[RESOLVED] Conditional showing based on choice in a dropdown menu

phpzero

Hey all,

I've got a question for you all. I would like to repeat a form a number of times based on what the user filled out in a form on the previous page.

This is the first coding page:

selecttravel.php

<html>
<body>
<form method="post" action="definetrip.php">
<h4>Will the trip include one or more stop-overs?</h4>

 <table>
  <tr>
    <td>No</td>
	<td> <input type="radio" name="numberofstops" value="0" checked> </td>
  </tr>
  <tr>
    <td> Yes, </td> 
		<td> 1 stop</td>
		<td> 2 stops</td>
		<td> 3 stops</td>
		<td> 4 stops</td>
		<td> 5 stops</td>
    </tr>
  <tr>
    <td> </td> 
		<td> <input type="radio" name="numberofstops" value="1"> </td>
		<td> <input type="radio" name="numberofstops" value="2"> </td>
		<td> <input type="radio" name="numberofstops" value="3"> </td>
		<td> <input type="radio" name="numberofstops" value="4"> </td>
		<td> <input type="radio" name="numberofstops" value="5"> </td>
    </tr>

</table>
</body>
</html>

the definetrip.php page I'm currently looking at is the following:

<html>
<head><title>Definition of the trip</title></head>
<body>
<table width="650">
    <tr>
	  <td width="250">Please select the stopover:</td>
      <td><select name="transit" size="1">
                   <?
                    $result=mysql_query("SELECT name, country FROM cities.cities");
					$row = mysql_fetch_array($result);
                    while ($row) {
                         ?>
                         <option value="<?print($row['name']);?>"><?print($row['name']);?>
                         <?
                         $row = mysql_fetch_array($result);
                    }
                    ?>
		 </select><br></td>
       <br>
    </tr>
</table>
</body>
</html>

I tried to include the whole table in some while functions (I mean "while ($numberofstops>0, and so on...)") but never had any result that even came close to what I wanted to obtain.

Anyone having an idea on how to get me closer to what I'd like to do?

Thanks in advance for your tips!

johanafm

You lose the value inbetween page requests as there is nothing telling the server that one request is a followup unless you keep sending that info back and forth between browser and server. Three options:
1. Start a session, then use session variables (there will be a cookie keeping track of the session). Have a look at session_start()
2. Set a cookie
3. Keep the info in a hidden input:

echo '<input type="hidden" name="numberofstops"
value="'.(intval($_POST['numberofstops']) - 1).'" />';

rincewind456

First I would create a variable with your city list select box like

$result=mysql_query("SELECT name, country FROM cities.cities");

$cityDropDown = '<td><select name="transit" size="1">';

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

$cityDropDown .='<option value="'.$row['name'].'">'.$row['name'].'</option>';
}

$cityDropDown .=  '</select></td>';

to stop the need to run the sql query more than once.

Then use a for loop like

$count =  (int)$_POST['numberofstops'];
for ($i=1;$i<=$count;$i++){
	print $cityDropDown;
}

to print out the correct number of drop downs within your table.

phpzero

Alright! That works out just fine! You guys rock! 🙂