Updating an HTML combo box from MySQL DB using PHP

So,

what is the problem? Is your script not working or do you want a active Combobox on one screen to be updated, when on another screen a new entry for supplier is entered?

Ok, some hints:

1. The "select"-Tag should be placed outside the loop.
2. Put the column name in "
3. Add a display-Value for the options

So, without checking it:

/*echo '<form action="">';*/
echo "<select name='supplier'>";    
while($row = mysql_fetch_array($result)) 
{ 
    echo "<option value = '".$row["supplier"]."'>".$row["supplier"]."</option>"; 
}                        
echo "</select>"; 

Your problem is here:

            $result = mysql_query("SELECT supplier FROM supplier"); 
            while($row = mysql_fetch_array($result)) 
                    { 
                    /*echo '<form action="">';*/ 
                    echo "<select name='supplier'>"; 
            echo "<option value = '$row[supplier]'>""</option>"; 
                    echo "</select>"; 

You're creating the drop down box (the Select) inside of the mysql data loop. As @Hitesh has explained. You need to create this outside of the loop and only echo out the data results within. For example:

$result = mysql_query("SELECT supplier FROM supplier");
echo "<select name='supplier'>";
while($row = mysql_fetch_array($result))
{
    echo "<option value=".$row['supplier'].">".$row['supplier']."</option>";
}
echo "</select>";

This will output your drop down box, with all your supplier names as the value and the displayed text options.

If you attempted to do the following:

            $result = mysql_query("SELECT supplier FROM supplier"); 
            while($row = mysql_fetch_array($result)) 
                    { 
                    /*echo '<form action="">';*/ 
                    echo "<select name='supplier'>"; 
            echo "<option value = '$row[supplier]'>""</option>"; 
                    echo "</select>"; 
                    }

You would just end up with as many drop down boxes as you had suppliers in your database. This is because you're creating a new Select box for each record found.

Also, without specifying the second $row['supplier'] between the option tags, you'd just end up with a blank (empty) drop down box.

Hope this helps.

Here is complete code:

PHP Code:

<?php
// select box open tag
$selectBoxOpen =  "<select name='supplier'>"; 
// select box close tag
$selectBoxClose =  "</select>";
// select box option tag
$selectBoxOption = ''; 

// connect mysql server
$con = mysql_connect("localhost","user","pass"); 
if (!$con) { 
    die('Could not connect: ' . mysql_error()); 
} 

// select database
mysql_select_db("rtgs", $con); 
// fire mysql query
$result = mysql_query("SELECT supplier FROM supplier");
// play with return result array 
while($row = mysql_fetch_array($result)){   
    $selectBoxOption .="<option value = '".$row['supplier']."'>".$row['supplier']."</option>"; 
}
// create select box tag with mysql result
$selectBox =  $selectBoxOpen.$selectBoxOption.$selectBoxClose;   
?>

HTML Code:

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

<table>
    <tr>
        <td>Supplier Name:</td>
        <td> 
            <?php echo $selectBox;?>
        </td>
    </tr>

    <tr>
        <td>Bill No. :</td>
        <td><input type ="text" name="billno"/></td>
    </tr>

    <tr>
        <td>Bill Date : </td>
        <td><input type="date" name="billdate"/></td>
    </tr>

    <tr>
        <td>Bill Amount : </td>
        <td><input type="text" name="billamt"/></td>
    </tr>

    <tr>
        <td>NEFT / RTGS No. :</td>
        <td><input type="text" name="rtgs"/></td>
    </tr>

    <tr>
        <td><input type="submit" name="Save"/></td>
    </tr>
</table>

</form>