php/mySQL help
| From: | Stinsman, Scott | Date: | Wed, 07 Feb 2001 21:00:29 +0000 |
| Subject: | php/mySQL help | ||
| Groups: | php.db | ||
| Request: | Send a blank email to php-db+get-6453@lists.php.net to get a copy of this message | ||
Hi everyone..been away and off the list for awhile.i am back to work and
need some help. I am not sure if I am having a problem with my PHP or
mySQL.
I have a table called "Restaurant" with a field called "Cuisine". There are
5 records in this table. When I run my code, it only returns one cuisine
when there are 5. I am trying to get all the Cuisines in the array called
CuisineMenuArray with no duplicates so I can generate a drop-down menu of
cuisines "on the fly".
Any help would be greatly appreciated! Thanx.
Here is the code:
<?php
$db=mysql_connect("localhost", "*******", "********");
mysql_select_db("quickcit", $db) OR DIE ("died at connect");
$query = "SELECT Restaurant.Cuisine ";
$query .= "FROM Restaurant ";
$query .= "ORDER BY Restaurant.Cuisine ASC ";
$mysql_result=mysql_query($query, $db) OR DIE ("died at query");
if(mysql_num_rows($mysql_result))
{
$h = 0;
$i = 1;
while ($row=mysql_fetch_array($mysql_result));
{
if ($i == 1)
{
//this only happens once since $i has the value 1 here only.
$CuisineMenuArray[0] = $row[0];
}
//this checks the next instance in $row against the previous instance in
$CuisineMenuArray. If they are not the same, it adds what is in $row to
thenext position in $CuisineMenuArray. If theyare the same it just skips to
the next instance in $row and checks it against the last instance in
$CuisineMenuArray again.
if ($row[$i] != $CuisineMenuArray[$h])
{
$CuisineMenuArray[$h+1] = $row[$i];
echo "$CuisineMenuArray[$h]";
echo "$row[$i]";
$h = ($h + 1);
$i = ($i + 1);
}
else
{
$i = ($i + 1);
}
}
}
?>
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) scott.stinsman@uphs.upenn.edu