hiii all,
i've posted this question some days back. but didnt find answer. so posting
it again.
in my phpscript, some users are allowed to del/modify contents of the
database. for this they have been provided login name and password, after
user loggs into the secure area, session starts.
Hi Alex,
I believe your database-design is not good ... think about a re-design or
send us some info
about your tables. Maybe we find a better solution...
Regards,
Marcel
-Original Message-
From: Alex Shi [mailto:[EMAIL PROTECTED]]
Sent: 06 October 2002 22:43
To: [EMAIL PROTECTED]
Hi John W. Holmes,
You can find out why with mysql_error(), but it's probably because you
haven't selected a database with mysql_select_db().
now I have this code, but with the same result
?
mysql_connect(localhost,root);
mysql_select_db(finance);
mysql_db_query(finace,INSERT INTO
Hy,
I'm having problems with Zend development tool..
anyone can help me in order to run functions like oci* and ora*
It not recognize these functions
where sholud I have to make changes in order to run zend debugger...otherwise the php
is running from the browser but in the debugger it does
On Monday 07 October 2002 16:16, Ruprecht Helms wrote:
Hi John W. Holmes,
You can find out why with mysql_error(), but it's probably because you
haven't selected a database with mysql_select_db().
now I have this code, but with the same result
?
mysql_connect(localhost,root);
Use mysql_error() in your die() message to find out what the problem
was...
---John Holmes...
-Original Message-
From: root@linux [mailto:root@linux] On Behalf Of Ruprecht Helms
Sent: Monday, October 07, 2002 4:17 AM
To: John W. Holmes
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP-DB]
Hi John W. Holmes,
Use mysql_error() in your die() message to find out what the problem
was...
ok, typemissmatch was the problem. Thank for the tip.
Regards,
Ruprecht
Ruprecht Helms IT-Service und Softwareentwicklung
Tel/Fax:
Hi there everyone,
I have a problem i'm trying to figure out but i'm not too good with arrays, just know
the basics, if anyone could help me out on this it would be wonderful :-)
I have two sets of data in columns of a MySQL DB, these items are size and price.
Size is seperate by comma's in
What does everyone typically use to display the results of a query. For
example, I have a database that has a series of subjects and grades. If I
select * from the table, I want a nice way for the data to be displayed. In
cold fusion, I can simply use a grid that dynamically fills in. Can I
It depends what you want to do with it.
To display it in HTML, you can write them in a 2-D array: myarray[$i][$j],
where $i is the row nb and $j the column nb, and write two nested loops.
To export them to Access, CSV, ... you can issue your query through Access
ODBC. You can then save as a
Will this work (assume you're using mysql):
function printresult ($rs)
{
if ($a = mysql_fetch_assoc($rs))
{
// Print header
echo 'tabletr';
foreach (array_keys ($a) as $v)
echo 'th' . $v . '/th';
What does everyone typically use to display the results of a query. For
example, I have a database that has a series of subjects and grades. If I
select * from the table, I want a nice way for the data to be displayed.
In
cold fusion, I can simply use a grid that dynamically fills in. Can
Hi All,
Although I am not so new to PHP (not expert either), this is the first time
for me to use PHP class.
I am starting a project, and plan to use partially OOP and partially
traditional programming. I use PHP/MySQL. I have one class, DB_Do, which
does every thing to do with database, and
Hi peoples,
Just wondering if it's possible with mysql to start an auto-increment field
at a specified number instead of 0001?
I could always just code it in php to add another number onto it to get it
to the right start point, but leaves areas for human error to creep in!
Any help would be
Well, I'll tell you what I do.
I'm not a OOP expert so if someone feel like I'm doing it wrong I'll welcome
any comment about .
I've got a main class which I use to generate every page in my website (some
are actually generated by inherited classes but that's the main idea). As I
need a
Gavin,
You can specify the next auto_increment id by using
SET INSERT_ID = #;
Where # is your desired number. Will only work for the next INSERT/ALTER.
Have a look down the bottom of 7.33 in the MySQL language reference.
Cheers,
Owen
Gavin Nouwens wrote:
Hi peoples,
Just wondering if
On October 8, 2002 01:49 am, Gavin Nouwens wrote:
Hi peoples,
Just wondering if it's possible with mysql to start an auto-increment field
at a specified number instead of 0001?
I could always just code it in php to add another number onto it to get it
to the right start point, but leaves
BTW... you only need to do it once. Subsequent INSERTS will follow on from
the previous insert.
Owen Prime wrote:
Gavin,
You can specify the next auto_increment id by using
SET INSERT_ID = #;
Where # is your desired number. Will only work for the next INSERT/ALTER.
Have a look down
CREATE TABLE table_name (ID INT NOT NULL AUTO_INCREMENT = 1 PRIMARY
KEY, ... );
---John Holmes...
-Original Message-
From: Gavin Nouwens [mailto:[EMAIL PROTECTED]]
Sent: Monday, October 07, 2002 9:50 PM
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Auto Increment
Hi peoples,
Thanks Wilmar, for sharing your experience with me. My problem now is that
I have already got a class to do every thing related to database. If I
don't use this DB class within other classes, I will have a lot of
duplicated code. If I do use the DB class within other classes, I am not
sure if
Hi,
If anyone can help me with this I'd be most appreciative.
I'm constructing 2 arrays of IDs from 2 different tables. Then I want to
compare one to the other and only use the results that both arrays share
in common to do a query loop that pulls the information for the related
IDs to
Having a blonde moment
the code to this post does work...there was no error handling in the
code for the array_intersect() incase either of the arrays contained no
data.
My mistake.
--
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It may be because your not initialising your array. ie:
$partner_type = array();
$partner_region = array();
PHP should automatically make it an array (if its not previously a different
type) as soon as you do $partner_type[] but if you get no results from your
query then it wont
In article 000801c26e10$a2128510$f7fea8c0@chris, [EMAIL PROTECTED]
says...
Hi there everyone,
I have a problem i'm trying to figure out but i'm not too good with arrays, just
know the basics, if anyone could help me out on this it would be wonderful :-)
I have two sets of data in columns
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