Sorry for resending the same question. Actually,
I sent the following message a few days before.
However, it seems that no one has the experience
on LDAP. It's a headache for me. Hope this time
some experts can take a look at the question.
I appreciate your kind helps!
*
Sorry for resending the same question. Actually,
I sent the following message a few days before.
However, it seems that no one has the experience
on LDAP. It's a headache for me. Hope this time
some experts can take a look at the question.
I appreciate your kind helps!
***
Dear friends:
I have a problem of installation of php.
I'm using Linux 7.3. I have installed
openldap2.1.22 successfully.
Now I'm compiling php with the following
command:
./configure --with-apache2=/usr/local/apache2 \
--with-mysql=/usr/local/mysql \
--with-ldap=/usr/l
Hi, All:
I have some information stored in the database.
When I try to show them, sometimes it give me the
following warning:
Warning: printf(): Too few arguments ...
Here is the problematic code:
printf("$des ");
I figured out that if there are "%" in the
string $des, it will give me this war
Yes. I'll do that. In this example, the only thing
that I can think about input checking is to use
is_int() function. Is there any other ways to do
the input checking?
Thanks.
Zhan Xu
EECS Department
Case Western Reserve University
- Original Message -
From: Jason Wong <[EMAIL PROTECTED]
Hi, Jason and Shahmat:
I figured it out! The register_globals is turned
off. To parse the id with this parameter off, I
just need to judge whether $_GET["id"] is set or not. The problem is that I wrote the
SQL
statement as:
$SQLstr="SELECT * FROM employees WHERE id=$_GET["id"]";
That's a wrong
Hi, Shahmat:
Thank you very much!
Yes, I use "echo phpinfo()" and found out that
"register_globals" has been turned off. How can
I turned it on? Do I need to recompile the Php?
I also tried "if (isset($_GET["id"])) and
found out that $id is NULL. So, there will
be an error in the SQL statement.
Hi, Jason:
Thank you very much for your quick reply. I tried
the if (isset($id)) but it doesn't work. Any
thoughts?
By the way, I'm using Php4.3.3 and MySQL4.0.15a.
The tutorial I'm reading is about Php3. Will this
be the problem?
Many thanks.
Zhan Xu
EECS Department
Case Western Reserve Univer
Hi, All:
I'm a beginner of PHP. While trying the code from
a tutorial, I encountered the following problem.
The variable $id can not be transfered to my server.
You can find the code at the end of this email. When
I visit http://mydomain.com/test.php?id=1
it always shows the list of the database
