RE: [PHP] PHP Command line script
Check the error from mysqli:
http://fi.php.net/manual/en/function.mysqli-error.php
Best regards,
Peter Lauri
www.dwsasia.com - company web site
www.lauri.se - personal web site
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-Original Message-
From: Nathaniel Hall [mailto:[EMAIL PROTECTED]
Sent: Tuesday, May 01, 2007 10:13 PM
To: php-general@lists.php.net
Subject: [PHP] PHP Command line script
I am attempting to run a script that will run from the command line
nightly to update a field in a database. I already created a script
that would access the database and insert most of the information when a
webpage is visited and I had no problems with it. The command line
script appears to fail on the prepare. I have echo'ed the SQL statement
to the screen, copied it, and run it on the MySQL server with no
problems. Any ideas?
?php
$mysqli = new mysqli('localhost', 'root', 'abc123', 'mydb');
if (mysqli_connect_errno()) {
echo Unable to connect to database.\n;
exit;
} else {
$login = date('m\-d\-Y');
if ($logout = $mysqli-prepare(UPDATE `mydb`.`authlog`
SET `logout` = ? WHERE `login` LIKE '$login%')) { // --- Will not go
any further than here, even when hard coding the information.
$logout-bind_param(s, date('m\-d\-
Y\TH\:i\:s'));
$logout-execute();
$logout-close();
}
}
$mysqli-close();
?
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Re: [PHP] PHP Command line script
Greg Donald wrote:
On 5/1/07, Nathaniel Hall [EMAIL PROTECTED] wrote:
I am attempting to run a script that will run from the command line
nightly to update a field in a database. I already created a script
that would access the database and insert most of the information when a
webpage is visited and I had no problems with it. The command line
script appears to fail on the prepare. I have echo'ed the SQL statement
to the screen, copied it, and run it on the MySQL server with no
problems. Any ideas?
?php
$mysqli = new mysqli('localhost', 'root', 'abc123', 'mydb');
if (mysqli_connect_errno()) {
echo Unable to connect to database.\n;
exit;
} else {
$login = date('m\-d\-Y');
if ($logout = $mysqli-prepare(UPDATE `mydb`.`authlog`
SET `logout` = ? WHERE `login` LIKE '$login%')) { // --- Will not go
any further than here, even when hard coding the information.
$logout-bind_param(s,
date('m\-d\-Y\TH\:i\:s'));
$logout-execute();
$logout-close();
}
}
$mysqli-close();
?
Add full error reporting, then make sure you can see the errors, then
test to see if you have the mysqli extension:
error_reporting( E_ALL );
ini_set( 'display_errors', 1 );
ini_set( 'log_errors', 1 );
if( !in_array( 'mysqli', get_loaded_extensions() ) )
{
die( 'no mysqli found' );
}
I get no errors and I have verified that mysqli is loaded.
Also, why do you need to escape the dashes in the date() calls?
php -r 'echo date(Y-m-d);'
2007-05-01
Habit.
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Re: [PHP] PHP Command line script
On Tue, May 1, 2007 3:13 pm, Nathaniel Hall wrote:
?php
$mysqli = new mysqli('localhost', 'root', 'abc123', 'mydb');
if (mysqli_connect_errno()) {
echo Unable to connect to database.\n;
exit;
} else {
$login = date('m\-d\-Y');
if ($logout = $mysqli-prepare(UPDATE
`mydb`.`authlog`
SET `logout` = ? WHERE `login` LIKE '$login%')) { // --- Will not
go
any further than here, even when hard coding the information.
$logout-bind_param(s,
date('m\-d\-Y\TH\:i\:s'));
$logout-execute();
$logout-close();
}
//ALWAYS do your error-checking!!!
else{
die(mysqli_error());
}
//You may want to put the query in $query so you can print that
//as part of the 'die' output
}
$mysqli-close();
?
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[PHP] PHP Command line script
I am attempting to run a script that will run from the command line
nightly to update a field in a database. I already created a script
that would access the database and insert most of the information when a
webpage is visited and I had no problems with it. The command line
script appears to fail on the prepare. I have echo'ed the SQL statement
to the screen, copied it, and run it on the MySQL server with no
problems. Any ideas?
?php
$mysqli = new mysqli('localhost', 'root', 'abc123', 'mydb');
if (mysqli_connect_errno()) {
echo Unable to connect to database.\n;
exit;
} else {
$login = date('m\-d\-Y');
if ($logout = $mysqli-prepare(UPDATE `mydb`.`authlog`
SET `logout` = ? WHERE `login` LIKE '$login%')) { // --- Will not go
any further than here, even when hard coding the information.
