[PHP] help me please!

2003-07-15 Thread pehepe php
My system is windows 98 second edition
im using pws
i built php 4.2.2
when ý struggle to reach any .php file for example 127.0.0.1/sample.php
this error occur:


Security Alert! The PHP CGI cannot be accessed directly.
This PHP CGI binary was compiled with force-cgi-redirect enabled. This means 
that a page will only be served up if the REDIRECT_STATUS CGI variable is 
set, e.g. via an Apache Action directive.

please help me!!

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[PHP] help me please

2003-03-16 Thread Blair Robinson
Hello,
 I am making a web site i just want to be able to view my php outputs On my Computer 
how do i do this i have downlaoded the 1mb windows php installer. and have the zipped 
version(4.3.0) that is 4mb I have windows xp. Can you please point me in the right 
direction or give me some answers on how to view my pages.
 
 Thanks
 Blair.


RE: [PHP] help me please

2003-03-16 Thread Bryan Brannigan
download and install the apache web server.  that along with php should do
the trick.  if you're doing any database work you may want to pick up the
windows version of that too.

some decent instructions are here:
http://www.thesitewizard.com/archive/php4install.shtml

-Original Message-
From: Blair Robinson [mailto:[EMAIL PROTECTED]
Sent: Sunday, March 16, 2003 7:47 PM
To: [EMAIL PROTECTED]
Subject: [PHP] help me please


Hello,
 I am making a web site i just want to be able to view my php outputs On my
Computer how do i do this i have downlaoded the 1mb windows php installer.
and have the zipped version(4.3.0) that is 4mb I have windows xp. Can you
please point me in the right direction or give me some answers on how to
view my pages.

 Thanks
 Blair.


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[PHP] help me please

2003-02-25 Thread bionicegg
Hello all,


I am having problems with a script.  The script works great on one of my
sights, however, when I transferred it to another site, it does not
function.  I contacted my hosting company (both sites are on the same
server), they told me that I needed to check the script and that it has
nothing to do with them, since they ran a  test.php and it worked fine.
   Here is the script:

?PHP

$ToEmail = [EMAIL PROTECTED];

$ToSubject = Flash Contact Form;

$EmailBody = Sent By: $FirstName\nSenders Email: $Email\n\nMessage
Sent:\n$ToComments\n;

mail($ToName. .$ToEmail.,$ToSubject, $EmailBody, From: .$FirstName.
.$Email.);

?

Please if anyone knows anything about this stuff (Im a newbie to this) You
help is greatly appreciated!

Thanks in advance,
mark johnson






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Fw: [PHP] help me please

2003-02-25 Thread Rick Emery
You don't have to break up the strings:
mail($ToName $ToEmail,$ToSubject, $EmailBody, From: $FirstName $Email);

What are values of $ToName and $FirstName?

why did you submit tghis 3 times?
- Original Message - 
From: bionicegg [EMAIL PROTECTED]
To: 
Sent: Tuesday, February 25, 2003 11:38 AM
Subject: [PHP] help me please


Hello all,


I am having problems with a script.  The script works great on one of my
sights, however, when I transferred it to another site, it does not
function.  I contacted my hosting company (both sites are on the same
server), they told me that I needed to check the script and that it has
nothing to do with them, since they ran a  test.php and it worked fine.
   Here is the script:

?PHP

$ToEmail = [EMAIL PROTECTED];

$ToSubject = Flash Contact Form;

$EmailBody = Sent By: $FirstName\nSenders Email: $Email\n\nMessage
Sent:\n$ToComments\n;

mail($ToName. .$ToEmail.,$ToSubject, $EmailBody, From: .$FirstName.
.$Email.);

?

Please if anyone knows anything about this stuff (Im a newbie to this) You
help is greatly appreciated!

Thanks in advance,
mark johnson






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[PHP] help me please.

2002-10-04 Thread Steve Jackson

got a problem here people.
What I want to do is upload a file and store its name in a database.
For some reason I cant.

I can upload the file fine but can't pass it's name to the mysql DB.
Any ideas:

Upload form:
?
require( ../Cart.php);
Brand();
DBInfo();


echo bYou can now upload a picture.../bbrbr;
echo  form enctype='multipart/form-data' action='uploadItem.php'
method=post;
echo  INPUT TYPE='hidden' name='MAX_FILE_SIZE' value='25000';
echo  INPUT TYPE='hidden' name='ID' value='$II';
echo  INPUT name='userfile' type='file';
blueFont( Arial, Send this file: );
echo  INPUT TYPE='submit' VALUE='Send File'/form;

echo pbThis file MUST be a JPEG (jpg) image./b;

//adminFooter($Relative);
?

