RE: [PHP] Question about dynamically creating a form

Read the PHP manual section on variable variables http://www.php.net/manual/en/language.variables.variable.php Try $var = $search_result['list_serv_cd']; echo $var; if ($$var == 1){ } > -----Original Message----- > From: Amanda McComb [mailto:amccomb@indiana.edu] > Sent: Wednesday, April 16, 2003 1:05 PM > To: php-general@lists.php.net > Subject: [PHP] Question about dynamically creating a form > > > ok...so... > > I am creating a form using fields from a database. > > For one table, I select everything, loop through, and for each row, I > create a checkbox where the name = primary key, the value = 1, and the > label = whatever the name of the field is. It looks like this: > > <?$query = "SELECT * FROM list_serv_t WHERE retired_flag is NULL OR > retired_flag = 0 ORDER BY list_serv_name"; > $search_results = mysql_query($query) or die("Select Failed!"); ?> > > <?while ($search_result = mysql_fetch_array($search_results)) {?> > <INPUT > TYPE="checkbox" > NAME="<? echo $search_result['list_serv_cd']; > ?>" > SIZE="20" > MAXLENGTH="50" > VALUE="1"> <? echo > $search_result['list_serv_name'] ?> <BR> > <?}?> > > Now, that works fine, but how do I get the value out of that so I can > error check it or put it in the database? I tried re-creating the name by > looping through the table again and making a new variable that contains > the name like this: > > $var = "$"; > $var .= $search_result['list_serv_cd']; > echo $var; > > But when I try to test the value: > > if ($var == 1){ > } > > Obviously, it won't work, because $var == "$22" or whatever the number > happens to be. > > What's the best way to do this? > > Thanks, > Amanda > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php >