php-windows Digest 18 Jul 2001 14:04:35 -0000 Issue 657
Topics (messages 8362 through 8373):
phpinfo()
8362 by: Stuart Hunter
8363 by: Robin Bolton
8369 by: Philippe Saladin
please help, a strange problem I met in form subimt.
8364 by: tttk
Re: In Win2K, how to set the file upload permission, thank you.
8365 by: tttk
Access Violation on WinNT4
8366 by: Jorge B. Ribarr
8367 by: Jorge B. Ribarr
8371 by: elias
how to update the cookie?
8368 by: science
Printing text on a printer in PHP4 Using PRINTER function
8370 by: William Schipper
Re: I cant install php on my computer
8372 by: Phil Driscoll
mysql_fetch_array
8373 by: Andrew.Martin
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----------------------------------------------------------------------
I have just installed version 4.06 on my Windows 2000 machine, but when
I open the web page I have that contains phpinfo() it still reports
4.05. Is that a known issue or have I done something wrong in my
upgrade?
Stuart
My guess is that you probably have an outdated .dll in you winnt\system32
folder. Try copying all the necessary .dll files from the 4.06 zip file
again.
> -----Original Message-----
> From: Stuart Hunter [mailto:[EMAIL PROTECTED]]
> Sent: July 17, 2001 3:40 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-WIN] phpinfo()
>
>
> I have just installed version 4.06 on my Windows 2000 machine, but when
> I open the web page I have that contains phpinfo() it still reports
> 4.05. Is that a known issue or have I done something wrong in my
> upgrade?
>
> Stuart
>
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>
>
Is php406 installed instead of 405 or have you installed it in an other
directory? in the second case, you will have to configure your web server
again.
Philippe
"Stuart Hunter" <[EMAIL PROTECTED]> a �crit dans le message
news: [EMAIL PROTECTED]
> I have just installed version 4.06 on my Windows 2000 machine, but when
> I open the web page I have that contains phpinfo() it still reports
> 4.05. Is that a known issue or have I done something wrong in my
> upgrade?
>
> Stuart
I use win2000, IIS, PHP4.04.
I made a form file which included a "file" object for uploading files, and
it's named "upload.php". The "action" parameter of the file object is set to
another file called "submit.php".
In submit.php, I expect it return back to it's caller "upload.php" after
finishing it's task. So I added code as: { header("Location: upload.php"); }
in the end. After I submit my request, an error happened, which said "can't
find server or DNS error".
But, if I add a code as: { echo "<body
onload=\"location.assign('upload.php')\"></body>"; }
which product a javascript redirect indication, oH, it's OK.
The same case, if the form doesn't include a file object, the header()
function will execute correctly.
Who can tell me what's the reason? Thank you for all.
Thank you all for your help.
I have already find the reason. It's simply for a wrong path spelling
mistake.
Hi,
When a user stop a process, with the stop button or click in another link.
Php popup a mensage "Access Violation".
I use php4.0.5, apache 1.3.9, ms sql server 7, odbc , winnt4.
All most this occours in a query execution. I use a class phpodb.php (from
www.zend.com - code galery)to access the odbc conection.
I would like to now, if this is the cause of the access violation.
see this sample:
$dbBook = new PHPDBObject;
$dbBook->opendb(); // conect with database
$sSQL = "select AUTHOR,TITLE from BOOKS order by TITLE";
$dbBook->openrs($sSQL); // execute query
$dbBook->movetofirst();
while (!$dbBook->eof()){
$sAuthor = $dbBook->("AUTHOR"); // get colum value from table
$sTitle = $dbBook->("TITLE");
$dbBook->movetonext();
print "$sTitle - $sAuthor <br>\n";
}
$dbBook->free(); // clear result set
$dbBook->close(); // close dbconection
Please, i dont understand that is happen here.
Excuse-me my bad english.
