The display will look like
1
2 2 2 2 2 2 2 2 2
3 3 3 3 3 3 3 3 3 3 3 3 3
4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4
Everybody on level 1 will be on same row
and so on
On 5/7/08, A. Joseph [EMAIL PROTECTED] wrote:
Please i knew this not the best place to post this
Rainer Bendig Aka Ny wrote:
I want to build an array containing two arrays $navm[] and $navs[],
$navs[] should go to $navm[items].
- -sourcecode:start8--
$resultm = $db-query(SELECT * FROM .$p._cats WHERE ms='m' \
ORDER BY sortorder ASC);
Miguel Guirao [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi!!
I have a table with many options from when the user can select them,
after that, his/her selections will be stored in a database.
In the mean time, I want to store his/her selections in an array like
this one:
if($access ==2)
{
echo do something;
}
Peppe wrote:
Hi
I have a variable cold $access and there are values 1,2,3,4,5
How can I check for example
If ($access =='2'){
echo go further;
}
How can I make this work
Thanx in advance
--
PHP Database Mailing List (http://www.php.net/)
To
Hi
Pete I forgot to write that $acces is a field from Db and there are values
1,2,3,4
a need to split those values to make that IF
Brettking I think that is solution
Peppe [EMAIL PROTECTED] wrote in message
news:[EMAIL PROTECTED]
Hi
I have a variable cold $access and there are values 1,2,3,4,5
Hi there. here ya go.
$array_of_exams = Array();
$array_of_exams['Full Head'] = 'Head Examination History';
$array_of_exams['Full Body'] = 'Body Examination History';
$array_of_exams['Full Butt'] = 'Butt Examination History';
$exam_dates = Array();
foreach ($array_of_exams as $exam =
Thanks that works like a charm and I learned some tricks that I can apply to
some other code!!! Ozzie
Hi there. here ya go.
$array_of_exams = Array();
$array_of_exams['Full Head'] = 'Head Examination History';
$array_of_exams['Full Body'] = 'Body Examination History';
$array_of_exams['Full
Much simpler:
foreach ($arrText as $text)
{
print $text;
}
Use that code. Just add in your HTML tags and it will be much cleaner and
faster.
Michael Conway [EMAIL PROTECTED] wrote in message
news:!~!UENERkVCMDkAAQACABgAeI4MkiQKXEijmrrs5/EPIcKA
[EMAIL PROTECTED]
I am
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: Array help needed :-(
In article 000801c26e10$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
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
(i am new to these groups, but shouldn't there be a follow-up-to on a
X-post?)
César aracena wrote:
I have this form in which one Administrator can insert new members and
after that, in another page, a form where he/she can insert the new
members sons daughters. I want to display a table
-Original Message-
From: Andy [mailto:[EMAIL PROTECTED]]
Sent: 04 January 2002 20:24
To: [EMAIL PROTECTED]
Subject: [PHP-DB] Re: Array not supported for strings???
$stmt=
SELECT country
from $DB2.$geo_T1
where country_code = '$country_id[$i
On Saturday 05 January 2002 04:24, Andy wrote:
Here is the full code:
###
# Get the name of the country:
if (isset($country_id)){ //only if there are results
for($i=0; $i count($country_id); $i++){
$stmt=
SELECT country
Hi,
I have a drop-down menu that has been prepopulated with an array from a
column name CompanyName. What I need to do is take the CompanyName lookup
the CompanyID in the company table and insert the CompanyID into a row in
a
table called contacts with a CompnayID column.
[...]
Is this a
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