and the php engine unless I have to so I usually echo a lot of html;

$query = "SELECT userid, username FROM users ORDER BY username ASC";

$result = mysql_query($query,$link);
$count=0;
echo "<table><tr><td>User ID</td><td> Username </td></tr>";
while ($report_row = mysql_fetch_assoc($result)){

$userID = $report_row['userid'];
$userName = $report_row['username'];

if ($count % 2 == 0) {
echo "<tr><td class= 'alt_row'>" . $userID . "</td><td class= 'alt_row'>" . $userName . "</td></tr>";
}else {echo "<td>" . $userID . "</td><td>" . $userName . "</td></tr>";}
$count++;
}
echo "</table>"; 

[Dasher]

Opps I left out a <tr> in the } else { echo "<tr><td>... line.

[riskmod]

Opps I left out a <tr> in the } else { echo "<tr><td>... line.

I tried that but it gave me nothing...Is there a way i can show the whole code here? Anyone else have any ideas?

[Dasher]

Change this line;

$result = mysql_query($query,$link); 

to this

$result = mysql_query($query); 

or better to this;

$result = mysql_query($query)or die("<b>A fatal MySQL error occurred</b>.\n<br />Query: " . $query . "<br /><br />\nError: (" . mysql_errno() . ") " . mysql_error()); 

Have you tried to execute the MySQL query using phpMyAdmin?