$logout-bind_param(s, date('m\-d\-Y\TH\:i\:s'));
$logout-execute();
$logout-close();
}
}
$mysqli-close();
?
--
Nathaniel Hall, GSEC GCFW GCIA GCIH GCFA
Spider Security
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Re: [PHP] PHP Command line script
First and foremost, it's a VERY BAD idea to use root for MySQL. If your
code isn't perfect (and even sometimes if it is), arbitrary commands and SQL
injection attacks could lead to migraines that no Tylenol will ever be able
to alleviate.
Secondly, what error is the CLI kicking out when you run it from the
command line?
On 5/1/07, Nathaniel Hall [EMAIL PROTECTED] wrote:
I am attempting to run a script that will run from the command line
nightly to update a field in a database. I already created a script
that would access the database and insert most of the information when a
webpage is visited and I had no problems with it. The command line
script appears to fail on the prepare. I have echo'ed the SQL statement
to the screen, copied it, and run it on the MySQL server with no
problems. Any ideas?
?php
$mysqli = new mysqli('localhost', 'root', 'abc123', 'mydb');
if (mysqli_connect_errno()) {
echo Unable to connect to database.\n;
exit;
} else {
$login = date('m\-d\-Y');
if ($logout = $mysqli-prepare(UPDATE `mydb`.`authlog`
SET `logout` = ? WHERE `login` LIKE '$login%')) { // --- Will not go
any further than here, even when hard coding the information.
$logout-bind_param(s,
date('m\-d\-Y\TH\:i\:s'));
$logout-execute();
$logout-close();
}
}
$mysqli-close();
?
--
Nathaniel Hall, GSEC GCFW GCIA GCIH GCFA
Spider Security
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[mobile] (570-) 766-8107
Re: [PHP] PHP Command line script
Daniel Brown wrote:
First and foremost, it's a VERY BAD idea to use root for MySQL.
If your code isn't perfect (and even sometimes if it is), arbitrary
commands and SQL injection attacks could lead to migraines that no
Tylenol will ever be able to alleviate.
I changed the user I was connecting as in order to post. I don't use
root in the real code.
Secondly, what error is the CLI kicking out when you run it from
the command line?
It doesn't give an error. The only thing it does is continue on through
the IF statement, which goes nowhere. I have added an ELSE to the
script and run it. It ends up running the code in the ELSE.
$login = date('m\-d\-Y');
if ($logout = $mysqli-prepare(UPDATE `mydb`.`authlog` SET
`logout` = ? WHERE `login` LIKE '$login%')) { // --- Will not
go any further than here, even when hard coding the information.
$logout-bind_param(s, date('m\-d\-Y\TH\:i\:s'));
$logout-execute();
$logout-close();
}
--
Nathaniel Hall, GSEC GCFW GCIA GCIH GCFA
Spider Security
--
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Re: [PHP] PHP Command line script
On 5/1/07, Nathaniel Hall [EMAIL PROTECTED] wrote:
I am attempting to run a script that will run from the command line
nightly to update a field in a database. I already created a script
that would access the database and insert most of the information when a
webpage is visited and I had no problems with it. The command line
script appears to fail on the prepare. I have echo'ed the SQL statement
to the screen, copied it, and run it on the MySQL server with no
problems. Any ideas?
?php
$mysqli = new mysqli('localhost', 'root', 'abc123', 'mydb');
if (mysqli_connect_errno()) {
echo Unable to connect to database.\n;
exit;
} else {
$login = date('m\-d\-Y');
if ($logout = $mysqli-prepare(UPDATE `mydb`.`authlog`
SET `logout` = ? WHERE `login` LIKE '$login%')) { // --- Will not go
any further than here, even when hard coding the information.
$logout-bind_param(s, date('m\-d\-Y\TH\:i\:s'));
$logout-execute();
$logout-close();
}
}
$mysqli-close();
?
Add full error reporting, then make sure you can see the errors, then
test to see if you have the mysqli extension:
error_reporting( E_ALL );
ini_set( 'display_errors', 1 );
ini_set( 'log_errors', 1 );
if( !in_array( 'mysqli', get_loaded_extensions() ) )
{
die( 'no mysqli found' );
}
Also, why do you need to escape the dashes in the date() calls?
php -r 'echo date(Y-m-d);'
2007-05-01
--
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http://destiney.com/
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Re: [PHP] PHP Command Line Scripts 'Aborting' at end ...
