Finally got the pieces to work together. For those of you who know what
you're doing, the following falls into the 'duh!' class. For anybody
else new to the game, I hope this helps.
Todo:
Disconnect the MySQL session.
Figure out a way to pass the db connection to the 2nd file rather than
reconnecting.
Convert everything to functions where it makes sense.
Play with other other form construction - embed php in html rather than
embedding html in php, figure out other methods to pass data between php
and html.
Dogu
Two files, both php.
First one returns a list of sites and launches the second form.
The second one uses the selected site name to return the password.
File 1:
<html>
<body>
<form action="getpw.php" method="get">
<?php
// connect to db
$db = mysql_connect("localhost");
mysql_select_db("test",$db);
$dbtbl = mysql_select_db("test",$db);
// build query
$sitequery = "SELECT site from sitedata order by site";
// generate result set
$siteresult = mysql_query($sitequery);
// use results
echo "Select a site from the list below: ";
echo "<br><br>";
echo "<select name='website'>";
while($siterow = mysql_fetch_array($siteresult))
echo "<option
value='".$siterow["site"]."'>".$siterow["site"]."</option>";
echo "</select></td>";
?>
<br>
<br>
<input type="submit" />
</form>
</body>
</html>
File 2 - getpw.php
The site you selected is <?php echo $_GET['website']; ?>.
<br>
<br>
The password is
<?php
// connect to db
$db = mysql_connect("localhost");
mysql_select_db("test",$db);
// build query
$pwquery = "SELECT password from sitedata where
site="."\"".$_GET['website']."\"";
// generate result set
$pwresult = mysql_query($pwquery);
$sitepw = mysql_data_seek($pwresult,0);
while($sitepw = mysql_fetch_array($pwresult))
echo $sitepw["password"];
echo "<br>";
echo "query is ".$pwquery;
?>
dogu wrote:
I'd like to have a form that does the following:
User picks a site from a dropdown list from a MySQL db (that works -
code below, thanks to all those on the web who posted code for this).
Once the value is selected, a second query is run using the site as the
'where' clause and the results are displayed back to the user.
Any help for this mysql, php, htlm noob. I think once I get this bit,
I'll be able to move forward with more interesting and complex code.
TIA
Doug
ps - if this message shows up 3 times, my apologies but I've been having
trouble posting.
<html>
<body>
<?php
// connect to db
$db = mysql_connect("localhost");
mysql_select_db("test",$db);
// build query
$sitequery = "SELECT site from sitedata order by site";
// generate result set
$siteresult = mysql_query($sitequery);
// use results
echo "Select a site from the list below: <br><br><select name='site'>";
while($siterow = mysql_fetch_array($siteresult))
echo "<option
value='".$siterow["site"]."'>".$siterow["site"]."</option>";
echo "</select></td>";
?>
</body>
</html>
