Comment/Uncomment code based on condition

Question

I have been using the gulp-remove-code plugin to remove the specific code from the source files, when I need to change the environment from staging to production or vice-versa. Most of the changed code is API keys and boolean flags. This is for JS, Python, and Yaml files.

The problem with the approach is that, I need to keep the original file at a separate place, since it removes the other environment code.

Instead of removing the code based on a variable, I want to comment or uncomment code, so that I can do an in-place gulp.dest() without having a separate file.

Is there any way to do it via 'gulp' or using the 'gulp-remove-code' plugin or any other plugin?

Answer 1

I use npm module config for that purposes.

All you have to do is:

• set NODE_ENV env variable
• add config files with the same names as NODE_ENV variable
• set NODE_CONFIG_DIR env variable

This approach is very convenient.