Re: [PHP-DB] Help in passing multiple HTML form options to SQL query -- Newbie

2001-10-28 Thread TorrentUK

Richard,

Many thanks for that.
I was reading about the for statement in a book earlier today and wondered
if I could implement it in some way.

Looks so simple now I've been shown how :)
Thanks again.

Richard Chauvaux [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 first, the select tag in your form should be setup as an array: select
 name=country[] ... 
 then the array $country[] will be available to your script on submittal.
 a simple way to build your sql where clause would be to iterate through
 the array using a subscript variable:
 $sql = select * from tbl where ctry='$country[0]';
 for ($n=1; count($country)  $n; $n++) {
 $sql.=  or ctry='$country[$n]';
 }

 richard

 TorrentUK wrote:

  I am designing  a ski web site and am presently trying to put together a
  resort database. I started the search script tonight and am quite
pleased
  that it works. Where I am struggling is that I would like my visitors to
be
  able to select multiple countries from the drop down list and for my
query
  to pull back info on all those selected. When I try it it only pulls
back
  the last one (naturally).
 
  Please advise on what would be the most efficient way of constructing
the
  query. Please talk in easy speak as I am real new at this and am not a
  developer by any stretch of the imagination.
 
  Many Thanks
  Torrent
 
  Here is a cut down version of my code, the html form is at the bottom.
 
  ?php
 
  if ($search) {
 
   $sql = SELECT rst_name, cty_name FROM resort_tbl WHERE
  cty_name='$country';
 
   $result = mysql_query($sql);
 
   if ($row = mysql_fetch_array($result)) {
echo centertable border=1\n;
  echo trtdName/tdtdCountry/td/tr\n;
 
do {
   printf(trtd%s/tdtd%s/td\n, $row[rst_name],
  $row[cty_name]);
   } while ($myrow = mysql_fetch_array($result));
echo /table/center\n;
   }
  }
 
  ?
 
  form name=resort method=post action=?php echo $PHP_SELF?
   p Select country /fontbr
  select name=Country size=5 multiple
option value=AndorraAndorra/option
option value=AustriaAustria/option
option value=BulgariaBulgaria/option
option value=FinlandFinland/option
option value=FranceFrance/option
option value=ItalyItaly/option
option value=North AmericaNorth America/option
option value=NorwayNorway/option
option value=RomaniaRomania/option
option value=SloveniaSlovenia/option
option value=SpainSpain/option
option value=SwedenSweden/option
option value=SwitzerlandSwitzerland/option
  /select
  br
 input type=submit name=search value=Search
/p
  /form
 
 
 




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[PHP-DB] Mathematics Examples MySQL and PHP

2001-10-28 Thread Mark Weber

Hello Everybody,

Does anybody have some examples about howto make mathematic functions in 
PHP with values from any
MySQL table such as:

Value 1 + Value 2 = Value 3
Value 1 - Value 2 = Value 3
Value 1 * Value 2 = Value 3
Value 1 / Value 2 = Value 3

With kind regards,


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[PHP-DB] Resource id #2

2001-10-28 Thread Andrew Duck

I am trying to select a message from a table in a database. the message will will be 
in column 'msg' and will be in same row as id='0'.
I need that msg put to the screen. With the code below I get the error: Resource id #2

Can someone please explain how I can fix this..

Thankyou in advance.

?
$msgcre = SELECT msg FROM creators WHERE id='0';
$msg = mysql_query($msgcre);
echo FONT color='#00' style='font-size: 10pt; font-weight: bold'$msg/font;
 ?





Re: [PHP-DB] Resource id #2

2001-10-28 Thread DL Neil

I am trying to select a message from a table in a database. the message will will be 
in column 'msg' and will be
in same row as id='0'.
I need that msg put to the screen. With the code below I get the error: Resource id #2

Can someone please explain how I can fix this..

Thankyou in advance.

?
$msgcre = SELECT msg FROM creators WHERE id='0';
$msg = mysql_query($msgcre);
echo FONT color='#00' style='font-size: 10pt; font-weight: bold'$msg/font;
 ?


