OK, I am stumped!?

gordon7

I am trying to populate a drop down box with data from mySQL. The box shows up at about the right length (based on the data available), but it is empty. Something else is an additional output which makes no sense. I have spent a couple days on this now and I am extremely frustrated.

Here is the code:

<?php
// setup variables for db access
$db_name = "rvr";
$db_table = "rvr_service_employee";
$dbconnect = @mysql_connect("localhost") or die(mysql_error());
//connect to the db
$db = @mysql_select_db($db_name, $dbconnect) or die(mysql_error());
// select db criteria
$sql = "SELECT * FROM $db_table ORDER BY lname";
$results = @($sql, $dbconnect) or die (mysql_error());
// create variables for db items
while ($row = mysql_fetch_array($results)) {
$last_name = $row["lname"];
$worker_class = $row["worker_class"];

if ($worker_class == "server") {
$option_server[$row] = $last_name;
$option_block_server .= "<option value=\"$option_server[$row]\">$last_name</option>" ;

}
}
?>

<!doctype html public "-//W3C//DTD HTML 4.01//EN">

<html>

<head>
<title>Untitled</title>
<meta http-equiv="generator" content="PHP Designer 2005" />
</head>

<body bgcolor="#FFFFFF" text="#000000" link="#0000FF" vlink="#800080" alink="#FF0000">
<?php echo "$option_server[$row]";

echo "$last_name<br />"; ?>
<select>
<?php echo "$option_block_server";?>
</select>

</body>
</html>

Here is the strange output:

$last_name";}}?>

I have noticed tha if I remove the \" from

$option_block_server .= "<option value=\"$option_server[$row]\">$last_name</option>" ;

the strange output goes away, but the drop down is still empty.

Somebody please help me...

Gordon

halojoy

<?php 
// setup variables for db access 
$db_name = "rvr"; 
$db_table = "rvr_service_employee"; 
$dbconnect = @mysql_connect("localhost") or die(mysql_error()); 
//connect to the db 
$db = @mysql_select_db($db_name, $dbconnect) or die(mysql_error()); 
// select db criteria 
$sql = "SELECT * FROM $db_table ORDER BY lname"; 
$results = @mysql_query($sql, $dbconnect) or die (mysql_error()); 
// create variables for db items 
while ($row = mysql_fetch_array($results)) { 
$last_name = $row["lname"]; 
$worker_class = $row["worker_class"]; 

if ($worker_class == "server") { 
$option_server[$row] = $last_name; 
$option_block_server .= "<option value=\"$option_server[$row]\">$last_name</option>" ; 

} 
} 
?> 

<!doctype html public "-//W3C//DTD HTML 4.01//EN"> 

<html> 

<head> 
<title>Untitled</title> 
<meta http-equiv="generator" content="PHP Designer 2005" /> 
</head> 

<body bgcolor="#FFFFFF" text="#000000" link="#0000FF" vlink="#800080" alink="#FF0000"> 
<?php echo "$option_server[$row]"; 

echo "$last_name<br />"; ?> 
<select> 
<?php echo "$option_block_server";?> 
</select> 

</body> 
</html>

We are used 'php'-bb-tag colors
so here is same script
for php tag, click in editor bar - next right-est icon = php

<?php

$option_block_server .= "<option value=\"$option_server[$row]\">$last_name</option>" ;

// you might try this
$option_block_server .= "<option value='{$option_server[ 'row' ]}'>$last_name</option>" ;

?>

It is a bit trickier with ---- indexed['variables'] ---- in echo or in strings
may need {curly backets around}

gordon7

Thanks,
That took care of the wierd output, but the drop down box is still not populating???

gordon

halojoy

gordon7 wrote:
Thanks,
That took care of the wierd output, but the drop down box is still not populating???

gordon

what does the HTML Source output look like?