Process the form:
?
require(../Cart.php);
Root();
DBinfo();

mysql_connect($DBHost,$DBUser,$DBPass);
$UploadURL = ../images/;
if ($usefile==none)
{
echo Error: no file uploaded;
exit;
}
//Connect to DB
$mysql = mysql_connect($DBHost,$DBUser,$DBPass);
if(!$mysql)
{
echo Cannot connect to database.;
exit;
}

mysql_query(INSERT INTO Items (ImageName) WHERE ItemID = '$II'
VALUE('test'));

$upfile = $UploadURL.$userfile_name;

if ( !copy($userfile, $upfile))
{
echo Error: Could not move file into directory;
exit;
}
? 

Steve Jackson
Web Developer
Viola Systems Ltd.
http://www.violasystems.com
[EMAIL PROTECTED]
Mobile +358 50 343 5159


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[PHP] help me please! :)

2002-06-10 Thread Veronica Ghezzi

Hi,
   i must get the information saved in a several select list named
liv_1 select name=liv_1
liv_2   select name=liv_2
liv_3   select name=liv_3

liv_n   select name=liv_n

To get the value i work in this way...

$n = 50;
for ($i=1; $i=$n;i++){
...
echo $liv_$i;   in asp i do:   response.write 
(request(liv_ +
i))
...
}
But i get only
1
2
3
...
50

What can i do to get $liv_1 ... $liv_2...  ???
Thank you a lot!

Veronica Ghezzi


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RE: [PHP] help me please! :)

2002-06-10 Thread Kevin Porter

You need to escape the dollar sign with a backslash to prevent PHP trying to
interpolate the variable $liv (which presumably doesn't exist).

echo \$liv_$i;

HTH,

- Kev

 -Original Message-
 From: Veronica Ghezzi [SMTP:[EMAIL PROTECTED]]
 Sent: 10 June 2002 09:24
 To:   Php-General
 Subject:  [PHP] help me please! :)
 
 Hi,
i must get the information saved in a several select list named
   liv_1 select name=liv_1
   liv_2   select name=liv_2
   liv_3   select name=liv_3
   
   liv_n   select name=liv_n
 
 To get the value i work in this way...
 
   $n = 50;
   for ($i=1; $i=$n;i++){
   ...
   echo $liv_$i;   in asp i do:   response.write
 (request(liv_ +
 i))
   ...
   }
 But i get only
   1
   2
   3
   ...
   50
 
 What can i do to get $liv_1 ... $liv_2...  ???
 Thank you a lot!
 
 Veronica Ghezzi
 
 
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Re: [PHP] help me please! :)

2002-06-10 Thread PHPCoder

?php
$i = 1;
$liv_1 = one;
$liv_2 = two;
$liv_3 = three;
$liv_4 = four;
$liv_5 = five;
while ($i = 5 ) {
$do = echo \$liv_{$i};;
eval($do);
echo br;
 $i ++;
}
?

Simply echoing the \$liv_$i will not work, need to evaluate the string 
after parsed as in example above...


Kevin Porter wrote:

You need to escape the dollar sign with a backslash to prevent PHP trying to
interpolate the variable $liv (which presumably doesn't exist).

echo \$liv_$i;

HTH,

- Kev

-Original Message-
From: Veronica Ghezzi [SMTP:[EMAIL PROTECTED]]
Sent: 10 June 2002 09:24
To:   Php-General
Subject:  [PHP] help me please! :)

Hi,
   i must get the information saved in a several select list named
  liv_1 select name=liv_1
  liv_2   select name=liv_2
  liv_3   select name=liv_3
  
  liv_n   select name=liv_n

To get the value i work in this way...

  $n = 50;
  for ($i=1; $i=$n;i++){
  ...
  echo $liv_$i;   in asp i do:   response.write
(request(liv_ +
i))
  ...
  }
But i get only
  1
  2
  3
  ...
  50

What can i do to get $liv_1 ... $liv_2...  ???
Thank you a lot!

Veronica Ghezzi


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Re: [PHP] help me please! :)

2002-06-10 Thread Marek Kilimajer

Try:

for ($i=1; $i=$n;i++){
...
$tmp=liv_$i;
echo {$$tmp};
...
}


Veronica Ghezzi wrote:

Hi,
   i must get the information saved in a several select list named
   liv_1 select name=liv_1
   liv_2   select name=liv_2
   liv_3   select name=liv_3
   
   liv_n   select name=liv_n

To get the value i work in this way...

