Re: [PHP] Big problem....need help

2003-07-09 Thread Nicolas Costes
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Hash: SHA1

Le Mercredi 9 Juillet 2003 15:13, phpu a écrit :
 I have a table like this:

 ..
 $sql = SELECT * FROM products, category WHERE
 category.name_category=products.name_category and
 category_id=.$_GET['category_id'] ;


 When i choose the category catg 1 the result is
 Name: aaa
 Name: aaa
 Name: ddd

 As you can see the name aaa appears 2 times insted 1.
 How to make the result look like this:
 Name: aaa
 Name: ddd

Use GROUP BY 

$sql = SELECT * FROM products, category WHERE 
category.name_category=products.name_category and 
category_id=.$_GET['category_id'] GROUP BY products.mark_name;


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   ,,
  (°   Nicolas Costes
  /|\   IUT de La Roche / Yon
 ( ^ )  Clé publique: http://www.keyserver.net/
  ^ ^   http://www.gnu.org/philosophy/can-you-trust.fr.html


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Re: [PHP] Big problem....need help

2003-07-09 Thread phpu
 It isn't work. it gives me an error at this line.

- Original Message -
From: Nicolas Costes [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, July 09, 2003 4:49 PM
Subject: Re: [PHP] Big problemneed help


-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Le Mercredi 9 Juillet 2003 15:13, phpu a écrit :
 I have a table like this:

 ..
 $sql = SELECT * FROM products, category WHERE
 category.name_category=products.name_category and
 category_id=.$_GET['category_id'] ;


 When i choose the category catg 1 the result is
 Name: aaa
 Name: aaa
 Name: ddd

 As you can see the name aaa appears 2 times insted 1.
 How to make the result look like this:
 Name: aaa
 Name: ddd

Use GROUP BY 

$sql = SELECT * FROM products, category WHERE
category.name_category=products.name_category and
category_id=.$_GET['category_id'] GROUP BY products.mark_name;


- --
   ,,
  (°   Nicolas Costes
  /|\   IUT de La Roche / Yon
 ( ^ )  Clé publique: http://www.keyserver.net/
  ^ ^   http://www.gnu.org/philosophy/can-you-trust.fr.html


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Re: [PHP] Big problem....need help

2003-07-09 Thread Haseeb







try like this


$sql = "SELECT * FROM products, category WHERE category.name_category=products.name_category and category_id=".$_GET['category_id']." GROUP BY products.mark_name";





---Original Message---


From: phpu
Date: Wednesday, July 09, 2003 09:51:08 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] Big problemneed help

 It isn't work. it gives me an error at this line.

- Original Message -
From: "Nicolas Costes" [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, July 09, 2003 4:49 PM
Subject: Re: [PHP] Big problemneed help


-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Le Mercredi 9 Juillet 2003 15:13, phpu a écrit :
 I have a table like this:

 ..
 $sql = "SELECT * FROM products, category WHERE
 category.name_category=products.name_category and
 category_id=".$_GET['category_id'] ;


 When i choose the category catg 1 the result is
 Name: aaa
 Name: aaa
 Name: ddd

 As you can see the name aaa appears 2 times insted 1.
 How to make the result look like this:
 Name: aaa
 Name: ddd

Use "GROUP BY" 

$sql = "SELECT * FROM products, category WHERE
category.name_category=products.name_category and
category_id=".$_GET['category_id'] GROUP BY products.mark_name;


- --
 ,,
 (° Nicolas Costes
 /|\ IUT de La Roche / Yon
 ( ^ ) Clé publique: http://www.keyserver.net/
 ^ ^ http://www.gnu.org/philosophy/can-you-trust.fr.html


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Re: [PHP] Big problem....need help

2003-07-09 Thread phpu
10x a lot. it worked.
Now I have another problem. How to find the number of the records in the table.
this is the way i want (in this case}

Name: aaa   2 times
Name  ddd   1 times

Please help with this one




- Original Message - 
  From: Haseeb 
  To: [EMAIL PROTECTED] ; [EMAIL PROTECTED] 
  Cc: [EMAIL PROTECTED] 
  Sent: Thursday, July 10, 2003 8:18 AM
  Subject: Re: [PHP] Big problemneed help


 try like this

$sql = SELECT * FROM products, category WHERE 
category.name_category=products.name_category and category_id=.$_GET['category_id']. 
GROUP BY products.mark_name;
  




---Original Message---

From: phpu
Date: Wednesday, July 09, 2003 09:51:08 PM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [PHP] Big problemneed help

  It isn't work. it gives me an error at this line.

