php-install Digest 27 Jul 2001 07:57:36 -0000 Issue 383
Topics (messages 3716 through 3723):
Re: Installing PHP on WinNT and Apache 1.3
3716 by: aLmAgEdOn
PHP datatype vs Oracle Varchar2
3717 by: Victoria Zhou
3718 by: Thies C. Arntzen
3719 by: Victoria Zhou
3723 by: Benjamin Heckmann
storage size of 'buf' isn't known - Slackware
3720 by: David Schanen
zlib version error
3721 by: David Minor
Re: Symbol not fount: pow
3722 by: Andrey Zhiblovsky
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----------------------------------------------------------------------
really thanks for your help...
I did what you said... my httpd.conf has now these lines (may i have an
incorrect apache version? i've got 1.3 and php asks for 1.3.x):
ScriptAlias /php/ "C:/servers/server1/php/"
AddType application/x-httpd-php .php
Action application/x-httpd-php "/php/php.exe"
when i run php hola.php, it shows the right lines, but when i run it with my
browser, it gives a 500 error... an in the error log, appear these lines:
[Thu Jul 26 11:10:17 2001] [error] [client 127.0.0.1]
c:/servers/server1/php/hola.php is not executable; ensure interpreted
scripts have "#!" first line
[Thu Jul 26 11:10:17 2001] [error] [client 127.0.0.1] (2)No such file or
directory: couldn't spawn child process: c:/servers/server1/php/hola.php
It asks for a #! line... i've got perl... does it matters?
if i add #!/php/php.exe, the script runs, but it prints into the browser
this line...
do you know what's wrong?
REALLY THANKS FOR YOUR HELP
aLmAgEdOn
Phil Driscoll <[EMAIL PROTECTED]> escribió en el mensaje de noticias
[EMAIL PROTECTED]
> On Thursday 26 July 2001 02:15, aLmAgEdOn wrote:
> > Hi everyone...
> >
> > I'm really a newbie in PHP. I've got Perl running in Apache (NT 4,
> > workstation), and i want to install PHP 4 (binary) too.
> >
> > These are the lines i added to httpd.conf
> >
> > ScriptAlias /php/ "C:/servers/server1/php/"
> > AddType application/x-httpd-php .php
> > LoadModule php4_module c:/php/sapi/php4apache.dll
> >
> > And configured c:\winnt\php.ini as install help file explains...
> That's a mixture of configuration for the cgi and teh module version.
> If you want to run the cgi version, you need to get rid of the loadmodule
> line and add:
> Action application/x-httpd-php "/php/php.exe"
>
> If you want to run the module version, get rid of the script alias and
make
> sure you copy php4ts.dll to winnt/system32.
>
> Cheers
>
> --
> Phil Driscoll
Hi,
The comparison between Oracle VARCHAR2 and a PHP string doesn't seems to
work. Detail as follows:
I have one PHP variable and one Oracle column:
. $form_user_id: a PHP string obtained from web form input, such as 'user1'.
. Username: an oracle table column of type VARCHAR2(20).
The following sql statement failed:
"select * from mytable where Username='$form_user_id'"
It failed because Username has a strlen of 20, where $form_user_id has a
strlen of 5. If I changed the Username from VARCHAR2(20) to CHAR(20), then
this sql worked fine. But I think with Oracle VARCHAR2, the strlen of
Username in this case should be 5 instead of 20.
Victoria Zhou
On Thu, Jul 26, 2001 at 05:08:42PM -0500, Victoria Zhou wrote:
>
> Hi,
>
> The comparison between Oracle VARCHAR2 and a PHP string doesn't seems to
> work. Detail as follows:
>
> I have one PHP variable and one Oracle column:
> . $form_user_id: a PHP string obtained from web form input, such as 'user1'.
> . Username: an oracle table column of type VARCHAR2(20).
>
> The following sql statement failed:
> "select * from mytable where Username='$form_user_id'"
>
> It failed because Username has a strlen of 20, where $form_user_id has a
> strlen of 5. If I changed the Username from VARCHAR2(20) to CHAR(20), then
> this sql worked fine. But I think with Oracle VARCHAR2, the strlen of
> Username in this case should be 5 instead of 20.
you must be doing something wrong here;-)
please try to create a _tiny_ testcase (< 20 lines) and post
it to the list, i'm sure you're overseeing something
important.
tc
Here is a description of the testcase for the problem:
Oracle:
Create table Mytable (
Username varchar2(20),
Passwd varchar2(20),
Usertype varchar2(50))
Insert into Mytable (Username, Passwd, Usertype)
values('tester1','tester1','Operator');
PHP:
$sql = "select * from Mytable where Username='tester1'";
$sql_statement = OCIParse($conn, $sql) or die ("Couldn't parse statement");
OCIExecute($sql_statement) or die ("Couldn't execute query");
Problem:
The string comaparison in the $sql (Username='tester1') is never true,
because Username is a variable length string and 'tester1' is considered as
a fixed length string.
I don't have the flexibility to change VARCHAR2 to CHAR in order to make
this work, because a fixed length string will cause a lot of trouble for
string comparisons for other code in the program.
Thanks for your help!