Andrew,

Although the assignment statement looks as though the LHS (left-hand side) should 
contain the result of the
query, it does not (as you have discovered). Instead of being an integer or string, it 
is something called a
'resource'. You have already 'met' a resource in connection with the db connect 
process. After issuing a query,
you must then use one of the mysql_fetch_... functions to extract the data, one field, 
or one row at a time,
from the 'resource'. Please check out the manual at 
http://uk2.php.net/manual/en/ref.mysql.php. Also you might
consider working through one of the many excellent texts and/or online articles 
discussing the PHP/MySQL
combination.

Regards,
=dn



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[PHP-DB] PHP 4 Book

2001-10-28 Thread PHP Newbie

Hello!  =)

I'm just starting to learn PHP but I don't know what book shall I use.  Can you 
recommend to me a good PHP book that is FREE for download except the PHP Manual?  
Please include the complete URL of that book.  Thanks.

Any help will be much appreciated.  =)

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[PHP-DB] dbase help!!

2001-10-28 Thread Alejandro Viana

I'm trying to read a record from a dbase data base so I've previously
opened it successfully. The problem is that I call the function
dbase_get_record, but it returns no records. The database is ok because I
call the funcion dbase_numfields and dbase_numrecords and they give me
correct information. Would you help me, please?. Here is the source code
that fails.

[EMAIL PROTECTED]


Estoy intentando leer un registro de una base de datos dbase, por lo que
antes la he abierto con éxito. el problema residen en que llamo a la función
dbase_get_record, pero no me devuelve ningún registro. La base de datos está
bien, porque llamo a la función dbase_numfields y dbase_numrecords y me dan
información correcta. ¿Me podríais ayudar, por favor?. Ahí va el código
fuente:

[EMAIL PROTECTED]



Source code- Código fuente:

?
$based=articulo.dbf;
if (($descriptor=dbase_open ($based, 0))==0){
   printf (brError al abrir la base de datos);
}else{
   printf (brBase de datos abierta);
   $num_registros=dbase_numrecords($descriptor);
   $num_campos=dbase_numfields($descriptor);
   for ($i=1;$i=$num_registros;$i++){
  $registro= dbase_get_record ($descriptor, $i);
  for ($j=0;$j$num_campos;$j++){
 printf (brFila %d,Campo %d vale %S, $i, $j, $registro[$j]);
  }
   }
   dbase_close($descriptor);
   printf (brBase de datos cerrada);
}
?











[PHP-DB] Help with PHP CSS selector

2001-10-28 Thread Brian Tegtmeier

Currently I have PHP code that runs a server side check on your browser (placed above 
the html tags) and assigns a certain CSS stylesheet to it. The problem I have with 
the current code is it only goes by the browser and not both the browser and platform. 
I have the source code (listed below) that shows what the selector does. I was 
wondering if any of you could take a look at this and give some suggestions to have it 
run multiplatform (Windows and Mac) and have stylesheets for each like ie_win.css for 
IE on Windows and ie_mac.css for IE on Mac and so on. Thanks :)


?php
$browser = getenv(HTTP_USER_AGENT);
if (preg_match(/MSIE/i, $browser)) {
$css = link rel=stylesheet href=ie_win.css type=text/css;
} elseif (preg_match(/Mozilla/i, $browser)) {
$css = link rel=stylesheet href=ns4_win.css type=text/css;
} if (preg_match(/Mozilla\/5.0/i, $browser)) {
$css = link rel=stylesheet href=ns6_win.css type=text/css;
} elseif  (preg_match(/opera/i, $browser)) {
$css = link rel=stylesheet href=op_win.css type=text/css;
}
?



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[PHP-DB] create table with php -whats wrong?

2001-10-28 Thread Saulius Jankauskas

Hello all.

I am trying to create simple test table with php, but I get:

Parse error: parse error in /home/vanesaw/public_html/create.php on line 5

Line 5 is command to create table (see code below).

