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.
When I run this code, the output contains the system's version of sqlite3:
After installing sqlite v3.8.6 under ${HOME}/opt{include,lib} and setting this in my .bashrc:
I get my desired result:
Notice the SQLite version changes from
3.7.13
to3.8.6
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
andmake
commands with:so that the built
python3
by default is linked to your sqlite3. This worked for me.