- Original Message -
From: Nicolas Costes [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Sent: Wednesday, July 09, 2003 4:49 PM
Subject: Re: [PHP] Big problemneed help


-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1

Le Mercredi 9 Juillet 2003 15:13, phpu a écrit :
 I have a table like this:

 ..
 $sql = SELECT * FROM products, category WHERE
 category.name_category=products.name_category and
 category_id=.$_GET['category_id'] ;


 When i choose the category catg 1 the result is
 Name: aaa
 Name: aaa
 Name: ddd

 As you can see the name aaa appears 2 times insted 1.
 How to make the result look like this:
 Name: aaa
 Name: ddd

Use GROUP BY 

$sql = SELECT * FROM products, category WHERE
category.name_category=products.name_category and
category_id=.$_GET['category_id'] GROUP BY products.mark_name;


- --
,,
   (° Nicolas Costes
   /|\ IUT de La Roche / Yon
  ( ^ ) Clé publique: http://www.keyserver.net/
   ^ ^ http://www.gnu.org/philosophy/can-you-trust.fr.html


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=vH6U
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Re: [PHP] Big problem....need help

2003-07-09 Thread Mark
First, you might want to use more descriptive subjects. That might
describe your urgency, but it doesn't tell us what you need.

As for the question, try this:

$sql = SELECT *, count(products.mark_name) as count 
FROM products, category 
WHERE category.name_category=products.name_category
AND category_id=.$_GET['category_id']. 
GROUP BY products.mark_name;


but if you don't need all the other data in the rows, just do this:

$sql = SELECT p.mark_name, count(p.mark_name) as count 
FROM products as p, category as c
WHERE c.name_category=p.name_category
AND category_id=.$_GET['category_id']. 
GROUP BY p.mark_name;

--- phpu [EMAIL PROTECTED] wrote:
 10x a lot. it worked.
 Now I have another problem. How to find the number of the records
 in the table.
 this is the way i want (in this case}
 
 Name: aaa   2 times
 Name  ddd   1 times
 
 Please help with this one
 
 
 
 
 - Original Message - 
   From: Haseeb 
   To: [EMAIL PROTECTED] ; [EMAIL PROTECTED] 
   Cc: [EMAIL PROTECTED] 
   Sent: Thursday, July 10, 2003 8:18 AM
   Subject: Re: [PHP] Big problemneed help
 
 
  try like this
 
 $sql = SELECT * FROM products, category WHERE
 category.name_category=products.name_category and
 category_id=.$_GET['category_id']. GROUP BY products.mark_name;
   
 
 
 
 
 ---Original Message---
 
 From: phpu
 Date: Wednesday, July 09, 2003 09:51:08 PM
 To: [EMAIL PROTECTED]
 Cc: [EMAIL PROTECTED]
 Subject: Re: [PHP] Big problemneed help
 
   It isn't work. it gives me an error at this line.
 
 - Original Message -
 From: Nicolas Costes
 [EMAIL PROTECTED]
 To: [EMAIL PROTECTED]
 Sent: Wednesday, July 09, 2003 4:49 PM
 Subject: Re: [PHP] Big problemneed help
 
 
 -BEGIN PGP SIGNED MESSAGE-
 Hash: SHA1
 
 Le Mercredi 9 Juillet 2003 15:13, phpu a écrit :
  I have a table like this:
 
  ..
  $sql = SELECT * FROM products, category WHERE
  category.name_category=products.name_category and
  category_id=.$_GET['category_id'] ;
 
 
  When i choose the category catg 1 the result is
  Name: aaa
  Name: aaa
  Name: ddd
 
  As you can see the name aaa appears 2 times insted 1.
  How to make the result look like this:
  Name: aaa
  Name: ddd
 
 Use GROUP BY 
 
 $sql = SELECT * FROM products, category WHERE
 category.name_category=products.name_category and
 category_id=.$_GET['category_id'] GROUP BY
 products.mark_name;
 
 
 - --
 ,,
(° Nicolas Costes
/|\ IUT de La Roche / Yon
   ( ^ ) Clé publique: http://www.keyserver.net/
^ ^ http://www.gnu.org/philosophy/can-you-trust.fr.html
 
 
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Re: [PHP] BIG PROBLEm with headers

2002-03-08 Thread Analysis Solutions

Hi Krzysiek:

 I have problem with downloading any file from my site using IE 5.5 and 
 HTTPS protocol. My php script works fine with Netscape 6.2 and 
 HTTP/HTTPS protocols and with IE 5.5 but only with HTTP protocol.

