Hi,
one thing u can do is u can have a primary key the table you are
inserting so that duplication wont be done.
If thats not possible redirect the page to the results page
after updating the post data than showing him the results in the action page ( action i mean where the form
action has
On Mon, 20 Feb 2006, Mark Bomgardner wrote:
Is there a way in php to execute a SELECT statement in the future? What I am
wanting to do is use a page to parse a database at a future time. I want to
send out some email about an event I am putting on, but I don't want to
execute the select
i think its best to use php when both can perform the same opreration :).
there will less stress on server when u use php than mysql .
raj
On Sun, 29 Jan 2006, daniel smith wrote:
I am playing around with Mysql and php and I am not sure about when it
is best do something in mysql or php when
u can have the query as
$query = select * from $table where nameId = 31 ;
$result = mysql_query ( $query ) ;
$i = 0 ;
while ( $row = mysql_fetch_array( $result , MYSQL_NUM ) )
{
$array[$i][0] = $row[0] ;
$array[$i][1] = $row[1] ;
$array[$i][2] = $row[2] ;
think this is wrong ... as it gives unique alias even if the query is not
executed ..
rajesh
On Sun, 2 Oct 2005, Bastien Koert wrote:
you need to check the num of rows returned, since the query always(if it
works) returns true
$result = mysql_query($query);
if
i think for unique alias u have keep
alias='$alias' ??
and also if statement will always be true as the mysql statement is
always correct and will be executed .
U have to check the result of query for whether there
r ne aliases or not ...
On Sat, 1 Oct 2005, Ron Piggott wrote:
Take a
session_start() should be given at the starting of the file before others
u can include anthing after the session_satrt() functon only
On Fri, 26 Aug 2005, bo wrote:
here is the code for accesscontrol.php which control the access to protected
page, the server gives an error as of
Notice:
hai ..
if u r seeing http://www.google.com/blog/myblog.php
then
$_SERVER[SERVER_NAME] would give u www.google.com
and
$_SERVER[PHP_SELF] would give u /blog/myblog.php
for details just print the array $_SERVER and see it ..
On Sat, 6 Aug 2005, Ron Piggott wrote:
