quick question-query

bimmer528

How can I query info from table, then have that info post to a drop down menu where those values post to a differant table other then the original one queried?

sarahk

you have two scripts (or 2 parts to the same script).

the first queries the database and uses the output to build the text string that the browser identifies as an html dropdown.

the second uses the output of an html form to insert into a table.

neither script cares how it's output will be used or the exact source of it's input.

bimmer528

any example of this anywhere?

ereptur

here is a simplified version of what I think you are getting at.

print "<select name=\"select\">";

$query = "select value, displayname from table order by value";
$result = mysql_query($query);

while(list($value,$displayname) = mysql_fetch_array($resul)) {
   print "<option value=\"$value\">$displayname</option>";
}

bimmer528

// display form

print "<select name=\"uid\">";

$query = "SELECT uid, uname FROM xoops_users ORDER by uname";
$result = mysql_query($query);

while(list($uid) = mysql_fetch_array($result)) {
print "<option value=\"$uname\">$uid</option>";
}

k, I would like to have the value of uid inserted into a new table but have the script print the uname instead of uid so I know who is who.

So then the scipt would consist of this

//query and insert

if ($submit) {

// process form

$sql = "INSERT INTO users(user_id) VALUES ('$user_id')";

$result = mysql_query($sql);

echo "thank you";

} else{

// display form

print "<select name=\"uid\">";

$query = "SELECT uid, uname FROM users ORDER by uname";
$result = mysql_query($query);

while(list($uid) = mysql_fetch_array($result)) {
print "<option value=\"$uname\">$uid</option>";
}