use zerofill for the column when you create the table!
--- Dallas K. [EMAIL PROTECTED] wrote:
I am having the problem that Mysql is truncating the
zeros from both the int and char fields when I try to
insert How can I fix this.
example:
0123 in PHP = 123 in mysql
3210 in PHP =
Hello,
You may want to check out connecticut-web.com
I use their $36/year plan for several domains. They are
always up and running.
They provide mysql+php support.
--- Jason Bell [EMAIL PROTECTED] wrote:
I recently moved all of my domains to www.prohosters.com
and have been
nothing but
I have seen people using
include (config.php);
in the begining of the page for that purpose.
--- Aaron Moore [EMAIL PROTECTED] wrote:
Sorry about the newbie question. I have a config.php file
which contains a
string which I would like a function to access. How do i
make it global so
that
HEllo,
Did anybody have a luck with a good Chat Program.?
Thanks,
Mehmet.
=
Mehmet Erisen
http://www.erisen.com
__
Do You Yahoo!?
Get email alerts NEW webcam video instant messaging with Yahoo! Messenger
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--
PHP General
Please try
select dest.dest_name,
air.name as airport1,
air2.name as airport2
from destination,
airport air,
airport air2
where dest.airport_dep_id_id=air.airport_id and
dest.airport_arr_id=air2.airport_id;
you can use the same table as manu times as you like in
sql.
$sql = SELECT COUNT(*) AS myCount FROM Products;
$result = mysql_query($sql);
$count = mysql_result($result,0);
--- Kevin P [EMAIL PROTECTED] wrote:
I have counted some rows in MySQL and I need to know how
to pull out the
number of rows.
SELECT COUNT(*) AS myCount FROM Products
--
Hello Jeremy,
Two things:
FROM users LEFT OUTER JOIN picks using(id)
Where users.uid picks.user_id
1- when you do OUTER join, you do NOT have to use the .
That's what outher join means.
2- what is id? Is it a column? shouldn't it be uid?
Maybe that's your problem.
--- Jeremy Morano
Hi it may sound funny, but you may want to add a space at
the end of lines before .
I mean
$query = SELECT mpn_job.ID, job, assigned_to, status_date,
category ;
$query .= FROM mpn_job, mpn_job_category WHERE CID =
mpn_job_category.ID ;
$query .= AND category =
Yes, there is a problem.
--
SELECT *
FROM links
WHERE 1=1
and ( name LIKE %te%
OR description LIKE %te%
OR url LIKE %te% )
AND approved=1 LIMIT 5;
--
--- Jeff Lewis [EMAIL PROTECTED] wrote:
Guys, why isn't this working? :)
SELECT * FROM links WHERE name LIKE %te% OR description
LIKE
Hello Jeremy,
you are using two tables, but you are not joining them.
What's you goal?
--- Jeremy Morano [EMAIL PROTECTED] wrote:
Hi,
Can somebody help me out?
My where clause is completely being ignored.
More specifically the . I tried to use != and that
didn't work either.
However,
HEllo,
I think this will better work. What Jeremy wants to do is
to OUTER joing the tables:
SELECT distinct users.uid , users.username
FROM users LEFT OUTER JOIN picks using(uid)
may do the trick.
If you say letf join and the use ids not equal in the
where, then I think it's same as saying:
Hello All,
I think for a thin page with PHP+mySQL you can go as low as
$3 a month (35MB)
I have seen decent packages, with Vserver privileges at
around $18/mnt.
--- Alnisa Allgood [EMAIL PROTECTED] wrote:
At 4:09 PM +1000 8/24/01, Chris Aitken wrote:
Hi All,
Just a quick non-technical
This is a sql question:
Limit is limiting number of records that will come out of
query. So, it would not work.
SELECT users.names
FROM users, selection
Where users.name = selection.name
and setting = 'on' ;
you could use
SELECT DISTINCT users.names
FROM users, selection
Where
Hello,
Chris, with all due respect, DESC is a reserved word for
SQL. is english ORDER BY ID DESC meand sort by id in
descending order.
Martin,
your SQL seems to be fine. Are you sure that you are
connected to your database.
I have run the following SQL against my database
SELECT *
FROM users
instead of method=GET use method=POST
--- Seb Frost [EMAIL PROTECTED] wrote:
how do I prevent getting submit=shop%21 in the url
window?
form method=GET action=page.php4
pEmail address: input type=text name=email
size=20
input type=submit value=Shop!
name=submit
Hello All,
How expensive are the image functions as far as the system
resources go. Would it be too expensive to use php to
create simple gifs like buttons, tabs etc.
Thanks,
Erisen
ulrAslan.
=
Mehmet Erisen
http://www.erisen.com
__
Do
Hello All,
Another question.
Here is the problem:
- One SQL query
- 1 result.
I have a form that has more than one drop down lists. I
like to use the mysql_fetch_array($result) to populate the
select options.
The problem is that after the first use, the
mysql_fetch_array($result) does not
you are right.
?php
if (6448234192168122==6448234192168125) {
echo 'A' ;
} else {
echo 'B' ;
}
?
I tested with 99644xxx and 964 It printed A.
However, if you are down to 644xxx is prints B which is
correct. Perhaps a limitation of 16 digits.
?
good catch
--- Ratko [EMAIL PROTECTED]
Hello All,
I receive the following warning message. When I reload the page, the message is the
error message does not show up again. Funny thing is, that there are other parts of
the page generated from the same database using the same mysql_pconnect and they are
fine.
I am using a standard
In adition to that, you may want to check out the online
documentation on the www.php.net website. Find the section
under mySQL.
If you know how to manipulate data using SQL, php apis are
very easy to use ..
www.mysql.com could be good resource for you as well.
if you have a specific
Hello,
I have implemented a user login with PHP Session functions.
THe problem I have is that if the user is logged in the
site using
http://mysite.com/login.php and starts a session, the
session is not recognized when user is trying to utilize
the site with
http://www.mysite.com/dosomething.php
Hello,
I am using the PHPMyAdmin for that. There is a website
www.phpwizard.net that you can download phpMyAdmin.
Hope this helps.
Mehmet.
--- Dave.O [EMAIL PROTECTED] wrote:
I have read the documentation already.
I am trying to create a directory, which information can
be added and
Here is how this is done:
$sql = insert into employees
values ('10','John','Smith','Engineer') ;
$result = mysql_query($sql) or die ('Insert failed');
with php, you can execute any SQL statement as long as you
have access...
to update a table
$sql = update employees
set job =
I use a javascript to do that for my forms:
if (window.RegExp) {
var reg = new RegExp(^[_0-9a-z]*$,gi);
if (!reg.test(f.elements['username'].value)) {
alert(Kullanici adi harf ve/veya rakamlardan
olusmalidir. Lutfen Duzeltiniz!);
f.elements['username'].focus();
return(false);
}
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