ok, here is the code that runs after the submit button is clicked. i now
suspect that it has something to do with my "if" statements.....
$db=mysql_connect("localhost", "******", "*******");
mysql_select_db("*******", $db) OR DIE ("died at connect");
if ($AreaTemp == 'ALL' && $CuisineTemp != 'ALL') {
echo ($AreaTemp);
echo ($CuisineTemp);
$query = "SELECT * ";
$query .= "FROM Restaurant ";
$query .= "WHERE '$CuisineTemp' = Restaurant.Cuisine ";
$query .= "ORDER BY Restaurant.Price ";
$mysql_result=mysql_query($query, $db) OR DIE ("died at query");
------code here--------
}
elseif ($AreaTemp != 'ALL' && $CuisineTemp = 'ALL')
{
echo ($AreaTemp);
echo ($CuisineTemp);
$query = "SELECT * ";
$query .= "FROM Restaurant ";
$query .= "WHERE '$AreaTemp' = Restaurant.Area ";
$query .= "ORDER BY Restaurant.Price ";
$mysql_result=mysql_query($query, $db) OR DIE ("died at query");
--------code here------------
}
elseif (($AreaTemp == 'ALL') AND ($CuisineTemp == 'ALL')) {
$query = "SELECT * ";
$query .= "FROM Restaurant ";
$query .= "ORDER BY Restaurant.Price ";
$mysql_result=mysql_query($query, $db) OR DIE ("died at query");
-------code here-----------
} else{
$query = "SELECT * ";
$query .= "FROM Restaurant ";
$query .= "WHERE '$CuisineTemp' = Restaurant.Cuisine AND '$AreaTemp' =
Restaurant.Area ";
$query .= "ORDER BY Restaurant.Price ";
$mysql_result=mysql_query($query, $db) OR DIE ("died at query");
---------code here-------------
-----Original Message-----
From: Rolf Hopkins [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, February 14, 2001 12:14 PM
To: Stinsman, Scott
Cc: [EMAIL PROTECTED]
Subject: Re: mySQL/php query help
No, this does not help.
All you have provided was the code that does work. From your original
description, you are able select the various items from the list. That
means you are actually selecting items in a list. Now, if you said that the
only item in either or both lists was ALL then you would have a problem with
the code you gave us.
However you said you have a problem after you have selected ALL for both
lists and tried to get results. This is the code we need to see!! Ie. the
code that activates after pressing the submit button.
----- Original Message -----
From: "Stinsman, Scott" <[EMAIL PROTECTED]>
To: "'Rolf Hopkins'" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Wednesday, February 14, 2001 23:29
Subject: RE: mySQL/php query help
> hey rolf--
>
> everything works fine if the user chooses "ALL" in both pop-up menus and
> also works if they choose "ALL" in at least one of the pop-up menus. But
if
> the user does not choose the "ALL" option in both pop-up menus, it does
not
> work properly. i stuck a couple echo statements in the code to see what
> data was actually in the 2 variables. below is what happened.
>
> SELECTIONS VARIABLES
>
> ALL/ALL ALL/ALL
> ALL/Not ALL ALL/Not ALL
> Not ALL/ALL Not ALL/ALL
> Not ALL/Not ALL Not ALL/ALL
>
> It seems the AreaTemp variable takes the value "ALL" when the CuisineTemp
> variable is something other than "ALL".
>
> so,in english, the code works in 3 instances and fails in 1. The code
works
> if the user wants all cuisines in all areas, all cuisines in a specific
> area, and a specific cuisine in all areas. but it does not work if the
user
> wants a specific cuisine in a specific area and this is because the
AreaTemp
> variable remains "ALL".
>
> does this help? any suggestions?
>
> --scott
> -----Original Message-----
> From: Rolf Hopkins [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, February 13, 2001 8:30 PM
> To: Stinsman, Scott; [EMAIL PROTECTED]
> Subject: Re: mySQL/php query help
>
>
> I can't see anything wrong with your code? What exactly is the problem?
> Aren't the values, from the select statement printing out on your web page
> or something? Are you sure you provided the right code?
>
> ----- Original Message -----
> From: "Stinsman, Scott" <[EMAIL PROTECTED]>
> To: <[EMAIL PROTECTED]>
> Sent: Tuesday, February 13, 2001 23:17
> Subject: mySQL/php query help
>
>
> > can anyone figure out why the following code works fine when both select
> > variables = "ALL" and also when only 1 of the 2 ="ALL" but NOT when BOTH
> > select variables equal something other than "ALL"?
> >
> > here is the code:
> >
> > What area of the city would you like to dine in?
> > <?php
> > $db=mysql_connect("localhost", "******", "*********");
> >
> > mysql_select_db("*******", $db) OR DIE ("died at connect");
> >
> > $query = "SELECT DISTINCT Restaurant.Area ";
> > $query .= "FROM Restaurant ";
> > $query .= "ORDER BY Restaurant.Area ASC ";
> > $mysql_result=mysql_query($query, $db) OR DIE ("died at query");
> >
> > $select="<select name=\"AreaTemp\">\n";
> > $select.="<option value=\"ALL\">ALL</option>\n";
> > while(list($AreaTemp)=mysql_fetch_array($mysql_result)) {
> >
> > $select.="<option value=\"$AreaTemp\">$AreaTemp</option>\n";
> >
> > }
> >
> > $select.="</select>";
> > echo "$select";
> > ?>
> > <br>
> > What type of cuisine do you feel like?
> >
> >
> > <?php
> > $query = "SELECT DISTINCT Restaurant.Cuisine ";
> > $query .= "FROM Restaurant ";
> > $query .= "ORDER BY Restaurant.Cuisine ASC ";
> > $mysql_result=mysql_query($query, $db) OR DIE ("died at query");
> >
> > $select="<select name=\"CuisineTemp\">\n";
> > $select.="<option value=\"ALL\">ALL</option>\n";
> > while(list($CuisineTemp)=mysql_fetch_array($mysql_result)) {
> >
> > $select.="<option
> > value=\"$CuisineTemp\">$CuisineTemp</option>\n";
> >
> > }
> >
> > $select.="</select>";
> > echo "$select";
> >
> > ?>
> >
> > <br>
> >
> > <input type ="submit" value="Go!" name="submit">
> > <input type="RESET" value="Clear" name="RESET">
> >
> > Scott Stinsman
> > Academic Coordinator
> > General Internal Medicine
> > 1215 Blockley Hall
> > 423 Guardian Drive
> > Philadelphia, PA 19104-6021
> > (phone) 215-662-7623
> > (fax) 215-349-5091
> > (email) [EMAIL PROTECTED]
> >
> >
> > ---------------------------------------------------------------------
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> >
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