I suspect your script has nothing to do with it.  

A while back, I had problems serving my SSL'd pages to IE 5 users.  If I
recall correctly, it turned out to be a bug in IE not being able to
process certificiates with key lengths that were not the ones expected.  
Thawte or my ISP inadvertently used a 1028 bit key on the certificate 
rather than 1024.

Got a new certificate and everything's worked fine since then.

--Dan

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Re: [PHP] BIG PROBLEm with headers

2002-03-08 Thread Erik Price


On Friday, March 8, 2002, at 01:06  PM, Analysis  Solutions wrote:

 I have problem with downloading any file from my site using IE 5.5 and
 HTTPS protocol. My php script works fine with Netscape 6.2 and
 HTTP/HTTPS protocols and with IE 5.5 but only with HTTP protocol.

 I suspect your script has nothing to do with it.

 A while back, I had problems serving my SSL'd pages to IE 5 users.  If I
 recall correctly, it turned out to be a bug in IE not being able to
 process certificiates with key lengths that were not the ones expected.
 Thawte or my ISP inadvertently used a 1028 bit key on the certificate
 rather than 1024.

IE5 has known problems with SSL.  Some sites (sourceforge.net) use SSL 
only to authenticate the user, letting them back out of secure-mode 
after authentication succeeds, if the useragent is IE5.  It reduces the 
potential for problems.

Erik







Erik Price
Web Developer Temp
Media Lab, H.H. Brown
[EMAIL PROTECTED]


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RE: [PHP] BIG problem...

2001-09-14 Thread Nael Mohammad

You need to install the flex libraries from www.gnu.org: you locate it :
http://www.gnu.org/software/flex/flex.html

Nael Mohammad
Neomar, Inc.
[EMAIL PROTECTED]
415-403-7300 x314 (Work)
415-793-0609 (Mobile)
When Wireless Means Business


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Thank you.






-Original Message-
From: Joaquin [mailto:[EMAIL PROTECTED]]
Sent: Friday, September 14, 2001 4:38 PM
To: [EMAIL PROTECTED]
Subject: [PHP] BIG problem...


Hello.. i speak spanish so.. i prefear the answer to this question will in 
spanixh:

I have a problem... i do all correctily  but.. iy happens.. a lot of times..
:

bash-2.04# cd apache_1.3.19
bash-2.04# ./configure --prefix=/www
Configuring for Apache, Version 1.3.19
 + using installation path layout: Apache (config.layout)
Creating Makefile
Creating Configuration.apaci in src
Creating Makefile in src
 + configured for Linux platform
 + setting C compiler to gcc
 + setting C pre-processor to gcc -E
 + checking for system header files
 + adding selected modules
 + checking sizeof various data types
 + doing sanity check on compiler and options
Creating Makefile in src/support
Creating Makefile in src/regex
Creating Makefile in src/os/unix
Creating Makefile in src/ap
Creating Makefile in src/main
Creating Makefile in src/lib/expat-lite
Creating Makefile in src/modules/standardb
bash-2.04# cd php-4.0.6
bash-2.04# ./configure --with-mysql --with-apache=../apache_1.3.x 
--enable-track-vars
loading cache ./config.cache
checking for a BSD compatible install... /usr/bin/install -c
checking whether build environment is sane... yes
checking whether make sets ${MAKE}... yes
checking for working aclocal... missing
checking for working autoconf... missing
checking for working automake... missing
checking for working autoheader... missing
checking for working makeinfo... found
checking whether to enable maintainer-specific portions of Makefiles... no
checking host system type... i686-pc-linux-gnu
checking for gawk... gawk
checking for bison... no
checking for byacc... no
configure: warning: You will need bison if you want to regenerate the PHP 
parsers.
checking for gcc... gcc
checking whether the C compiler (gcc  ) works... yes
checking whether the C compiler (gcc  ) is a cross-compiler... no
checking whether we are using GNU C... yes
checking whether gcc accepts -g... yes
checking how to run the C preprocessor... gcc -E
checking for AIX... no
checking for gcc option to accept ANSI C... none needed
checking for ranlib... ranlib
checking whether gcc and cc understand -c and -o together... yes
checking whether ln -s works... yes
checking for flex... lex
checking for yywrap in -ll... no
checking lex output file root... ./configure: lex: command not found
configure: error: cannot find output from lex; giving up
bash-2.04#

well .. this is my problem.. i dont now ... if you can get me same help.. 
thank you

