At 02:23 PM 4/8/2002 +0200, you wrote:
>Hy,
>
>I try to do a SELECT but when I test my code I Have a parse error.
>If someone can explain to me why.
>
>Code:
>
>$db = mysql_connect(newsmanga_db);
>$req = "SELECT * FROM dvds WHERE nomdvd='".$nom."".$i"' "; /// PARSE
>ERROR IN THIS LINE
Your miss
You need to put {} in the while function like this
while {
do this
}
$query = "SELECT name FROM users";
$ret = mysql_query($query);
while(list($name) = mysql_fetch_row($ret)) or die("Error: ".mysql_error()) {
print ("your name is $name);
$query_update["name"] = ("UPDATE my_other_table set
I am a dork!
And you answered that almost before I sent it, you're like Mighty Mouse
on Ginseng!
[EMAIL PROTECTED] wrote:
>This doesn't appear to be a PHP question. Are you sure you are on the
>right list?
>
>On Sun, 13 Jan 2002, sundogcurt wrote:
>
>>Hi! I'm storing a date in this format :
>>
This doesn't appear to be a PHP question. Are you sure you are on the
right list?
On Sun, 13 Jan 2002, sundogcurt wrote:
> Hi! I'm storing a date in this format :
>
> 2002-01-13 11:08:40
> -mm-dd hh:mm:ss
>
> This is great because it gives me the date and the time, which I need.
> I want to
Did you forget the tags?
Jerry
-Original Message-
From: Martin Schichl [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, January 02, 2002 9:57 AM
To: Edwin Boersma; [EMAIL PROTECTED]
Subject: Re: [PHP] Select box won't display in Netscape 4.xx
Dear Edwin!
Yes, I have had this pr
Dear Edwin!
Yes, I have had this problem before ...
My problem was, that I a line in my style sheets, which
could not be solved by NS.
(I think it was "border?)
Yours,
Martin
>I'm developing a website for multiple browsers. In Netscape 4.xx (both
>Win98 and Linux versions), the php-scripts
Have you double and triple checked to be sure your form elements are
nested properly?
[EMAIL PROTECTED] wrote:
>Hi,
>
>I'm developing a website for multiple browsers. In Netscape 4.xx (both
>Win98 and Linux versions), the php-scripts display the select boxes in
>my forms only as plain text. I c
check and make sure the you have the correct opening and closing form tags.
NS 4.x will "choke" if you don't have an opening and closing tag. While IE
5, 5.5 and 6 will work fine.
Jim
- Original Message -
From: "Edwin Boersma" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, Dec
Miles ---
Here's the link to W3C: HTML/XHTML and CSS validation page.
It's a excellent tool for debuging pages.
http://validator.w3.org/
David
>
> Sounds like a problem with HTML syntax. NS is notoriously fussy and IE
> remarkably lax. Capture your source and compare against the HTML
> standar
Sounds like a problem with HTML syntax. NS is notoriously fussy and IE
remarkably lax. Capture your source and compare against the HTML standard.
I'm surprised NS6 is accepting what NS4 is choking on.
Happy New Year - Miles
PS I'd bet a cup of coffee that a closing tag is missing.
At 12:06 P
Hi Martin,
Excellent, thanks for your help :)
Regards
--
Shane
On Monday 17 Dec 2001 10:15 pm, Martin Towell wrote:
> you _can_ reference it by name - try this:
>
>
> function my_select(type)
> {
> frm = document.forms.test;
> ele = frm["mycheckbox[]"];
> len = ele.length;
>
you _can_ reference it by name - try this:
function my_select(type)
{
frm = document.forms.test;
ele = frm["mycheckbox[]"];
len = ele.length;
for (i = 0; i < len; i++)
ele[i].checked = type;
}
Martin
-Original Message-
From: Shane Wright [mailto:[E
Oh, one more thing.. The latest MSIE suddenly doesn't seem to support the function to
be named 'selection' . change that and it'll work ;)
bvr.
>
>
>I know *exactly* what you mean, the solution (took me at least an hour) is attached.