* Thus wrote Marc G. Fournier: Note that the following is based on php installed via the FreeBSD ports system ... I'm getting a core file, but if I try: gdb /usr/local/bin/php php.core ... its definitely not looking good: s# gdb /usr/local/bin/php php.core ... Core was generated by `php'. Program terminated with signal 6, Abort trap. Reading symbols from /usr/lib/libcrypt.so.2...done. Reading symbols from /usr/lib/libm.so.2...done. Reading symbols from /usr/lib/libc.so.4...done. Reading symbols from /usr/local/lib/php/20020429/interbase.so...done. Reading symbols from /usr/local/firebird/lib/libfbembed.so.1...Deprecated bfd_read called at /usr/src/gnu/usr.bin/binutils/gdb/../../../../contrib/gdb/gdb/dwarf2read.c line 3049 in dwarf2_read_section Error while reading shared library symbols: Dwarf Error: Cannot handle DW_FORM_strp in DWARF reader. Reading symbols from /usr/lib/libncurses.so.5...done. Error while reading shared library symbols: ì: No such file or directory. Error while reading shared library symbols: ynamic: No such file or directory. Segmentation fault (core dumped) Interesting core dump :) Try running php directly from gdb instead of reading the core file: % gdb /usr/loca/bin/php ... run script.php ... bt See if that produces a better backtrace. mod_php4 appears to work fine, just the command line version seems to be off ... and its running, producing expected output, its just that last 'Abort' that tends to screw things up a bit ... My bet is you have a buggy library somewhere, ncurses being a viable candidate since it does not get loaded in mod_php4. Curt -- Quoth the Raven, Nevermore. -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] PHP Command Line Scripts 'Aborting' at end ...
Note that the following is based on php installed via the FreeBSD ports system ... I have a really simple PHP script that, when you run it, generates an Abort at the end of it: ams# /tmp/test.php testAbort (core dumped) ams# cat /tmp/test.php #!/usr/local/bin/php ?php echo test; ? Even if I change the script slightly, so that last line isn't being run via php, it does the same: ams# cat /tmp/test.php #!/usr/local/bin/php ?php echo test; ? test ams# /tmp/test.php test test Abort (core dumped) I'm getting a core file, but if I try: gdb /usr/local/bin/php php.core ... its definitely not looking good: s# gdb /usr/local/bin/php php.core GNU gdb 4.18 (FreeBSD) Copyright 1998 Free Software Foundation, Inc. GDB is free software, covered by the GNU General Public License, and you are welcome to change it and/or distribute copies of it under certain conditions. Type show copying to see the conditions. There is absolutely no warranty for GDB. Type show warranty for details. This GDB was configured as i386-unknown-freebsd...Deprecated bfd_read called at /usr/src/gnu/usr.bin/binutils/gdb/../../../../contrib/gdb/gdb/dbxread.c line 2627 in elfstab_build_psymtabs Deprecated bfd_read called at /usr/src/gnu/usr.bin/binutils/gdb/../../../../contrib/gdb/gdb/dbxread.c line 933 in fill_symbuf Core was generated by `php'. Program terminated with signal 6, Abort trap. Reading symbols from /usr/lib/libcrypt.so.2...done. Reading symbols from /usr/lib/libm.so.2...done. Reading symbols from /usr/lib/libc.so.4...done. Reading symbols from /usr/local/lib/php/20020429/interbase.so...done. Reading symbols from /usr/local/firebird/lib/libfbembed.so.1...Deprecated bfd_read called at /usr/src/gnu/usr.bin/binutils/gdb/../../../../contrib/gdb/gdb/dwarf2read.c line 3049 in dwarf2_read_section Error while reading shared library symbols: Dwarf Error: Cannot handle DW_FORM_strp in DWARF reader. Reading symbols from /usr/lib/libncurses.so.5...done. Error while reading shared library symbols: ì: No such file or directory. Error while reading shared library symbols: ynamic: No such file or directory. Segmentation fault (core dumped) mod_php4 appears to work fine, just the command line version seems to be off ... and its running, producing expected output, its just that last 'Abort' that tends to screw things up a bit ... Not sure how to debug ... help? Thanks ... Marc G. Fournier Hub.Org Networking Services (http://www.hub.org) Email: [EMAIL PROTECTED] Yahoo!: yscrappy ICQ: 7615664 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] PHP Command Line Scripts 'Aborting' at end ...
On Wed, 27 Oct 2004 13:44:54 -0300 (ADT), Marc G. Fournier [EMAIL PROTECTED] wrote: Not sure how to debug ... help? You said you installed in via ports, so did you happen to check the 'debug' option when you installed it? [ ] DEBUG Enable debug Just a question.. as that might help explain more what's going on. -- Greg Donald Zend Certified Engineer http://gdconsultants.com/ http://destiney.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] PHP Command Line Scripts 'Aborting' at end ...
