Hi,
I have a bit offtopic question: we have a problem enabling
the PHP safe mode under Solaris 2.6 even though it works
under Linux. This must be something very simple: everything
works fine, but the uid for files is always 1 under Solaris...
I tried looking into the source code, but haven't found the
reason yet. I would like to step through the source code
using the gdb. So I have recompiled the httpd like this:
cd ~/src/mod_perl-1.24_01/
perl Makefile.PL APACHE_SRC=/home/eedalf/src/apache_1.3.14 \
APACHE_PREFIX=/home/eedalf/apache PREFIX=/home/eedalf \
DO_HTTPD=1 USE_APACI=1 EVERYTHING=1 PERL_DEBUG=1
make
make test (worked fine)
make install
Then I go to:
cd ~/apache/bin
gdb ./httpd
GNU gdb 4.17
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 "sparc-sun-solaris2.6"...(no debugging symbols found)...
(gdb) source /home/eedalf/src/mod_perl-1.24_01/.gdbinit
(gdb) set args -X
(gdb) list
No symbol table is loaded. Use the "file" command.
(gdb) file ./httpd
Reading symbols from ./httpd...(no debugging symbols found)...done.
I know that PHP is not compiled in yet... I wanted to try executing
"httpd -X" step by step first - just to see if it works. What am I
doing wrong? Doesn't PERL_DEBUG=1 add the debugging symbols?
Thank you
Alex
