I've noticed that the odbc_exec call appears to call SELECT statements as
updateable in db2. Is there then a way to update each row of the response
(an updateable cursor)?
Any thoughts would be appreciated.
thanks
Rob
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On Tuesday 17 June 2003 14:16, Martin E. Koss wrote:
> Error found when testing the code below:
> Warning: Variable passed to each() is not an array or object in
> apache/htdocs/speeddials.php on line 11
> Line 11 is: "while ($record_set_array = each($result_set)) {"
> > $result_set = mysql_query(
Ok, I put in the environment variables
LIBPATH=/home/db2/sqllib/lib
LD_LIBRARY_PATH=$LD_LIBRARY_PATH:/home/db2/sqllib/lib
VWSPATH=/home/db2/sqllib
VWS_LOGGING=/home/db2/sqllib/logging
VWS_TEMPLATES=/home/db2/sqllib/templates
DB2INSTANCE=db2
(basically, I ran the ~db/sqllib/db2profile command
do a var_dump($result_set) to see what $result_set contains (if it's
actually and array.
(did you implement the code given as is? if not AND you have put line-11
in a function and $result_set as a global variable then you need to
define $result_set as a global inside the function.)
ramil
On Tu
Error found when testing the code below:
Warning: Variable passed to each() is not an array or object in
apache/htdocs/speeddials.php on line 11
Line 11 is: "while ($record_set_array = each($result_set)) {"
What did I do wrong?
Thanks
Martin
-Original Message-
From: [EMAIL PROTECTED] [ma
Hi, I don't read this often so please reply to my email:
[EMAIL PROTECTED]
I'm using PHP 4.3.1 with DB2 8.1 on RedHat 8.0. I compiled against DB2
fine, with no problems. I connect using db2 with no problems - the 'test'
database is perfectly catalogued.
However when I try to run the below page
Check that mysql is still the owner of its subdirectories.
You might also look on the mysql list for other possibilities as this is a frequent
topic.
Doug
On 17 Jun 2003 10:02:29 +0800, Ramil G. Sagum wrote:
>
>On Tue, 2003-06-17 at 03:08, Nathan wrote:
>
>> # ./bin/mysqld_safe &
>> Starting
pg_last_oid returns the last object id (an id used in the system
catalogs) and not the ID/index of the row.
if you need to get the last ID, you need to execute another select or
write a pl/sql function.
On Tue, 2003-06-17 at 10:13, Norma Ramirez - TECNOSOFT wrote:
> Hi all,
> I´m starting with
Hi all,
I´m starting with posgresql and i trying to use pg_last_oid to get the last id in a
insert query, but this function returns me a big number
like 21318 and bigger, but the table just have 3 or 4 records. Could some one help?
Any ideas?
This is the script i´m using
$insert = "insert into t
On Tue, 2003-06-17 at 03:08, Nathan wrote:
> # ./bin/mysqld_safe &
> Starting mysqld daemon with databases from /var/lib/mysql
> 030616 12:04:24 mysqld ended
>
> I'm running the command as root so I don't think it's a permission issue. Does
> anyone know anything about the PHP install that wo
It's always something obvious. I was not checking the session variable's value
at the correct point in the script, hence the display was not being altered by
its results. I've managed to get my code to display properly now, but I can't
retain state between queries.
My reason for using session va
> the first list that is displayed has an ID for each product. But when I
> click on it, it goes to the next page but does not display the rest of the
> products that are listed in the second table.
>
> $query = "SELECT * FROM product_list WHERE productid = id";
>
> Id being the description ID t
Hello,
I just installed PHP 4.3.2 on my machine running redhat 7.2. The install seems like
it went well. However, I've been running MySQL (Ver 11.18) just fine until I
installed PHP and now it won't start at all. When I try to start the mysql daemon
from the command line, I just get...
# ./
Sorry forgot to state the question.
No bids. Looking for help.
the first list that is displayed has an ID for each product. But when I
click on it, it goes to the next page but does not display the rest of the
products that are listed in the second table.
$query = "SELECT * FROM produ
Okay... and? Do you have a question? Are you wanting us to write this for
free? Do you want bids on the project?
---John Holmes...
- Original Message -
From: "Ryan Holowaychuk" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, June 16, 2003 2:30 PM
Subject: [PHP-DB] query selecti
I have a databae built, at the moment it has two tables.
The first table has 2 fields, an ID and description. This is for a listing
of all the different samples that we have at the company.
When the user clicks on the description they want, it will open another page
that will show all the item
hi!
i know that it sends everything to itself if you omitt the
action-attribute, but it isn't propper html strict anymore... (ok -
thats not a big problem *g*)
maybe it helps if you put the testing in a place of the script where
there was no output yet (also no whitespaces)? at least this is a
> maybe try the invers (anyways you havn't defined an action-attribute -
> maybe this is a problem?):
Not declaring an action attribute automatically makes the form submit to itself.
When a URL query string is involved, this works much better than
$_SERVER['PHP_SELF'], which reduces the URL to its
> Are you certain that you have issued the session_start() call at the
> beginning of the page?
