Thank you for your quick reply.
So I try to rewrite code again below;
$handle = popen(/usr/bin/pdftotext \$original_name\ - -layout ,'r');
$read = fread($handle, 2048000);
echo $read;
pclose($handle);
But I can not see anything!!
Please do help me!
Yui
Micah Stevens wrote:
Just read the
type=password
bastien
From: Ron Piggott [EMAIL PROTECTED]
Reply-To: Ron Piggott [EMAIL PROTECTED]
To: PHP DB php-db@lists.php.net
Subject: [PHP-DB] Password field in a web form
Date: Sat, 17 Sep 2005 19:20:49 -0500
How do you make *'s come up on the screen in a password field instead of
what
Dunno if its relevant, but the recommended way to use sessions is to assign
stuff to a $_SESSION['varname']...I don't really see anything wrong with the
code...
Also, what arrays are causing the problems?
bastien
From: NIPP, SCOTT V (SBCSI) [EMAIL PROTECTED]
To: php-db@lists.php.net
Well, of course, this assumes that your pdftotext program is located
at /usr/bin/. You're getting some error, and stderr is not being shown using
this. Change the line to this:
$handle = popen(/usr/bin/pdftotext \$original_name\ - -layout 21, 'r');
That 'should' redirect stderr to the fread
Actually, this also assumes you're on a *nix type server. If you're running on
windows, you likely won't ever get this to work unless you compile pdftotext
using cygwin, and install it that way, which considering your skill level,
may be a bit much.
Otherwise I'm not sure I know of a windows
hi,
can some one enlighten me on what is wrong with my query?
I want to set the fax number of all the employees of a particular company to
the company's fax number. This is because the employees do not have personal
fax machines, thus instead of just leaving T020FaxNo blank, I want to update
it
i hope it's helping
sql code:
UPDATE T020Employee SET T020FaxNo = T010FaxNo
FROM T020Employee a JOIN T010Company b ON a.T020CoCd =b.T010CoCd
- Original Message -
From: Ng Hwee Hwee [EMAIL PROTECTED]
To: PHP DB List php-db@lists.php.net
Sent: Monday, September 19, 2005 9:14 AM
Subject:
only versions 4.1 and above support subqueries...
bastien
From: Ng Hwee Hwee [EMAIL PROTECTED]
To: PHP DB List php-db@lists.php.net
Subject: [PHP-DB] Subquery
Date: Mon, 19 Sep 2005 10:14:34 +0800
hi,
can some one enlighten me on what is wrong with my query?
I want to set the fax number of
thanks but it doesn't work =(
is it because i'm on MySQL version 3.23?
i have an error 1064 when running it...
hwee
- Original Message -
From: Nandar [EMAIL PROTECTED]
To: Ng Hwee Hwee [EMAIL PROTECTED]; PHP DB List php-db@lists.php.net
Sent: Monday, September 19, 2005 11:02 AM
Subject:
hi all,
i found the solution! it's so easy actually.. i think my brain is not
functioning well on a Monday morning!! ;p
UPDATE T020Employee, T010Company
SET T020FaxNo = T010FaxNo
WHERE T020CoCd = T010CoCd
thanks bastien and nandar for your help! =)
hwee
- Original Message -
From:
It doesn't work because MySQL version 3.23.50 does not support sub-select
statement.
http://dev.mysql.com/doc/mysql/en/subqueries.html
And
How to re-write it in MySQL 4.1
http://dev.mysql.com/doc/mysql/en/rewriting-subqueries.html
Nandar wrote on Monday, 19 September 2005 1:03 PM:
i hope
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