用 sqlite3 编译 Python 3.4 [英] Compile Python 3.4 with sqlite3

问题描述

我已经编译了 SQLite3 3.8.6 并将其安装到 ${HOME}/opt 中:

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

我现在正在尝试编译 Python 3.4.2 以使用此版本而不是为整个系统安装的版本.我在这个系统上没有 root 访问权限.要编译 Python，我使用:

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

如果使用 SQLite3，我可以用我的较新版本编译 Python 3.3.5，但这些相同的步骤似乎不适用于 3.4.2.

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.

如何编译 Python 3.4.2 以包含我位于 ${HOME}/opt 中的 SQLite 3.8.6 版本?

How can I compile Python 3.4.2 to include my version of SQLite 3.8.6 which is located in ${HOME}/opt?

谢谢.

它编译&安装正常，除了使用旧的系统版本的 sqlite3 而不是我编译的版本 &自己安装.

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.

推荐答案

还可以选择将自定义 Python 构建与您自己构建的 sqlite3 预先链接.(我遇到了同样的问题:自定义 python 使用的是系统提供的 sqlite3，完全忽略了我构建的 sqlite3).

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).

在 configure 和 make 命令前加上:

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

这样内置的 python3 默认链接到你的 sqlite3.这对我有用.

so that the built python3 by default is linked to your sqlite3. This worked for me.

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