php不会将变量传递给php rss函数

Question

I have http://communitychessclub.com/test.php

I have a php function named "getPosts" that I want to send an url to from a form:

<form action="#" method = "post"> 
<input type="hidden" name="sub" value="1" /> 
<select name="url" id="url">
<option value="http://www.chess.com/rss/articles">chess.com</option>
<option value="http://chesscafe.com/feed/">Chess Cafe</option>
<option value="http://www.chessdom.com/rss">Chessdom</option>
<option value="http://chess-news.ru/rss-eng">Chess-news</option>

</select>
</form>

For some reason, the php code below doesn't grab the option selected and pass it to the function. Why is that?

<?php
$sub = intval( $_POST["sub"]);  
if ($sub == 1){ 
$url= $_POST["url"];
}
else{
$url = "http://www.theweekinchess.com/twic-rss-feed"; 
}

getPosts($url); ?>

Answer 1

its because there is no submit button. if you want to auto-submit form on change you need to do something like

<form name="myform" action="#" method = "post">
<select name="url" id="url" onchange="document.forms['myform'].submit()">

I just checked and it looks it works fine.

Also you don't really need $sub thing, you can just do

<?php

if(isset($_POST["url")){
$url = $_POST["url"];
}
else{
$url = "http://www.theweekinchess.com/twic-rss-feed"; 
}

getPosts($url); ?>