I have compiled SQLite3 3.8.6 and installed it to ${HOME}/opt with:
LDFLAGS="-L${HOME}/opt/lib" CFLAGS="-L${HOME}/opt/include" ./configure --prefix=$HOME/opt
make && make install
I am now trying to compile Python 3.4.2 to use this version instead of the version installed for the entire system. I do not have root access on this system. To compile Python, I am using:
LDFLAGS="-L${HOME}/opt/lib" CFLAGS="-L${HOME}/opt/include" ./configure --prefix=$HOME/opt
make && make install
I was able to compile Python 3.3.5 with my newer version if SQLite3, but these same steps don't seem to work for me for 3.4.2.
How can I compile Python 3.4.2 to include my version of SQLite 3.8.6 which is located in ${HOME}/opt?
Thanks.
EDIT: It compiles & installs OK except for the fact that is using the older, system version of sqlite3 instead of the version that I compiled & installed myself.
There is also the option of pre-linking your custom Python build with your own-built sqlite3. (I had the same issue: the custom python was using the system-provided sqlite3, completely ignoring the sqlite3 I built).
Prefix your configure
and make
commands with:
LD_RUN_PATH=$HOME/opt/lib configure LDFLAGS="-L$HOME/opt/lib" CPPFLAGS="-I$HOME/opt/include" …
LD_RUN_PATH=$HOME/opt/lib make
so that the built python3
by default is linked to your sqlite3.
This worked for me.
import platform,sqlite3
print("Oper Sys : %s %s" % (platform.system(), platform.release()))
print("Platform : %s %s" % (platform.python_implementation(),platform.python_version()))
print("SQLite : %s" % (sqlite3.sqlite_version))
When I run this code, the output contains the system's version of sqlite3:
Oper Sys : Linux 3.2.0-4-amd64
Platform : CPython 3.4.2
SQLite : 3.7.13
After installing sqlite v3.8.6 under ${HOME}/opt{include,lib} and setting this in my .bashrc:
export LD_LIBRARY_PATH="${HOME}/opt/lib"
I get my desired result:
Oper Sys : Linux 3.2.0-4-amd64
Platform : CPython 3.4.2
SQLite : 3.8.6
Notice the SQLite version changes from 3.7.13
to 3.8.6