On Wed, 13 Jun 2001 10:07:07 -0500
Gustavo Mejia <[EMAIL PROTECTED]> wrote:
> And now It works, but now what I wanto to do i to make variable the criteria of
> my query, yes, something like
>
> select * from employee where empno=MY_VARIABLE
>
> Do you know how can I do that ?
See my post on
ED]>
Sent: Thursday, June 14, 2001 3:52 AM
Subject: Re: Problem with my query
>
> Gustavo,
>
> if you want to send me your code (you can send it directly to me) I'll be
> glad to take a look. At a guess, though, I would think that your query
> just isn't returning
ng at the code is a
good way to begin to understand what is really going on when you write XSP,
and will help you make better guesses as to the proper syntax to use.
Good luck.
-Christopher
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Subject: Re: Problem with my query
C
Christopher sorry for all the troubles,
But, I didn't get it, I tried it, there is not any error message, but
there is not any result.
any other idea ??
Thanks so much !
Gustavo
Christopher Painter-Wakefield wrote:
Gustavo,
you want to use the request taglib, for which there isn't any documenta
Gustavo,
you want to use the request taglib, for which there isn't any documentation
(hopefully soon there will be). Anyway, try this:
select * from employee where empno=
You'll need to add a declaration for the request namespace in your xsp:page
element, like this:
xmlns:requ
Thanks again,
Right, I need to do the first thing that you told me, I tried but with no
results, I have something like this:
select * from employee
where empno=VAR
and in my browser the url is:
http://localhost/cocoon/servlet/sql/esql.xml?VAR="01";
but, doesn't show i
Gustavo,
you're welcome. There are probably many ways to do what you are asking;
you should probably get a book on servlet programming or web application
development. Anything that you can do with CGI or servlets you can do (I
believe) with XSP and Cocoon. So, for instance, you can pass param
Thanks Christopher,
But now, how can I assign the variable from other XML, how can I pass to
the XML with my query ?
I really apreciate your hellp !
Gustavo
Christopher Painter-Wakefield wrote:
> Gustavo,
>
> try this:
>
> select * from employee where empno=
> MY_VARIABLE
>
>
> You may ne
Gustavo,
try this:
select * from employee where empno=
MY_VARIABLE
You may need to specify a type, such as ,
otherwise it will assume a String.
-Christopher
Thanks,
I chanche my code to:
http://www.apache.org/1999/XSP/Core";
xmlns:esql="http://apache.org/cocoon/SQL/v2";
>
ora
Thanks,
I chanche my code to:
http://www.apache.org/1999/XSP/Core";
xmlns:esql="http://apache.org/cocoon/SQL/v2";
>
oracle.jdbc.driver.OracleDriver
jdbc:oracle:thin:@MyIP:MyPort:cta
UserID
Password
select * from employee
header info
Gustavo,
have you already added something like this in cocoon.xconf (under the
datasources tag) ?
jdbc:oracle:thin:@localhost:1521:ORA81
login
password
Remember to re-start Tomcat afterwards any change to cocoon.xcon
Your esql syntax doesn't look right. Look at the esql.xml sample in the
samples/sql directory.
-Christopher
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Subject: Problem with my query
Hi,
I am using Cocoon 1.8.2, and I am trying to run this query, but I don't
see a
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