100%. I went so far as to move session_start() before every other element on
the page, even removing it from my include file and directly inserting it into
the actual page.
> Furthermore, your registe
Funny you should ask. I actually didn't try it on anything else. So I went back and
tried a simple script and got the same message. Thanks for the idea. I haven't gotten
any other feedback on this.
<>< Ryan
-Original Message-
From: Jacob Marble [mailto:[EMAIL PROTECTED]
Sent: Friday, J
$qry = "SELECT * FROM TABLE_NAME";
$result_set = mysql_query($qry) or die(mysql_error());
***this has to be called before any headers are written to the browser.
***
header("Content-type: application/csv");
header("Content-Disposition: attachement; filename="YOURFILENAME.csv");
while ($record_se
I've heard or read about the headers somewhere but right now I sure could
use pointing in the direction of a step-by-step tutorial for extracting my
data into a CSV file and, of course, setting up the PHP file to handle it
correctly.
Thanks
Martin
-Original Message-
From: Dillon, John [m
> -Original Message-
> From: Hutchins, Richard [mailto:[EMAIL PROTECTED]
> Sent: 16 June 2003 17:08
>
> session_start();
> session_register("isRegistering");
>
> $_SESSION["isRegistering"] = "true";
B't! If you're using the $_SESSION array, then you MUST NOT use
s
Edward,
Are you certain that you have issued the session_start() call at the
beginning of the page? According to the session stuff on PHP.NET, you need
to make sure you have session_start() at the top of the page before you do
anything else with the session. For example, I have on one of my simple
hi
maybe try the invers (anyways you havn't defined an action-attribute -
maybe this is a problem?):
if(!isset($_SESSION['item_details']) || (
isset($_SESSION['item_details'] && $_SESSION['item_details']) {
echo ''.
''.
''.
'';
else
$_SESSION['item_details'] = 'yes';
hope i underst
Despite my best efforts, I can't seem to set session variables via a basic form.
I'm trying to use a button to alter display settings, which should be a simple
task, but apparently it is not. Here's the code I'm working with.
";
print "";
print "";
}
else
$_SESSION['item_detai
> I am connecting to MySQL (4.1) using ODBC (I have my reasons) and am
looking
> for an equivalent to the mysql_insert_id() function so that I can get the
ID
> of the record just created.
You could always just issue a SELECT LAST_INSERT_ID() query and get the
resulting number that way.
---John Ho
Hi,
I have a table with 1,008 columns in it and at it's peak it has up to 500
rows of data.
I want to enable my client to click on a link and the next thing they are
prompted with is 'Save or Open' (the CSV file).
The data will be compiled into a comma delimited CSV file but rather than be
printed
Sorry for the delayed response.
Not sure I quite understand your problem here. When you say:
>But when I turned register_globals On in PHP.ini and don't use
> $_GET['id'],
> echo $querystring in show.php file display empty:
> SELECT * FROM news_tb WHERE id=''
If you have indeed turned regis
Dear Mike and Nabil
Thank you for your help.
Regards
David
Anagram Systems
"David" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Dear All
>
> I am writing a support pages for my company at present. I would like to
> display 10 records per page, e.g. like google does it, and if t
Hi,
I am connecting to MySQL (4.1) using ODBC (I have my reasons) and am looking
for an equivalent to the mysql_insert_id() function so that I can get the ID
of the record just created.
Cheers
George
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check this out :))
http://www.phpfreaks.com/print.php?cmd=tutorial&tut_id=43
Nabil
"David" <[EMAIL PROTECTED]> wrote in message
news:[EMAIL PROTECTED]
> Dear All
>
> I am writing a support pages for my company at present. I would like to
> display 10 records per page, e.g. like google does it,
check the mktime() function
:-)
-Original Message-
From: Delz [mailto:[EMAIL PROTECTED]
Sent: maandag 16 juni 2003 11:21
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Calculating Time
Hi All,
How do I calculate time using php? Let's say I want to know the difference
between 12:30PM & 13:30P
Dear All
I am writing a support pages for my company at present. I would like to
display 10 records per page, e.g. like google does it, and if there are more
than 10 then a next | previous button appears at the bottom. Has anybody
have experience of this and be able to give me some ideas, e.g. usi
take a look at http://www.php.net/function.mktime
.ma
e: [EMAIL PROTECTED]
w: http://www.abendstille.at
# life would be easier if i knew the source code
Am Montag, 16.06.03 um 11:21 Uhr schrieb Delz:
Hi All,
How do I calculate time using php? Let's say I want to know the
difference
between 12
Hi All,
How do I calculate time using php? Let's say I want to know the difference
between 12:30PM & 13:30PM?
Hope someone can give me an idea?
Delz
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Louai wrote:
I got the following error when I tried to connect to MS SQL server:
Unknown () mssql: unable to initialize module
Please advise
You could try checking that your extension_dir path in php.ini is
set up correctly.
Peter.
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