>
>have fun,
>bvr.
>
--
PHP General Mailing List (http:
Hi Charles,
As a button, ideally two in fact - 'select all' and 'unselect all'.
I've done it easily with ASP sites - where the '[]' is not required to have
the values passed back as an array.
(I know I could find a non-array way of doing it but its a pain)
--
Shane
On Monday 17 Dec 2001
I know *exactly* what you mean, the solution (took me at least an hour) is attached.
have fun,
bvr.
On Mon, 17 Dec 2001 17:54:42 +, Shane Wright wrote:
>Hi
>
>I'm having somewhat of a problem :(
>
>I have my banks of checkboxes, all named 'mycheckbox[]' with the values as
>different IDs.
Shane,
is this meant for initial settings of the form only or as a button, or some
such, to allow someone to select all the boxs?
chuck
- Original Message -
From: "Shane Wright" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Monday, December 17, 2001 6:54 PM
Subject: [PHP] 'Select Al
Hi Shane.. :-)
thankyou for your help..
I was wondering myself about that.. and I surfed the net for many days just searching
for a solution...
but now I know what it is the limitation.. thanks again. =)
I think as you say.. consider a client-side applet.
best regards from Chile
--Patricio
Hi fitiux
I'm sure the list will correct me if I'm wrong, but what you're talking about
isnt possible, at least not in this way.
This isn't a limitation of PHP - its a limitation of the HTML form element
that handles the uploads.
You could consider a client-side applet to achieve the result a
2001 1:59 PM
> To: Jason Dulberg; [EMAIL PROTECTED]
> Subject: Re: [PHP] select based on time/date
>
>
> Timestamp only sets itself to Now() if it isn't explicitly set (or set to
> NULL). So try this:
>
> $out = "UPDATE logged_in SET session='',
Website Administrator
FoxJet, an ITW Company
www.foxjet.com
- Original Message -
From: "Jason Dulberg" <[EMAIL PROTECTED]>
To: "Sheridan Saint-Michel" <[EMAIL PROTECTED]>;
<[EMAIL PROTECTED]>
Sent: Wednesday, September 26, 2001 12:44 PM
Subject: RE: [PHP] sel
u again for your time. I've learned alot from this.
__
Jason Dulberg
Extreme MTB
http://extreme.nas.net
> -Original Message-
> From: Sheridan Saint-Michel [mailto:[EMAIL PROTECTED]]
> Sent: September 26, 2001 9:38 AM
> To: Jason Dulberg; [EMAIL PROTECTED]
&
f the date types.
Sheridan Saint-Michel
Website Administrator
FoxJet, an ITW Company
www.foxjet.com
- Original Message -
From: "Jason Dulberg" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Tuesday, September 25, 2001 6:24 PM
Subject: RE
TECTED]>;
<[EMAIL PROTECTED]>
Sent: Tuesday, September 25, 2001 5:08 PM
Subject: RE: [PHP] select based on time/date
> I tried basically what you have but I got an error:
>
> $sql="select * from logged_in where time_in + interval 1 hour <= now()";
> $result
To: 'Jason Dulberg'; 'Sheridan Saint-Michel'; [EMAIL PROTECTED]
> Subject: RE: [PHP] select based on time/date
>
>
> select * from logged_in where date_add(time_in,interval 1 hour) <= now()
>
> -jack
>
> -Original Message-
> From: Jason Du
select * from logged_in where date_add(time_in,interval 1 hour) <= now()
-jack
-Original Message-
From: Jason Dulberg [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, September 25, 2001 6:08 PM
To: Sheridan Saint-Michel; [EMAIL PROTECTED]
Subject: RE: [PHP] select based on time/date
I tr
>
> select * from sessions where $timein + interval 1 hour <= now();
- Original Message -
From: "Jason Dulberg" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Tuesday, September 25, 2001 2:35 PM
Subject: RE: [PHP] select based on time/date
- Original Message -
From: "Jason Dulberg" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Tuesday, September 25, 2001 2:35 PM
Subject: RE: [PHP] select based on time/date
> I had a look at the DATE_FORMAT info on the mysql doc
I had a look at the DATE_FORMAT info on the mysql doc page and its a bit
easier to understand than the CONCATS.