Whats wrong with it? Thanks.

?
$sql = mysql_connect(host, login, pass);
mysql_select_db(dbname);

CREATE TABLE test (id TINYINT not null AUTO_INCREMENT, process LONGTEXT not
null , title LONGTEXT not null , description LONGTEXT not null , PRIMARY KEY
(id));

?
done




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Re: [PHP-DB] create table with php -whats wrong?

2001-10-28 Thread DL Neil

 Hello all.
 
 I am trying to create simple test table with php, but I get:
 
 Parse error: parse error in /home/vanesaw/public_html/create.php on line 5
 
 Line 5 is command to create table (see code below).
 
 Whats wrong with it? Thanks.
 
 ?
 $sql = mysql_connect(host, login, pass);
 mysql_select_db(dbname);
 
 CREATE TABLE test (id TINYINT not null AUTO_INCREMENT, process LONGTEXT not
 null , title LONGTEXT not null , description LONGTEXT not null , PRIMARY KEY
 (id));
 
 ?
 done


Saulius,
http://www.php.net/manual/en/function.mysql-query.php
=dn



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Re: [PHP-DB] create table with php -whats wrong?

2001-10-28 Thread Richard S. Crawford

Place your query in a string variable, like $sSql, and use the mysql_query 
command to execute the query.  For example:

$sSql=CREATE TABLE test (id TINYINT not null AUTO_INCREMENT, process 
LONGTEXT not null , title LONGTEXT not null , description LONGTEXT not null 
, PRIMARY KEY (id));

if (!mysql_query($sSql)) die (Could not create table);




At 02:57 PM 10/28/2001, Saulius Jankauskas wrote:
Hello all.

I am trying to create simple test table with php, but I get:

Parse error: parse error in /home/vanesaw/public_html/create.php on line 5

Line 5 is command to create table (see code below).

Whats wrong with it? Thanks.

?
$sql = mysql_connect(host, login, pass);
mysql_select_db(dbname);

CREATE TABLE test (id TINYINT not null AUTO_INCREMENT, process LONGTEXT not
null , title LONGTEXT not null , description LONGTEXT not null , PRIMARY KEY
(id));

?
done




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Sliante,
Richard S. Crawford

http://www.mossroot.com
mailto:[EMAIL PROTECTED]
AIM: Buffalo2K   ICQ: 11646404  Y!: rscrawford
It is only with the heart that we see rightly; what is essential is 
invisible to the eye.  --Antoine de Saint Exupéry

Push the button, Max!


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[PHP-DB] counting categories

2001-10-28 Thread olinux

How would you get the count of each category?

Like this - http://www.scriptsearch.com 

Would you have to run a different select for each 
one? or would you do something like select category
from $table  and then count all of the matches for
each category perl php asp

thanks much,
Josh

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RE: [PHP-DB] Application variables

2001-10-28 Thread Beau Lebens

php doesn't have application variables as such, however there are a couple
options for implementing them...

1. have a single file with your application variables and manually include
that in all files of the project
2. use the session_*() functions to keep the vars alive on pages
3. as you said, store to db and update once in a while

(or a combination of these... i.e. 3 + 2, load from db, carry around in
session and then save to db again)

HTH

beau

// -Original Message-
// From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]]
// Sent: Sunday, 28 October 2001 3:15 AM
// To: [EMAIL PROTECTED]
// Subject: [PHP-DB] Application variables 
// 
// 
// Pls tell me how to implement application variables in PHP 
// like they are supported
// in ASP else i might have to have the value stored in the 
// database and update
// every now and then which is simply an overhead.
// 
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Re: [PHP-DB] counting categories

2001-10-28 Thread Frank M. Kromann

Hi,

Depending on the database you are selecting from you could do it in one select like 
this:

select category, count(*) from $table group by category order by category;

- Frank

 How would you get the count of each category?
 
