[RESOLVED] $_POST problem

twilliamsen

I can get one variable to pass from the form to a table to be able to print out. But I cannot get a different one to pass to a table on another page...
Am I messing something? No errors, no data, weird... I cannot see any differences

<form action="update_flr_plan.php" method="post" >

<? 
$connection = @mysql_connect("", "") or die("Could not Connect to Database Server"); 
$db = @mysql_select_db("move_db", $connection) or die("Could Not Select Database"); 
$sql = "SELECT ID, FirstName, LastName FROM employees  WHERE Position = 'Collector' ORDER BY LastName "; 
$result = @mysql_query($sql, $connection) or die("Could Not Execute Query"); 

echo '<select name = "employ1" ><option value="0">- Collector -</option>'; 
    while ($row = mysql_fetch_array($result)) { 
        $id = $row['ID']; 
		$fname1 = $row['FirstName'];
        $lname1 = $row['LastName']; 

    echo "<option value= \"$id1\">$lname1,$fname1</option>\n"; 
} 

echo '</select>';

And it gets passed to update_flr_plan.php

like so...


<td class="style1">Collector: <? echo $_POST['employ1'] ;  ?></td>

Piranha

You give the variable the name $id, then you use the variable $id1 as the option value:

$id = $row['ID']; 
echo "<option value= \"$id[color=red]1[/color]\">$lname1,$fname1</option>\n";

twilliamsen

ok... that helps, but now it only returns the ID number not the selected value?

twilliamsen

I cannot see why it is only returning the ID number, not the

$fname and $lastname variables.

Any ideas?

Piranha

Only what is in the value are returned.

value= \"$id1\"

It means that only the id will be transferred to the next page. I suggest that you do a tutorial to learn the basics of HTML if you don't understand this.

twilliamsen

that isn't a helpful answer. I am looking for an answer why, and how to fix it.

Piranha

I beg to differ. I answered your question, only what is in "value" are sent, and that is exactly what I wrote above.

In addition to that I gave you a very helpful answer. Do a tutorial or google to get easy and fast answers. I don't see any reason to spend my time to help people that don't want to help themselfes, and a simple search on google explained everything on the first hit. Of course I can do everything for you, but if you want me to do that then you have to pay me.

twilliamsen

I wouldn't pay a hack, besides I figured it out. I used trim() function to create a new variable