Would I want to use the CURTIME() function since I would want to select
sessions based on the hour? How would I go about combining CURTIME with the
rest of my query?
Theoretically, her
On Tue, 25 Sep 2001 13:38, Maxim Maletsky \(PHPBeginner.com\) wrote:
> Don't do it with PHP. Instead use the native functions of mySQL itself.
>
> Use the combination of CONCAT(), some string and date functions of
> mySQL.
> Something like CONCAT(SUBSTRING(), SUBSTRING(), ...) so you end up in
> t
Don't do it with PHP. Instead use the native functions of mySQL itself.
Use the combination of CONCAT(), some string and date functions of
mySQL.
Something like CONCAT(SUBSTRING(), SUBSTRING(), ...) so you end up in
the right time format i.e.: -00-00. With that string you should be
able to do
You're doing something wrong! Indexes are only for performance and
possibly they may be used for helping in ensuring uniquenes. sorry cant be
any more help without code.
"Marcos Mathias" <[EMAIL PROTECTED]> on 13/09/2001 20:40:40
Please respond to <[EMAIL PROTECTED]>
To: <[EMAIL PROTE
Couple of other tips for you.
1. SELECT * is bad unless you really need every field. If you don't, specify
which ones you want - the query will perform better.
$s = "SELECT * FROM News"; // bad
$s = "SELECT NewsHeadline, NewsTeaser FROM News"; // better
2. You appear to be connecting to your
In your query add ORDER BY "field name like date or ID" DESC.
That way it will put them in descending order and I do believe that is what
you're looking for :)
Jeff
- Original Message -
From: "Tarrant Costelloe" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, August 29, 2001
Hi,
if your script makes errors on mysql_query() try to printthe errormessages sent back
by MySql:
mysql_query(...) or die(mysql_error());
see: http://www.php.net/manual/en/function.mysql-error.php
- Original Message -
From: "Roman" <[EMAIL PROTECTED]>
To: "Php-General" <[EMAIL PROTE
Jeremy,
The reason you are only getting the first value is because you are only
retrieving the first value and assigning it to $data. You need to step
through the result and do something with it. For example,
if ($row = mysql_fetch_array($result)) {
do {
oops slight mistake i noticed there..
should be
thats because mysql_fetch_row only returns one row.
to get all the rows do
while ($row = mysql_fetch_array($result)) {
$firstcloumn = $row[0];
}
This will run through all the rows putting the colunms into the array $row[]
Regards
Jo
Jeremy Morano wrote:
>
> ...
>
> $data=mysql_fetch_row($result);
> $count=$data[0];
>
> echo $count;
>
> Yet the only result I get is the first value of the first
> column.
>
Of course, and $data[1] is the first value of second column, etc.
If you want to get all result, use w
thats because mysql_fetch_row only returns one row.
to get all the rows do
while ($row = mysql_fetch_array($result) {
//processing code here
}
This will run through all the rows putting the colums into the array $row[]
Regards
Jon
--
Jon Farmer
Systems Programmer, Entanet www.enta
> I have an array of id numbers ($catids).
> This syntax fails:
> select * from categories
> where cid in $catids
Never mind. I was being stupid. For the record, one needs to implode(",",
$catids) and then use that in the WHERE statement, i.e. "where cid in
($catids)"
c
--
PHP General Mailin
rewarding than reaching the goal you've set for
yourself"
- Original Message -
From: George E. Papadakis <[EMAIL PROTECTED]>
To: Jacky <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Thursday, May 24, 2001 2:56 AM
Subject: Re: [PHP] select the most repeated value in a f
select result,count(*) as cnt from table group by result order by cnt desc ;
- Original Message -
From: "Jacky" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, May 24, 2001 10:02 PM
Subject: [PHP] select the most repeated value in a field
Hi people
How do i query to select
use GROUP BY
SELECT
result,
COUNT(result) AS res_count
FROM
result
GROUP BY
result
ORDER BY
res_count
DESC
;
Sincerely,
Maxim Maletsky
Founder, Chief Developer
PHPBeginner.com (Where PHP Begins)
[EMAIL PROTECTED]
www.phpbeginner.com
---
On Wed, May 23, 2001 at 12:27:04AM +0200, Gyozo Papp wrote :
> Thanks for your fast reply.
