I need help figuring out how to remove all lines of code that start with #include. Here's an example:
#include 'utils/namespace.js'
#include 'utils/constants.js'
#include 'utils/helpers.js'
#include 'utils/json2.js'
#include 'utils/polifill.js'
If you have any suggestions, please let me know. Thank you!
I'm not entirely sure if this solution will be effective, but have you considered trying out code similar to this?
var replace = require('gulp-replace');
gulp.task('templates', function(){
gulp.src(['file.txt'])
.pipe(replace(/#include[^\n]*/g, ''))
.pipe(gulp.dest('build/file.txt'));
});
Sure, using gulp-replace like Ipg suggested is a great approach for this task.
Alternatively, you can take a more "hands-on" approach by utilizing through2 like so:
var gulp = require("gulp");
var through = require("through2");
gulp.task("processFiles", function() {
return gulp
.src("src/data.txt")
.pipe(through.obj(function(file, encoding, callback) {
var updatedContent = file.contents.toString().replace(/#include[^\n]*/g, "");
file.contents = new Buffer.from(updatedContent);
callback(null, file);
}))
.pipe(gulp.dest("output"));
});
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