Dropdown Menu used as a Filter

[pdlp]

Greetings,

I have searched these boards and haven't quite seen an answer to my dilemma.

I have a page that displays all messages preached(mostly in spanish). That works well.

Also, I have a drop down box on that page that dynamically displays the authors of those messages. That works well.

Question: I plan to have 30 messages displayed on the screen at most. How do I use the names in the dropdown box as a filter so that a user can view messages of that one author?

Code for drop down box and submit button

	echo '<form method="post" action="messages.php">';
	// creates drop down box form
	echo '<select name="author_txt">';
	echo '<option value="" selected>Choose Preacher</option>';

/*Generate List from mySql and chooses distinct names*/

	while ( $row = mysql_fetch_array( $dlist_ ) ) 
{

$info_ = $row['author_txt'];
echo "<option value='$info_'>$info_</option>";
}
echo '</select>';
echo ' <input type="submit" VALUE="Submit">';
echo '</form>';

After this code I am at a loss how to proceed. Do I need another query? More php code? I have used these forums for other php tasks and have found the answers I needed. I am thankful for that. As you can see this is my first post for help.
Thanks in advance for any assistance.

Paul

[ahundiak]

You do need another query. When the user presses the submit button, the value of whatever option was selected will be returned in author_txt.

So you need something like this in messages.txt

/* See if an author was picked */
$author_txt = $_POST['author_txt'];
if ($author_txt) {

  $set = mysql_query("Select * from messages where author='$author_txt'");
  if (mysql_errno()) die (mysql_error());
  while($rec = mysql_fetch_assoc($set)) {
    // Echo out each message
  }
}
else {
  // No author, use default query
}

[cgraz]

author_txt

Your select name is author_txt, so on the page
that processes this form, you run a select query that selects everything from the database
where the author_txt field matches the $author_txt variable (which is what they choose in the drop down)

Something like this:

$author_txt = $_POST['author_txt'] // if your php version is < 4.1.0 use $HTTP_POST_VARS
$query = mysql_query("SELECT * FROM your_table WHERE author_txt='$author_txt'");
// display your results

Cgraz

[pdlp]

Since, I am still new to php it took me a while to incorporate the code necessary to make it function. I took the first answer I saw. It makes sense now that I see it written and working.

Thanks much for your help.

Paul