> One more question related to this topic.
> Would it be valueable to extend the existing array_values()
> function to support this feature? just to balance with
> array_keys() and its optional value argum
t; <[EMAIL PROTECTED]>
To: "Gyozo Papp" <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: 2001. május 23. 00:14
Subject: Re: [PHP] select specified elements from an array
> On Wed, May 23, 2001 at 12:06:42AM +0200, Gyozo Papp wrote :
> > function array_part($fromar
On Wed, May 23, 2001 at 12:06:42AM +0200, Gyozo Papp wrote :
> function array_part($fromarray, $keys)
> {
> foreach($fromarray as $key => $val)
> {
> if (in_array($key, $keys)
>$anotherarray[$key] = $val;
> }
> return $anotherarray;
> }
foreach( $keys as $key)
select max(scorevalue) from score;
- Original Message -
From: "Jacky" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: 2001. május 21. 22:38
Subject: [PHP] select the max value
Hi all
Is this the corerct way of selecting the max value in a table?
"select * from score where MAX(scoreva
Hi
try:
select * from score order by scorevalue DESC LIMIT 1
or if your db supports sub selects:
select * from score where scorevalue=(select MAX(scorevalue) from score)
Tom
At 03:38 PM 21/05/01 -0500, Jacky wrote:
>Hi all
>Is this the corerct way of selecting the max value in a table?
>"selec
SELECT MAX(scorevalue) FROM score
Will display the maximum value of your scorevalue column.
Sincerely,
Craig Vincent
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrator
In MySQL you can do this:
$sql = "SHOW TABLES LIKE 'box%'";
$r = mysql_query($sql);
// then loop through the tables and populate
--Joe
On Tue, Apr 17, 2001 at 07:49:11PM -0400, Matt TrollBoy Wiseman wrote:
> How would I word a query to see what tables exist in a db that begin with
> box
>
>
On 17-Apr-01 Matt \"TrollBoy\" Wiseman wrote:
> How would I word a query to see what tables exist in a db that begin with
> box
>
> for example include
> boxTABLE1
> boxTABLE2
> boxTABLE3
> but exclude
> sphereTABLE1
>
> I'm basically trying lo populate a list box with the tables beginning with
"Matt "TrollBoy" Wiseman" <[EMAIL PROTECTED]> wrote:
> How would I word a query to see what tables exist in a db that begin with
> box
>
> for example include
> boxTABLE1
> boxTABLE2
> boxTABLE3
> but exclude
> sphereTABLE1
>
> I'm basically trying lo populate a list box with the tables beginning
On 05 Apr 2001 11:53:21 +0200, Christian Reiniger wrote:
switch ($foo) {
case "bar1":
$foobar="fooey";
break;
default:
$foobar = "NoBar";
}
> > I have a list of conditions I want to test against.
> > In ASP, I would do:
> > How do I do the equivelent in PHP?
> Try reading the m
On Wednesday 04 April 2001 20:07, you wrote:
> I have a list of conditions I want to test against.
> In ASP, I would do:
>
> Select Case foo
> case "bar1"
> foobar = "fooey"
> case "bar2"
> foobar = "fooey2"
> case "bar3"
> foobar = "fooey3"
> case else
>
See the "switch" statement.
http://www.php.net/manual/en/control-structures.switch.php
Kirk
> -Original Message-
> From: Allen May [mailto:[EMAIL PROTECTED]]
> Sent: Wednesday, April 04, 2001 12:07 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP] Select case equivelent
>
>
> I have a list
switch ($i) {
case 0:
print "i equals 0";
break;
case 1:
print "i equals 1";
break;
case 2:
print "i equals 2";
break;
}
- Original Message -
From: "Allen May" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, April
> select * from nmensagem m, nusuario u, nprefuser p where
> m.cdpreferencia=p.cdpreferencia and p.flag=1;
>
> With "m.cdpreferencia=p.cdpreferencia" I should get just the values there
> exists on m.preferencia and p.cdpreferencia?
