Re: Populate picklist from directory and return file name

In article <1126718878.324648.246160@g49g2000cwa.googlegroups.com>,
 "sunbum" <mbrashars@gmail.com> wrote:

> Sorry for replying to my own post.
>
> Here is the latest version and I think it's closer.
>
> Can someone help me clean up the syntax?
>
> Thanks!
> ------- Code begin (this is NOT in the script ------------------
>
> <?php
> if (isset($_POST['file']))
> {
> // process the data, see if it's valid.
> // if it's valid and you're satisfied, do whatever you want with
> the data
> // if not, start outputting the form
>
> $file = "/inetpub/wwwroot/templogs/"+$_POST['file'];
> // Display the Graph
> displaygraph();
> }
>
> else {
> // display the form
> displayform();
> }
>
> <?php
> function displaygraph() {
> echo "So far so good, this would draw graph with $file";
>
> }
>
>
> ?>
>
> <?php
> function displayform() {
>
>
> <form name="cooler" action="<?php $server['PHP_SELF']?>"
> method=\"post\">";
> $dir = "/inetpub/wwwroot/templogs/";
>
> echo "<select name='file'>";
> $dir = opendir($dir);
> while (false !== ($file = readdir($dir))){
> if (in_array($file, array(".", ".."))) continue;
> print "<option value='$file'>$file</option>";
> }
> </select>;
>
>
> <input type="submit" value="Submit">;
> </form>
>
> }
>
>
> ?>
>
>
> ?>
>
> ------- Code end (this is NOT in the script ------------------
>
> Mike

The above is a good start. But you can't just start typing "<form name..."
inside a function block. You need to use

    print "<form name='cooler' ....";

I always do my web app logic like this:

<?
    while (1){
        if ($_POST["file"]){
            # Do fancy stufff
            break;
        }

        # output form
        $dir = "/inetpub/wwwroot/templogs/";
        # ...

        break;
    }

-- 
Sandman[.net]