   $n = 50;
   for ($i=1; $i=$n;i++){
   ...
   echo $liv_$i;   in asp i do:   response.write 
(request(liv_ +
i))
   ...
   }
But i get only
   1
   2
   3
   ...
   50

What can i do to get $liv_1 ... $liv_2...  ???
Thank you a lot!

Veronica Ghezzi


  




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Re: [PHP] help me please! :)

2002-06-10 Thread Josep R. Raurell

This work for me:

liv_1select name=liv[1]
liv_2select name=liv[2]
liv_3select name=liv[3]

liv_nselect name=liv[n]

To get the value i work in this way...

$liv= $_REQUEST['liv'];

$n = 50;
for ($i=1; $i=$n;i++){
...
echo $liv[$i];


Josep R. Raurell



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[PHP] Re:[PHP] HELP ME PLEASE: php not run on IIS 4.0

2002-03-25 Thread Liam

25/03/2002 8:27:51 PM

hehehe
You could start with loosing IIS, it's bad news.
The port of Apache for windows is good, if not, Xitami's also quite good.

http://www.apache.org/dist/httpd/binaries/win32/
http://www.xitami.com/

But, no, sorry.  I can't help you.



Berlina [EMAIL PROTECTED] wrote on 25/03/2002 10:19:46 AM:

Hi,

Im trying to install PHP 4.1.2 under Windows 2000 Server and IIS 4.0

* If I configure PHP as a MODULE, all run but SESSIONS DON'T WORK
* If I configure PHP as a CGI, SESSIONS WORK but ORACLE MODULE for PHP not
load

Any ideas?
Anybody can help me?

Advanced thanks,
Berli

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[PHP] HELP ME PLEASE: php not run on IIS 4.0

2002-03-25 Thread Berlina

Hi,

Im trying to install PHP 4.1.2 under Windows 2000 Server and IIS 4.0

* If I configure PHP as a MODULE, all run but SESSIONS DON'T WORK
* If I configure PHP as a CGI, SESSIONS WORK but ORACLE MODULE for PHP not
load

Any ideas?
Anybody can help me?

Advanced thanks,
Berli

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Re: [PHP] Help me please.(Running stand-alone php)

2001-08-14 Thread miguel valero

Tina,
Thank you for the help Tina  :-)) 
I getting a good experience working with PHP  and HTML
(forms). Because of my full time job and personal
commitments, I haven't been able to spend  as much of
my free time as I could in my project (Free Gallery
Management System). After October I hope to dedicate
more of my free time on this project.

The project will be open source and public domain. It
will be developed with the help of PHP, a and
PostgreSQL. More information  about the project can be
seen at the following web site:
   
http://www.columbia.edu/~mv169/private/gallery_f.html 
 Login Name:  guest
 Password:guest
 
If you have any comments, suggestions or if you want
to participate, e-mail me at [EMAIL PROTECTED] Your
input is very valuable to me.
 
Adios, miguel
--- Tina [EMAIL PROTECTED] wrote:
 The solution:
 
 You have to configure php with --enable-discard-path
 as in:
 ./configure --without-apache --enable-discard-path
 before you compile it.
 
 This is documented here:
 http://www.php.net/manual/en/security.php
 in the section:
 Case 4: PHP parser outside of web tree
 
 Now when you do !--#include virtual='php.cgi'--
 the #!/path/to/php will not show at the top of the
 page.
 
 (The problem was indeed not local to windows).
 
 Tina.
 


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[PHP] Help me please.(Running stand-alone php)

2001-08-07 Thread miguel valero

Unfortunately the -q parameter didn't fix the problem.

Is anybody in the PHP mailing list that is familiar
with this kind of problem and can tell me how to  get
rid of the path line in the browser output?

Any help will be greatly appreciated. 
Thanks.
--- mike cullerton [EMAIL PROTECTED] wrote:
 try the -q parameter. i'm guessing here, since i
 don't run windows nor the
 cgi version of php.
 
 :)
 
 on 8/7/01 8:27 AM, miguel valero at [EMAIL PROTECTED]
 wrote:
 
  Mike, 
  
  I running the scrip from the browser
  (http://localhost/cgi-bin/test.html) When I
 comment
  the first line in the script
 (#-#!c:\php\php.exe)
  I get the following error message:
  
  Tue Aug 07 09:56:13 2001] [error] [client
 127.0.0.1]
  c:/program files/apache
 group/apache/cgi-bin/test.php
  is not executable; ensure interpreted scripts have
  #! first line
  
  I can not run the script w/o the line
 #!c:\php\php.exe
  at the top of the script.
  