No, because you have a record for every single nusario, since you
hanks ben it works :)
>
> Just one thing if i say enter just one letter .. how could i make it show
> more than just the one result?
>
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED]]On Behalf Of Ben Peter
> Sent: Thursday, February 15, 2
: PHP MAIL GROUP
Subject: Re: [PHP] SELECT BLAH WHERE BLAH LIKE BLAH not working
Peter,
after querying the database, you need to fetch the returned values from
the result before you can use them:
> $result = mysql_query("SELECT company FROM resellers WHERE Company LIKE
> '%$
Peter,
after querying the database, you need to fetch the returned values from
the result before you can use them:
> $result = mysql_query("SELECT company FROM resellers WHERE Company LIKE
> '%$query%'");
$row = mysql_fetch_row($result);
$company = $row[0];
OR (shorter):
list($company)=mysql_
In article <[EMAIL PROTECTED]>,
[EMAIL PROTECTED] ("Peter Houchin") wrote:
>
>
>
>
> echo " BGCOLOR=\"#CC\" ALIGN=\"CENTER\">\n\n bgcolor=\"#006699\"> sans-serif\" size=\"3\" color=\"#FF\">Company ";
> $result = mysql_query("SELECT company FROM resellers WHERE Company LIKE
> '%$quer
L PROTECTED]\">webmaster
>
> ");
> // if successful then do this
> if ($num == 1) {
>
>
>
> include "quote2.php"; //has a hidden field referencing the user & the email
> address .. user shows up but again email does not
> }
>
&g
ebmaster
");
// if successful then do this
if ($num == 1) {
include "quote2.php"; //has a hidden field referencing the user & the email
address .. user shows up but again email does not
}
?>
-Original Message-
From: Philip Olson [mailto:[EMAIL PROTEC
-Original Message-
From: Peter Houchin [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, February 13, 2001 2:51 PM
To: PHP MAIL GROUP
Subject: RE: [PHP] SELECT statement
ok I've changed my code to
$sql = "SELECT id, email FROM users WHERE user='$user' and pass='$pass&#x
Yeah, sure you can:
$sql="SELECT id, email FROM table WHERE user='$user' and pass='$pass'";
Sincerely,
Maxim Maletsky
Founder, Chief Developer
PHPBeginner.com (Where PHP Begins)
[EMAIL PROTECTED]
www.phpbeginner.com
-Original Message-
From: Peter Houchin [mailto:[EMAIL PROTEC
On Tue, 13 Feb 2001 16:20, Peter Houchin wrote:
> ok I've changed my code to
>
> $sql = "SELECT id, email FROM users WHERE user='$user' and
> pass='$pass'"; but still no joy can any one suggest why?
>
> (& Yes email is a field in the table)
>
> Peter
>
> > Hi,
> >
> > Can you have a SELECT stateme
A possible way to find out :
$result = mysql_query($sql) or die(mysql_error());
Does it say anything? mysql_error() is your friend, it can be printed
anywhere within the script and will print the last mysql error. So
perhaps :
print mysql_error();
Right before the query or ...
Regar
ok I've changed my code to
$sql = "SELECT id, email FROM users WHERE user='$user' and pass='$pass'";
but still no joy can any one suggest why?
(& Yes email is a field in the table)
Peter
> Hi,
>
> Can you have a SELECT statement (using mysql) that goes something like
>
> $sql="SELECT
On Tue, 13 Feb 2001 16:00, Peter Houchin wrote:
> > Hi,
>
> Can you have a SELECT statement (using mysql) that goes something like
>
> $sql="SELECT id && email FROM table WHERE user='$user' and
> pass='$pass'";
>
> and if you can't is there a away around this?
>
> Thanks
>
> Peter
In SQL queries
Assuming you want to select both id and email from table, use commas :
SELECT id,email FROM ...
