[PHP-DB] Re: Session across Multiple Hosts
Hi Hayan
Thechnically yes - for instance you could send it as a form hidden input.
But I wonder whether there is sense in it.
regards
Ivo
"Hayan Al Mamoun" <[EMAIL PROTECTED]> wrote in message
000401c1a8a9$838d4ee0$5e00a8c0@cybernation">news:000401c1a8a9$838d4ee0$5e00a8c0@cybernation...
> Dear all, I was wondering, can session variables be transfered across
> multiple hosts, i.e. if I send a request from http://host1/page1.phtml, to
> http://host2/page1.phtml where I started the session by SessionRegister()
or
> some other way and I gave the variable VAR1 some value, then I went back
> with header("Location: http://host1/page2.phtml";), does the VAR1 still
hold
> the value? if not, please advice with the way I can do this (transfere
> database results from one host to another without remote access).
>
> Best Regards
> Hayan
>
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Re: [PHP-DB] Multiple rows returned ?. . .
Why don't you try wqith cicle?
""Mrvball008"" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Hi I have little tiny script like so..
>
> -
> $dbName = "somedb";
> $dbPass = "somepass";
> $dbUserName = "someuser";
> $host = "localhost";
> $db = mysql_connect($host, $dbUserName, $dbPass) or die ("Could not
connect
> to database");
> mysql_select_db($dbName,$db);
> $sql = "SELECT * FROM downloads";
> $result = mysql_query($sql, $db);
> $row = mysql_fetch_array($result);
> $row1 = mysql_fetch_array($result);
> ?>
> ---
>
> no I know there has to be a more efficient way of returning the $row and
> $row1 without writing it down like so..
>
> anyone one can throw me a clue? :)
>
> Thank
>
>
> Andre
>
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Re: [PHP-DB] Evaluting T/F (3)
Try
if($myrow = pg_fetch_array($result,0)) {
do whatever
} else {
or die(include("../sqlerror.inc"));
}
and it will work accordingly
Ivo
""Julio Cuz, Jr."" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> HI-- Sorry again...aggg...
>
> Hi--
>
> I'm trying to evaluate the following lines to see if "$myrow" is false or
> true, can you help?
>
> -
> $db .;
> $sql = "SELECT * FROM Employees WHERE ID=$id";
> $result = pg_exec($db, $sql);
> $myrow = pg_fetch_array($result,0) or die(include("../sqlerror.inc"));
> -
>
> Julio Cuz, Jr.
> Riverside Community College
> [EMAIL PROTECTED]
>
>
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Re: [PHP-DB] PHP how to compare?
Hi Duky
Try to add this before if statement
echo("$PASS$PASSNOG");
it will show you whether your vars are correct and whether there is
something in at all!
if ($PASS != $PASSNOG) {
echo "password doesn't match! $back";
exit;
}
Good luck
Ivo
"Duky" <[EMAIL PROTECTED]> wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
> Can somebody tell me what the proper statement is for this? I want two
> variables to match with eachother but this doesn't seems to work. Help
> please? THNX
>
>if ($PASS != $PASSNOG) {
> echo "password doesn't match! $back";
> exit;
> }
>
> Duky
>
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Re: [PHP-DB] PHP-MySQL create table problem
Yes I did it on my test computer. It works fine. ""Cal Evans"" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > Maybe it's just me but I've never seen anyone create a table by using a > Select statement. Have you verified that you can do this? (like you've > checked the manual and this construct is supported?) > > Cal > http://www.calevans.com > > > -Original Message- > From: Ivo Stoykov [mailto:[EMAIL PROTECTED]] > Sent: Monday, March 12, 2001 3:37 PM > To: [EMAIL PROTECTED] > Subject: [PHP-DB] PHP-MySQL create table problem > > > Hello > > I have following problem and really need your help. > > I'm trying to create a new table using "CREATE TABLE partlist_back SELECT * > FROM partlist" from a php script. The new table have to be based on an > existing table. On the machiene I test it works fine when I upload it on the > server I receive a query error! > > Please help me! > > Ivo > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > To contact the list administrators, e-mail: [EMAIL PROTECTED] > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, e-mail: [EMAIL PROTECTED] > For additional commands, e-mail: [EMAIL PROTECTED] > To contact the list administrators, e-mail: [EMAIL PROTECTED] > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
[PHP-DB] PHP-MySQL create table problem
Hello I have following problem and really need your help. I'm trying to create a new table using "CREATE TABLE partlist_back SELECT * FROM partlist" from a php script. The new table have to be based on an existing table. On the machiene I test it works fine when I upload it on the server I receive a query error! Please help me! Ivo -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] To contact the list administrators, e-mail: [EMAIL PROTECTED]