Can you 'view source', copy and post it here?

gordon7

Here is the "view source" as requested:

if ($worker_class == "server") {
$option_server[$row] = $last_name;
$option_block_server .= "<option value='{$option_server[ 'row' ]}'>$last_name</option>" ;

}
}
?>

<!doctype html public "-//W3C//DTD HTML 4.01//EN">

<html>

<head>
<title>Untitled</title>
<meta http-equiv="generator" content="PHP Designer 2005" />
</head>

</body>
</html>

I didn't think the php could be seen - is there something wrong with it too???

Gordon

halojoy

Using our
PHP PHP PHP Tag
PHP TAG - is in Reply editor .... is in REPLY editor menu bar

Php can only be output by php self, as an echo of this php source

any page called 'pagename.php' and starting with
<?php

should only put out HTML code

So something is not right,
can be in:
- your script
- your page names
- your webserver settings
- your php.ini settings

<?php 
// setup variables for db access 
$db_name = "rvr"; 
$db_table = "rvr_service_employee"; 
$dbconnect = @mysql_connect("localhost") or die(mysql_error()); 
//connect to the db 
$db = @mysql_select_db($db_name, $dbconnect) or die(mysql_error()); 
// select db criteria 
$sql = "SELECT * FROM $db_table ORDER BY lname"; 
$results = @mysql_query($sql, $dbconnect) or die (mysql_error()); 
// create variables for db items 
while ($row = mysql_fetch_array($results)) { 
$last_name = $row["lname"]; 
$worker_class = $row["worker_class"]; 

if ($worker_class == "server") { 
$option_server[$row] = $last_name; 
$option_block_server .= "<option value='{$option_server[ 'row' ]}'>$last_name</option>" ; 

} 
} 
?> 

<!doctype html public "-//W3C//DTD HTML 4.01//EN"> 

<html> 

<head> 
<title>Untitled</title> 
<meta http-equiv="generator" content="PHP Designer 2005" /> 
</head> 

<body bgcolor="#FFFFFF" text="#000000" link="#0000FF" vlink="#800080" alink="#FF0000"> 

<select> 
<?php echo "$option_block_server";?> 
</select> 

</body> 
</html>

One last asking,
to maybe see for myself (and other helpers too can see)

Can you give a link to this script
and I can run it, and see what I get.
Probably same Browser 'View source' result, as you got.

/halojoy - wants to help to solve this, but think this is a mystery
PS.
I am sure you have run phpinfo.php at your server:

<?php

// this php page is called 'phpinfo.php'
// gives all setings of my server and of php.ini
phpinfo();

?>

If you havent, tells me
you Are a real nice beginner

Welcome to the phpinfo() gang!

[man]phpinfo[/man]
[man]phpinfo[/man]
[man]phpinfo[/man]
[man]phpinfo[/man]

🙂

gordon7

Thanks,
I had named the file .html. Remaming solved the list population problem, but gives the following message:

Warning: Illegal offset type in C:\Apache2\htdocs\RVR_2\modified_input_newcoce.php on line 17

Line 16 & 17 in the code are:

if ($worker_class == "server") {
$option_server[$row] = $last_name;

Yes, I have run phpinfo(), but truthfully, I'm not sure what I am looking at. Oh, I know it is a list of paramaters for Apache, PHP, ..., but beyond that...????

Again, thanks for the help,
Gordon

laserlight

$row is an array, so you cant use it as an array index (like what you're doing with $option_server[$row])

bradgrafelman

What exactly are you trying to do with $option_server? Create an array with row numbers as keys and the $last_name variable as values? If so, try this:

// create variables for db items 
$i = 0;
while ($row = mysql_fetch_array($results)) { 
$last_name = $row["lname"]; 
$worker_class = $row["worker_class"]; 

if ($worker_class == "server") { 
$option_server[$i] = $last_name; 
$option_block_server .= "<option value='{$option_server[$i]}'>$last_name</option>" ; 

}

$i++;
}