gulp-remove-code
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crissdev
Need function that remove comment without removing what inside.
It would be nice if you could add a function or parameters that remove the "removeIf/endRemoveIf" comments without removing what inside of them.
I use gulp-remove-code to remove some stuff from the minified version of my JS files that I don't want in production, but in the unminified version I keep those thing there, but the "removeIf/endRemoveIf" stay as well.
Would love to be able to keep them but remove the "removeIf/endRemoveIf" comments from the final build.
maybe something like .pipe(removeCode({ removeCommentsOnly: true }))
Thx.
Interesting suggestion. I'll have this implemented for the next version. Thanks.
@crissdev first, thanks for the module. Any chance of this getting implemented? You mentioned 'next version' back in 2015 and closed #11 last year (the title of which more closely matches what I'm thinking for this issue)
In the meantime, @Sugarcaen / @jfmmm how would you use gulp-strip-comments (or something else?) to only remove the empty/remaining '[rR]emoveIf lines? Seems like there's only an 'ignore' regex option there - no explicit 'remove only matches' option.
I'd love to be able to [at least somewhat] cleanly strip all 'true' removeIf sections as well as specifically the '[rR]emoveIf' comments from all 'false' removeIf sections...
Looking at it I don't see how you would do that.. It would need some way to pass a regex or something.
Don't remember how I ended up fixing that and don't have access to that project anymore sadly.
I just want to make sure I'm understanding you, @JoshuaSCochrane, its been a while since I've touched a project using gulp. You want to pipe something like:
//removeif(production)
app.doSomething()
//endremoveif(production)
app.doSomethingElse()
And get:
app.doSomething()
app.doSomethingElse()
Correct?
@Sugarcaen precisely, except ideally controlled by an additional option to the gulp removeCode part, like if you have .pipe(removeCode({ production: true }, options.removeInactiveMarkers: true )) (or however you specify options - not quite clear and no examples in the readme). Or perhaps an addition to the markers themselves, something like //removeIf(production, markerEvenIfFalse: true) \ //endRemoveIf(production) so you can control it on a marker-by-marker basis (though I can't imagine why you'd want to keep any inactive markers, maybe someone does)...
I honestly very briefly considered doing something like this (contrived example):
//removeIf(production)
//removeIf(config_A)
//endRemoveIf(production)
module.exports.configBThing = configBThing;
//removeIf(production)
//endRemoveIf(config_A)
//endRemoveIf(production)
(so it would remove configBThing if config_A but always remove the removeIfs) but then I realized how abysmally ugly that was going to get and may not even work anyway... :p
You can clean it up with gulp-delete-lines
// apply remove code
p = p.pipe(removeCode({ 'something': true }));
// remove lines starting with //removeIf and //endRemoveIf
p = p.pipe(deleteLines({'filters': [ /\/\/removeIf[^\n]*/g] }));
p = p.pipe(deleteLines({'filters': [ /\/\/endRemoveIf[^\n]*/g] }));
Note: This assumes that you use strictly //removeIf and //endRemoveIf comments with no space in-between.
Update: Using gulp-delete-lines instead of gulp-replace to remove whole lines.
I also encountered this issue/feature request. I have a feature toggle removeIf(!featureA) which correctly remove the code if the featureA is disabled, but if it's enabled I just want to keep the code without the removeIf comment being there.
The gulp-delete-lines solution sounds good, maybe it could be integrated in the core options as @JoshuaSCochrane mentioned.
I used this:
removeCode()
.pipe(gulpDeleteLines({
'filters': [
/\/\/(removeIf|endRemoveIf)[^\n]*/g,
]
}))
