RE: [PHP] PHP, JavaScript and populating DropDowns

[Creative Solutions New Media]

Thanks very much John.  It looks great.

Tim Winters
Manager, Creative Development
Sampling Technologies Incorporated (STI)
[EMAIL PROTECTED]
[EMAIL PROTECTED]
W: 902 450 5500
C:  902 430 8498

-----Original Message-----
From: John Manko [mailto:[EMAIL PROTECTED]
Sent: July 29, 2003 2:49 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED] Php. Net
Subject: Re: [PHP] PHP, JavaScript and populating DropDowns

assuming you have a table with colors, and each color has car associated
with it (yes, there might be dup color entries).
This should work (check spellings and case, etc..etc...)

$query="SELECT * from colors";
$results = mysql_query($query) or die ("DB ERROR:" . mysql_error());
$num = mysql_numrows($results);
$i=0;
echo "<script language=javascript>\n<!--\n";
while ($i < $num) {
    echo "\n ca[$i] = new Array();";
    echo "\n ca[$i][0] = \"" . mysql_result($results,$i,"CarID") . "\";";
    echo "\n ca[$i][1] = \"" . mysql_result($results,$i,"ColorID") . "\";";
    echo "\n ca[$i][2] = \"" . mysql_result($results,$i,"ColorName") .
"\";\n";
    ++$i;
}

echo "
function changeColor(ns)
{    var SelectColor, NumEntries, i;

    for (i = document.form1.selectcolor.length; i > 0; i--)
    {
        document.addform.selectcolor.options[i-1] = null;
    }

    document.form1.selectcolor.options[0] = new Option("-- Select Color
--",0);
    len = 0;
    if (ns != "0")
    {
        NumEntries = ca.length

        for (i = 0; i < NumEntries; i++)
        {
            if(ca[i][0] == ns){
                len++;
                document.form1.selectcolor.options[len] = new
Option(ca[i][2],ca[i][1]);
            }
        }
    }
    document.form1.selectcolor.selectedIndex = 0;
}

";
echo "\n//-->\n</script>\n";


this in later in the page...


$query="SELECT * from car";
$results = mysql_query($query) or die("<br>ERROR - Please Contact Site
Administrator<br>");
$num = mysql_numrows($results);
$i=0;
echo "<select onchange='changeColor(this.value);'
name='selectcar'><option value='0'>Select Car</option>";
while ($i < $num) {
          $id=mysql_result($results,$i,"CarID");
          $name=mysql_result($results,$i,"CarName");
         echo "<option value='$id'>$name</option>";
          ++$i;
 }
echo "</select>";


echo "<select  name="selectcolor"><option value='0'>-- Select Color
--</option></select>";



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Creative Solutions New Media wrote:

>Hello,
>
>I am using mySQL and PHP to pull some data into my page.
>
>I have 2 dropdown boxes.
>
>As an example lets say dropbox1 has a list of cars and dropbox2 has a list
>of colours.
>
>Here's the thing.  The colours for each car vary so I won't know what
>colours to populate the 2nd dropbox with until a car is chosen.  I really
>want this all contained on 1 page.
>
>Is there a standard way of dealing with this?  The only thing I can think
of
>is to reload the page once the make of car is selected which would give me
>the opportunity to populate the 2nd drop box at that time but that doesn't
>seem like a good solution.
>
>I don't know much in the way of JS but is there a way to do what I need
that
>way?
>
>Thanks.
>
>Tim Winters
>Manager, Creative Development
>Sampling Technologies Incorporated (STI)
>[EMAIL PROTECTED]
>[EMAIL PROTECTED]
>W: 902 450 5500
>C:  902 430 8498
>
>
>
>
>
>



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