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RE: [PHP] Big Problem

2001-04-26 Thread Michael Geier

in php.ini, put /var/qmail/bin/qmail-inject into sendmail_path

-Original Message-
From: Thomas Booms [mailto:[EMAIL PROTECTED]]
Sent: Wednesday, April 25, 2001 6:48 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Big Problem


Hi all,

I'm using on my new machine (SuSE 7.0) PHP4. In one PHP4-Script I use
the mail-Function. But my /var/log/mail (qmail) does not give any entry
for this mail.

Is there a bug?

In PHP3 it was all running good.

I have a serious problem here because this is a customers site!

Thanks in advance.

Thomas


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Re: [PHP] Big Problem

2001-04-25 Thread Mark Maggelet

On Thu, 26 Apr 2001 01:48:13 +0200, Thomas Booms
([EMAIL PROTECTED]) wrote:
Hi all,

I'm using on my new machine (SuSE 7.0) PHP4. In one PHP4-Script I
use
the mail-Function. But my /var/log/mail (qmail) does not give any
entry
for this mail.

Is there a bug?

In PHP3 it was all running good.

I have a serious problem here because this is a customers site!

Thanks in advance.

Thomas

maybe sendmail_path should be set, or else I think you can just check
that sendmail is in the path. I know that you're using qmail, but it
usually puts a wrapper called sendmail in the path which calls qmail.


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Re: [PHP] big problem

2001-03-22 Thread Tobias Talltorp

You need to set the permissions in that folder so that you can write.
PHP acts like IUSR(computernmae), so set IUSR(computername) permission to
write.
This is Windows, by the way...

// Tobias

""Salim Meethoo"" [EMAIL PROTECTED] wrote in message
99c90k$83u$[EMAIL PROTECTED]">news:99c90k$83u$[EMAIL PROTECTED]...
 hello there,
 can somebody help me out with my scripts while running this:
 ?
 ..
 $myfile = fopen("data.txt","w");
 .etc;
 ?
 i get the following error message:
 Warning: fopen("compteur","w") - Permission denied in
cmguestcount.inc.php3
 on line 11
 i'm using php3 and this scripts are for a counter of my page .
 thanks
 Salim




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Re: [PHP] BIG PROBLEM WITH SEARCH!!!!!!

2001-01-19 Thread Chris Lee

best way is this.

if ($CID)
$query[] = "CID = '$CID'";
if ($CUserName)
$query[] = "CUserName = '$CUserName'";
if ($CAge)
$query[] = "CAge = '$CAge'";
if ($CEducation)
$query[] = "CEducation = '$CEducation'";

if (isset($query))
$query = "SELECT * FROM clients WHERE CID = '$CID' AND CUserName =
'$CUserName' " . implode(' AND ', $query);
else
$query = "SELECT * FROM clients WHERE CID = '$CID' AND CUserName =
'$CUserName' ";


but this works too...

$query = "SELECT * FROM clients WHERE CID = '$CID' AND CUserName =
'$CUserName' ";
if ($CAge)
$query .= " AND CAge = '$CAge'";
if ($CEducation)
$query .= " AND CEducation = '$CEducation'";

the latter only works if there is at LEAST one variable to be used.

Is this what you meant? no? email me k.

Chris Lee
Mediawaveonline.com




Mike Yuen [EMAIL PROTECTED] wrote in message
[EMAIL PROTECTED]">news:[EMAIL PROTECTED]...
 I'm developing a site that requires people to sign up.  During this time,
 we get their name, email, city, etc.  There is also optional informational
 like (age, education) that they don't have to fill in.

 Now, the problem we have is we allow people search through the database
 using mostly the same criteria people filled in during their sign up.
 However, if someone does not have a preference in Education or age, how do
 we create a query that selects all the records.  Currently, the SQL
 statement has several "OR" clauses in between the optional fields
 (ie: select * form clients where (CID='$CID' AND CUserName='$CUserName')
 OR CAge='$CAge' OR CEducation='$CEducation').

 You can see from the query that if someone has no preference, it will take
 the value from the form and search just on that value - NOT everything.

 I'm stumped on how to solve this problem.  Can anyone help me out?

 Thanks,
 Mike


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