I want to make gulp work with files without extension also. I tried to do it the following ways:
function someTask() {
return gulp.src('./src/**/*.{html,,xml,txt}')
.pipe(gulp.dest('./build'));
}
function someTask() {
return gulp.src('./src/**/*.{html,!*,xml,txt}')
.pipe(gulp.dest('./build'));
}
function someTask() {
return gulp.src('./src/**/*.{html,[^*],xml,txt}')
.pipe(gulp.dest('./build'));
}
but I haven't got any results. Any suggestions?
I am still trying to find a better answer, but one route is to get all the files and then eliminate the extentions you don't want. Like
var debug = require('gulp-debug'); // nice, will list all the files in the stream
return gulp.src(['./src/**/*', '!./src/**/*.{js,png}'])
.pipe(debug())
.pipe(gulp.dest('./build'));
So here {js,png}
list the possible file extensions in your src
directories you don't want.
Hopefully there is a better answer.
[EDIT]: Maybe there is, try this:
return gulp.src(['./src/**/*.{html,xml,txt}', './src/**/!(*\.*)'])
This part ./src/**/!(*\.*)
appears to match only files with no .
in them, i.e., no extention.