 Like this - http://www.scriptsearch.com 
 
 Would you have to run a different select for each 
 one? or would you do something like select category
 from $table  and then count all of the matches for
 each category perl php asp
 
 thanks much,
 Josh
 
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[PHP-DB] Re: php command to create table

2001-10-28 Thread _lallous

mysql_query(
CREATE TABLE abcd (FieldName1 INT (3) UNSIGNED DEFAULT '0' AUTO_INCREMENT,
FieldName2 TINYINT (3) UNSIGNED DEFAULT '0', PRIMARY KEY(FieldName1),
UNIQUE(FieldName1))
);
Saulius Jankauskas [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 subj. Plz help.

 THanks.





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[PHP-DB] category structure

2001-10-28 Thread olinux

anyone have any ideas on this 

I have a table of directory categories 
id | parentid | category

so...
 1 | 0 | blah1   top
 2 | 1 | blah2   sub
 3 | 1 | blah3   sub2
 4 | 3 | blah4   sub2-sub

I want to hit the DB only once so i select all and
drop into array

Now i've created array with these like so:
$category[$parentid][$id]

But now what?

trouble is that 
$msub[2][4] is a sub category of $msub[1][2]
so i can't just run thru and print the array - need to
manipulate a little.

Any ideas? [or even a beeter way to do this?] 

thanks
olinux

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[PHP-DB] mySQL table update

2001-10-28 Thread Niklas Lampén

I have a large table with quite many dates in format dd.mm. (don't
ask why :) and I want to convert them into -mm-dd for obvious
reasons. How can I do this the smartest way? I know I could do it with
php and update it row by row but that doesn't seem too smart. Can I
somehow do a copy from table to table at the same time?
 
Like:
 
Table 1:
Name = John Doe
Date1 = 01.01.2001
Date2 = 02.02.2001
 
Table 2:
Name = NULL
Date1 = -00-00
Date2 = -00-00
 
And now I want to set Table 2 to this:
 
Name = John Doe
Date1 = 2001-01-01 // Value from Table 1
Date2 = 2001-02-02 // Value from Table 1
 
 
At least tell me what to look for in the manual. :)
 
Niklas



[PHP-DB] Oracle 9i

2001-10-28 Thread Juan Leon


 Hello,

 Has anyone managed to get PHP to link with Oracle9i?  I
suspect some Oracle 9 lib is OCI8-compatible, so it shouldn't
be too hard to get this to work -- possibly a small
change to the configure script.  Unfortunately I don't
know enough about Oracle to make this happen.

 A previous article on php-db suggested that someone had
succeeded and reported on the success to php-install, but
I haven't found any relevant solutions in the archives.

Thanks!

Juan
[EMAIL PROTECTED]




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[PHP-DB] Toronto php users group meeting.

2001-10-28 Thread Dariush

  www.realsoftstudio.com
  ... News Release ...


 PHP User Group Meeting Monday, Oct.29 at 7pm-10pm
PHP SPINAL INJECTION TUTORIAL
-

When : Monday, Oct.29 at 7pm-10pm

Where: Velocet Communications Inc.
   210 Dundas street East (At University)
   suite 8000 (8th floor)
   Toronto

   ** If anybody has a 1024x768 capable SVGA projector they could
   possibly bring to this, please email me: [EMAIL PROTECTED] **

Cost : $1,000,000 per person
   (If you can't afford to pay that, we will let you in for free)

Audience : HTML designers who want to get into writing dynamic web apps,
   weekend web warriors and experienced web developers who want
   to understand how PHP fits into the picture.

Abstract : PHP is a popular scripting language designed specifically for
   creating web-based applications.  The first half of this
fast-paced
   tutorial, will quickly run through the basic characteristics and
   then move on to demonstrate how to use PHP to create images,
pdf's
   and flash on the fly.  In the second half things will get more
   complex with topics such as SQL, LDAP, XML, XSL/XSLT plus a peek
   into the internals of PHP's extension API if there is time.

   Attendees should have a basic understanding of HTML and the basic
   HTTP client/server model.  For the latter topics a bit of
advanced
   programming experience would be helpful but certainly not
required.



www.realsoftstudio.com



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end


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