  How can I get rid of the path line on the browser
  output ?
  
  Thanks
  --- mike cullerton [EMAIL PROTECTED]
 wrote:
  i'm not sure i understand how you are running
  this...
  
  if it is a file you are running at the command
 line,
  try -q after the
  php.exe statement.
  
  if this is a web page, comment that line out.
  
  hope this helps,
  mike
  
  on 8/6/01 1:16 PM, miguel valero at
 [EMAIL PROTECTED]
  wrote:
  
  When I run the following script under an
  stand-alone
  php 4.0.6 and Apache 1.3 (Win95/98), the path to
  PHP
  (#!c:\php\php.exe) will show up on the first
 line
  of
  the browser output.
  
  #!c:\php\php.exe
  
  ?php
  
  phpinfo();
  
  ?
  
  Is there a way to fix this? How can I get rid of
  the
  path line, on the top of the page, on the
 browser
  output? Any help will be greatly appreciated.
  
  My apologies if this has been covered before, or
  if
  this is in the manual, but I couldn't see it
  
  thanks.
  
  
 


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Re: [PHP] Help me please.(Running stand-alone php)

2001-08-07 Thread Bjorn Van Simaeys

Miguel,


I see that your script has the extension .html
Rename that file to .php and it normally should work.


Greetz,
Bjorn Van Simaeys
www.bvsenterprises.com



--- miguel valero [EMAIL PROTECTED] wrote:
 Unfortunately the -q parameter didn't fix the
 problem.
 
 Is anybody in the PHP mailing list that is familiar
 with this kind of problem and can tell me how to 
 get
 rid of the path line in the browser output?
 
 Any help will be greatly appreciated. 
 Thanks.
 --- mike cullerton [EMAIL PROTECTED] wrote:
  try the -q parameter. i'm guessing here, since i
  don't run windows nor the
  cgi version of php.
  
  :)
  
  on 8/7/01 8:27 AM, miguel valero at
 [EMAIL PROTECTED]
  wrote:
  
   Mike, 
   
   I running the scrip from the browser
   (http://localhost/cgi-bin/test.html) When I
  comment
   the first line in the script
  (#-#!c:\php\php.exe)
   I get the following error message:
   
   Tue Aug 07 09:56:13 2001] [error] [client
  127.0.0.1]
   c:/program files/apache
  group/apache/cgi-bin/test.php
   is not executable; ensure interpreted scripts
 have
   #! first line
   
   I can not run the script w/o the line
  #!c:\php\php.exe
   at the top of the script.
   
   How can I get rid of the path line on the
 browser
   output ?
   
   Thanks
   --- mike cullerton [EMAIL PROTECTED]
  wrote:
   i'm not sure i understand how you are running
   this...
   
   if it is a file you are running at the command
  line,
   try -q after the
   php.exe statement.
   
   if this is a web page, comment that line out.
   
   hope this helps,
   mike
   
   on 8/6/01 1:16 PM, miguel valero at
  [EMAIL PROTECTED]
   wrote:
   
   When I run the following script under an
   stand-alone
   php 4.0.6 and Apache 1.3 (Win95/98), the path
 to
   PHP
   (#!c:\php\php.exe) will show up on the first
  line
   of
   the browser output.
   
   #!c:\php\php.exe
   
   ?php
   
   phpinfo();
   
   ?
   
   Is there a way to fix this? How can I get rid
 of
   the
   path line, on the top of the page, on the
  browser
   output? Any help will be greatly appreciated.
   
   My apologies if this has been covered before,
 or
   if
   this is in the manual, but I couldn't see it
   
   thanks.
   
   
  
 
 
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Re: [PHP] Help me please.(Running stand-alone php)

2001-08-07 Thread Tina

The solution:

You have to configure php with --enable-discard-path
as in:
./configure --without-apache --enable-discard-path
before you compile it.

This is documented here:
http://www.php.net/manual/en/security.php
in the section:
Case 4: PHP parser outside of web tree

Now when you do !--#include virtual='php.cgi'--
the #!/path/to/php will not show at the top of the page.

(The problem was indeed not local to windows).

Tina.


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[PHP] help me please :(

2001-04-06 Thread Nilesh Parmar

Hi :)
   Well, I have installed php on my Windows 98 machine. But i am unable
to use the imap funcationality. I have the php_imap.dll in my php.ini file
for which i have removed the semicolon (;extension = php_imap.dll) . But it
dosent seem to work .I suppose i should have (php_imap4r2.dll ). Can anyone
please send me a copy of the file if  u have it please :)
thanx in advance
bye
Nilesh



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