Jorge Ribarr
from Brazil
In the version 3.0.16 of php this dont ocours
Thanks
Jorge Ribarr
from Brazil
----- Original Message -----
From: Jorge B. Ribarr <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, July 18, 2001 12:37 AM
Subject: [PHP-WIN] Access Violation on WinNT4
Hi,
When a user stop a process, with the stop button or click in another link.
Php popup a mensage "Access Violation".
I use php4.0.5, apache 1.3.9, ms sql server 7, odbc , winnt4.
All most this occours in a query execution. I use a class phpodb.php (from
www.zend.com - code galery)to access the odbc conection.
I would like to now, if this is the cause of the access violation.
see this sample:
$dbBook = new PHPDBObject;
$dbBook->opendb(); // conect with database
$sSQL = "select AUTHOR,TITLE from BOOKS order by TITLE";
$dbBook->openrs($sSQL); // execute query
$dbBook->movetofirst();
while (!$dbBook->eof()){
$sAuthor = $dbBook->("AUTHOR"); // get colum value from table
$sTitle = $dbBook->("TITLE");
$dbBook->movetonext();
print "$sTitle - $sAuthor <br>\n";
}
$dbBook->free(); // clear result set
$dbBook->close(); // close dbconection
Please, i dont understand that is happen here.
Excuse-me my bad english.
Jorge Ribarr
from Brazil
Try using PHP as CGI instead of ISAPI...that if you're not already using
ISAPI
"Jorge B. Ribarr" <[EMAIL PROTECTED]> wrote in message
003101c10f3b$1ecca620$0101a8c0@rose">news:003101c10f3b$1ecca620$0101a8c0@rose...
Hi,
When a user stop a process, with the stop button or click in another link.
Php popup a mensage "Access Violation".
I use php4.0.5, apache 1.3.9, ms sql server 7, odbc , winnt4.
All most this occours in a query execution. I use a class phpodb.php (from
www.zend.com - code galery)to access the odbc conection.
I would like to now, if this is the cause of the access violation.
see this sample:
$dbBook = new PHPDBObject;
$dbBook->opendb(); // conect with database
$sSQL = "select AUTHOR,TITLE from BOOKS order by TITLE";
$dbBook->openrs($sSQL); // execute query
$dbBook->movetofirst();
while (!$dbBook->eof()){
$sAuthor = $dbBook->("AUTHOR"); // get colum value from table
$sTitle = $dbBook->("TITLE");
$dbBook->movetonext();
print "$sTitle - $sAuthor <br>\n";
}
$dbBook->free(); // clear result set
$dbBook->close(); // close dbconection
Please, i dont understand that is happen here.
Excuse-me my bad english.
Jorge Ribarr
from Brazil
Hi all,
Does anybody know how to update the value in the cookie?
Thanks
DP
Does anybody has a script that will print basic text to a printer that is on
the client side. In PHP4 it must be possible to write text to the printer.
I'm working with EasyPHP. Do i have to include the file 'php_printer.dll' in
the extensions dir and what lines do i have to add in the php.ini file? I
just want to print text. Can anybody help me please.
I can't find it anywhere.
William Schipper
[EMAIL PROTECTED]
On Tuesday 17 July 2001 22:22, Administrador wrote:
> I've tryed cgi version but php doesn't work anyway
Give us more details. As thousands of users will testify, it *does* work -
there will just be a problem with your setup.
Cheers
--
Phil Driscoll
I'm using this function to check the value of specific fileds per row in the
database.
such as....
$query = "SELECT * FROM $dbn WHERE status = 4";
$result = mysql_query($query) or die("failed to connect to DB");
while ($row = mysql_fetch_array ($result)) {
print "row id = $row[id]";
print "decription in row = $row[description]";
}
this works fine, it checks the DB for values with status = 4 and prints row
id and row description
My problem occurs when I want to creat variables based on these results.
I would ideally like to create dynamic variables based on row_id number e.g.
variable_row[id] = $row[description]
so later in the dynamically created page I could enter $variable_1 (1 being
the row[id] value) and the correct value would be selected. Is there anyway
of doing this? I tried using $$variable etc but no luck.
Any ideas appriciated.
Andrew