Victoria
> -----Original Message-----
> From: Thies C. Arntzen [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, July 26, 2001 6:10 PM
> To: Victoria Zhou
> Cc: [EMAIL PROTECTED]; [EMAIL PROTECTED]
> Subject: Re: [PHP-INST] PHP datatype vs Oracle Varchar2
>
>
> On Thu, Jul 26, 2001 at 05:08:42PM -0500, Victoria Zhou wrote:
> >
> > Hi,
> >
> > The comparison between Oracle VARCHAR2 and a PHP string doesn't seems to
> > work. Detail as follows:
> >
> > I have one PHP variable and one Oracle column:
> > . $form_user_id: a PHP string obtained from web form input,
> such as 'user1'.
> > . Username: an oracle table column of type VARCHAR2(20).
> >
> > The following sql statement failed:
> > "select * from mytable where Username='$form_user_id'"
> >
> > It failed because Username has a strlen of 20, where $form_user_id has a
> > strlen of 5. If I changed the Username from VARCHAR2(20) to
> CHAR(20), then
> > this sql worked fine. But I think with Oracle VARCHAR2, the strlen of
> > Username in this case should be 5 instead of 20.
>
> you must be doing something wrong here;-)
>
> please try to create a _tiny_ testcase (< 20 lines) and post
> it to the list, i'm sure you're overseeing something
> important.
>
> tc
>
>
> Here is a description of the testcase for the problem:
>
> Oracle:
> Create table Mytable (
> Username varchar2(20),
> Passwd varchar2(20),
> Usertype varchar2(50))
So, why are you using VARCHAR2 for this case? If you used
the simple VARCHAR for all of your columns this problem
would not occure?!
>
> Insert into Mytable (Username, Passwd, Usertype)
> values(\'tester1\',\'tester1\',\'Operator\');
>
> PHP:
> $sql = \"select * from Mytable where Username=\'tester1\'\";
> $sql_statement = OCIParse($conn, $sql) or die (\"Couldn\'t parse
statement\");
> OCIExecute($sql_statement) or die (\"Couldn\'t execute query\");
>
>
> Problem:
> The string comaparison in the $sql (Username=\'tester1\') is never true,
> because Username is a variable length string and \'tester1\' is considered
as
> a fixed length string.
>
> I don\'t have the flexibility to change VARCHAR2 to CHAR in order to make
> this work, because a fixed length string will cause a lot of trouble for
> string comparisons for other code in the program.
>
> Thanks for your help!
>
> Victoria
>
Benjamin
This is an older version of Slackware:
Linux www 2.0.33 #1 Tue Apr 7 20:12:12 PDT 1998 i586 unknown
Anyone have an idea what I can do to fix this?
/bin/sh /usr/local/src/php-4.0.6/libtool --silent --mode=compile gcc
-I. -I/usr/local/src/php-4.0.6/ext/standard
-I/usr/local/src/php-4.0.6/main -I/usr/local/src/php-4.0.6
-I/usr/local/apache/include -I/usr/local/src/php-4.0.6/Zend
-I/usr/local/src/php-4.0.6/ext/mysql/libmysql
-I/usr/local/src/php-4.0.6/ext/xml/expat/xmltok
-I/usr/local/src/php-4.0.6/ext/xml/expat/xmlparse
-I/usr/local/src/php-4.0.6/TSRM -DLINUX=20 -DUSE_HSREGEX -DUSE_EXPAT
-DNO_DL_NEEDED -DSUPPORT_UTF8 -DXML_BYTE_ORDER=12 -g -O2 -c filestat.c
filestat.c: In function `php_if_diskfreespace':
filestat.c:157: storage size of `buf' isn't known
make[3]: *** [filestat.lo] Error 1
make[3]: Leaving directory `/usr/local/src/php-4.0.6/ext/standard'
make[2]: *** [all-recursive] Error 1
make[2]: Leaving directory `/usr/local/src/php-4.0.6/ext/standard'
make[1]: *** [all-recursive] Error 1
make[1]: Leaving directory `/usr/local/src/php-4.0.6/ext'
make: *** [all-recursive] Error 1
I see this question asked a bunch in the archives, but don't ever really see
an answer. When configuring php, I keep getting:
checking whether to include zlib support... yes
checking for gzgets in -lz... no
configure: error: Zlib module requires zlib >= 1.0.9.
System: Debian 2.2/Apache 1.3.9/PHP 4.0.4pl1/Zlib 1.1.3
the libz.* files are in /usr/lib. zlib.h is in /usr/include.
My configure line is below:
./configure \
--with-mysql \
--with-zlib \
--with-gd \
--with-jpeg-dir \
--with-apxs=/usr/bin/apxs \
--enable-track-vars \
--enable-ftp
Please help,
David Minor
Sorry, now it's working. I didn't know that SCO Open Server hasn't DSO.
Andrey Zhiblovsky <[EMAIL PROTECTED]> ÓÏÏÂÝÉÌ × ÎÏ×ÏÓÔÑÈ
ÓÌÅÄÕÀÝÅÅ:000901c114a7$95d36e40$[EMAIL PROTECTED]
Hello,
I can't start apache server after compiling php-4.0.6
Here is attachment - messages from 'script' program.
Thank you,
Andrey Zhiblovsky.
