Re: [GENERAL] Installing module for 9.6, not 9.2, on Centos? - Mailing list pgsql-general
On Tue, May 23, 2017 at 5:34 PM, Adrian Klaver <adrian.klaver@aklaver.com> wrote:
On 05/23/2017 05:27 PM, Ken Tanzer wrote:
But the install still goes to 9.2:
PATH=/usr/local/pgsql96/bin/:$PATH make USE_PGXS=1 install
Did you do:
PATH=/usr/local/pgsql96/bin/:$PATH make USE_PGXS=1 clean all
I did.
Or is there some environment variable set that could be interfering?
I didn't see anything that would apply. I've attached a dump of environment variables and also a re-run of the compile/install process.
Try without the trailing:
PATH=/usr/local/pgsql96/bin:$PATH make USE_PGXS=1 clean all
Sorry, not sure I follow what is the trailing part here?
Does the compile stage specify where a module is to be installed? Or is it determined at install time?
Apart from the extra 96/bin directories I'm accumulating, I don't see anything suspicious in my path. (And my /usrl/ocal/(s)bin directories are empty.)
[root@hosting table_log-0.4.4]# PATH=/usr/local/pgsql96/bin:$PATH ; echo "Path is: $PATH" ; make USE_PGXS=1 clean all
Path is: /usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin:/root/bin
rm -f table_log.so table_log.o
...
[root@hosting table_log-0.4.4]# PATH=/usr/local/pgsql96/bin:$PATH ; echo "Path is: $PATH" ; make USE_PGXS=1 install
Path is: /usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/pgsql96/bin:/usr/local/sbin:/usr/local/bin:/sbin:/bin:/usr/sbin:/usr/bin:/root/bin
/bin/mkdir -p '/usr/pgsql-9.2/share/contrib'
/bin/mkdir -p '/usr/pgsql-9.2/lib'
/bin/mkdir -p '/usr/pgsql-9.2/doc/contrib'
/usr/bin/install -c -m 644 table_log.sql '/usr/pgsql-9.2/share/contrib/'
/usr/bin/install -c -m 755 table_log.so '/usr/pgsql-9.2/lib/'
/usr/bin/install -c -m 644 .//README.table_log '/usr/pgsql-9.2/doc/contrib/'
Thanks,
Ken
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