On Wed, 27 Oct 2004 13:44:54 -0300 (ADT), Marc G. Fournier [EMAIL PROTECTED] wrote: I have a really simple PHP script that, when you run it, generates an Abort at the end of it: ams# /tmp/test.php testAbort (core dumped) I just installed php4-cgi on a FreeBSD 4.10 system I have. The same test.php script runs fine for me. cat test.php #!/usr/local/bin/php ?php echo test; ? ./test.php Content-type: text/html X-Powered-By: PHP/4.3.9 test -- Greg Donald Zend Certified Engineer http://gdconsultants.com/ http://destiney.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] PHP Command Line Scripts 'Aborting' at end ...
I just did a reinstall from ports, and it works now as well ... maybe a stale library for one of hte modules :( thanks ... On Wed, 27 Oct 2004, Greg Donald wrote: On Wed, 27 Oct 2004 13:44:54 -0300 (ADT), Marc G. Fournier [EMAIL PROTECTED] wrote: I have a really simple PHP script that, when you run it, generates an Abort at the end of it: ams# /tmp/test.php testAbort (core dumped) I just installed php4-cgi on a FreeBSD 4.10 system I have. The same test.php script runs fine for me. cat test.php #!/usr/local/bin/php ?php echo test; ? ./test.php Content-type: text/html X-Powered-By: PHP/4.3.9 test -- Greg Donald Zend Certified Engineer http://gdconsultants.com/ http://destiney.com/ -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php Marc G. Fournier Hub.Org Networking Services (http://www.hub.org) Email: [EMAIL PROTECTED] Yahoo!: yscrappy ICQ: 7615664 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] PHP command line vars
I have a script that I want to run on the windows command line, and it is working fine. What I want to do is pass vars to the script when I run it, example: php script.php $var=hello I have read the docs but can't find the syntax to pass these in. Also, how to retrieve them in the script? argv other? TIA -Shawn -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] PHP command line vars
Shawn McKenzie wrote: I have a script that I want to run on the windows command line, and it is working fine. What I want to do is pass vars to the script when I run it, example: php script.php $var=hello The example will not work, command line arguments are traditionaly passed in the form of: php script.php --long-name hello or php script.php -n hello You will have to parse $argv array to get the command line arguments. I have read the docs but can't find the syntax to pass these in. Also, how to retrieve them in the script? argv other? http://www.php.net/manual/en/features.commandline.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
Re: [PHP] PHP command line vars
Thanks! The link was just what I needed. -Shawn Marek Kilimajer [EMAIL PROTECTED] wrote in message news:[EMAIL PROTECTED] Shawn McKenzie wrote: I have a script that I want to run on the windows command line, and it is working fine. What I want to do is pass vars to the script when I run it, example: php script.php $var=hello The example will not work, command line arguments are traditionaly passed in the form of: php script.php --long-name hello or php script.php -n hello You will have to parse $argv array to get the command line arguments. I have read the docs but can't find the syntax to pass these in. Also, how to retrieve them in the script? argv other? http://www.php.net/manual/en/features.commandline.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] php command line
I was trying to compile php command line and got some problem. Maybe some of you can help. version: 4.1.2 I did: ./configure --with-ldap --with-oracle --with-oci8 --with-mysql=/usr/src/mysql-3.23.43-pc-linux-gnu-i686 --enable-track-vars --disable-debug --prefix=/usr/local/apache/php --with-config-file-path=/usr/local/apache/lib --with-gd when tested, it gave: Failed loading /usr/local/apache/libexec/ZendOptimizer.so: /usr/local/apache/libexec/ZendOptimizer.so: cannot open shared object file: No such file or directory I couldn't find ZendOptimizer.so anywhere. Any clue? Thanks, Ezra Nugroho Web/Database Application Specialist Goshen College Information Technology Services Phone: (574) 535-7706 -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
[PHP] PHP command line scripting multiple process forking
I have been playing with this for a couple of days and have run into a few
issues:
I am wanting to run a set number of multiple processes (ie 25 at a time). I
can't use the exec command since it will not wait for the output (this is a
delayed response - network based).
$cmd = /path.to/some.cmd;
$options = --options;
$lines = file (/path.to/some.file);
while (list ($line_num, $line) = each ($lines)) {
$trimline = trim ($line);
shell_exec ($cmd $options $trimline );
}
This works but I want to fork it multiple times and have a set number of
lines running at a time. Anyone have any ideas?
Thank you,
Roy Walker
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