Also, check out this basic SQL tutorial :
http://www.sqlcourse.com/
It's fairly useful.
regards,
Philip
On Tue, 13 Feb 2001, Peter Houchin wrote:
> Hi,
>
> Can you have a SELECT stateme
SELECT options is null or not an object
> Does anyone know why I get this error when selecting an option in the box?
I think that's a JavaScript error message, no?... In a popup window rather
than in the middle of your page...
>
ittle time: http://chatmusic.com/volunteer.htm
- Original Message -
From: Gerry <[EMAIL PROTECTED]>
To: Richard Lynch <[EMAIL PROTECTED]>
Cc: <[EMAIL PROTECTED]>
Sent: Sunday, February 04, 2001 10:23 AM
Subject: Re: [PHP] Select list with PHP
> This worked!
> But, I
Well I messed up... I forgot to add the final ";} at the bottom of my
sample code but here it is. sorry!
Gerry wrote:
>
> This worked!
> But, I get this color selection increment on each subsequent menu where
> the colors from the row above mix with the ones below and so on. Assume
> that the nu
This worked!
But, I get this color selection increment on each subsequent menu where
the colors from the row above mix with the ones below and so on. Assume
that the numbers are the actual colors:
Select a color:
first menu1 next menu-> 1 next menu-> even longer
> while ($row = mysql_fetch_array($sql_result)) {
> echo"";
> echo $row["paint"];
> echo"";
> echo $row["bucket"];
> echo"";
> echo"http://www.\" method=\"POST\">";
> $Color = $row["Color"];
> if ($Color == $Color) {
Here is one problem: You can't use $Color for the current color you are
printi
At 3:24 PM -0700 2/3/01, Gerry wrote:
>Hello:
>
>I'm trying to create dinamic color SELECT lists with php. I have my
>database set up as follows:
>
>Paint = good
>bucket = good
>weight = 100kg
>Colors = green, blue, red
>
>and here is my php:
>
>while ($row = mysql_fetch_array($sql_result)) {
>ech
no prob,
--max
-Original Message-
From: Netbrain di M.L. [mailto:[EMAIL PROTECTED]]
Sent: Sunday, February 04, 2001 3:44 AM
To: PHPBeginner.com
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP] select prob.
Yes, it seems a good solution :)
thanks again
max
---
Please Help
EMAIL PROTECTED]
> www.phpbeginner.com
>
>
>
>
> -Original Message-
> From: Netbrain di M.L. [mailto:[EMAIL PROTECTED]]
> Sent: Sunday, February 04, 2001 3:34 AM
> To: PHPBeginner.com
> Cc: [EMAIL PROTECTED]
> Subject: RE: [PHP] select prob.
>
nner.com (Where PHP Begins)
[EMAIL PROTECTED]
www.phpbeginner.com
-Original Message-
From: Netbrain di M.L. [mailto:[EMAIL PROTECTED]]
Sent: Sunday, February 04, 2001 3:34 AM
To: PHPBeginner.com
Cc: [EMAIL PROTECTED]
Subject: RE: [PHP] select prob.
x
On Sun, 4 Feb 2001, PHPBeginner.com
x
On Sun, 4 Feb 2001, PHPBeginner.com wrote:
> then, why don't you instead have it NULL, like:
>
>
> Select your country
> ---
> United State
> United Kingdom
>
>
>
> then if it's empty it will be skipped ... I think ... I've never done so ...
>
> Ciao allo zio e la
On Saturday 03 February 2001 17:23, PHPBeginner.com wrote:
> then, why don't you instead have it NULL, like:
>
>
> Select your country
> ---
if no "value" is specified, the text in the field is taken as
value, i.e. $form['country'] == "Select your country" etc
--
Christian
then, why don't you instead have it NULL, like:
Select your country
---
United State
United Kingdom
then if it's empty it will be skipped ... I think ... I've never done so ...
Ciao allo zio e la lista italiana,
Maxim Maletsky
Founder, Chief Developer
PHPBeginner.co
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