Gulp, using current filename in `pipe()` function
I'm trying to render each file in my gulp source files with it's own json file, but I can't figure out how to access the current filename in the pipe function.
var gulp = require('gulp');
var handlebars = require('handlebars');
var gulpHandlebars = require('gulp-compile-handlebars');
gulp.task('compile-with-sample-data', function () {
var options = {}
return gulp.src('./src/**/*.html')
.pipe(gulpHandlebars({ data: require('./data/' + filename +'.json') }, options))
.pipe(gulp.dest('./build/'));
});
Where I can get it to work with the same file each time by just using require('./data/orders-complete.json') :
var gulp = require('gulp');
var handlebars = require('handlebars');
var gulpHandlebars = require('gulp-compile-handlebars');
gulp.task('compile-with-sample-data', function () {
var options = {}
return gulp.src('./src/**/*.html')
.pipe(gulpHandlebars({ data: require('./data/orders-complete.json') }, options))
.pipe(gulp.dest('./build/'));
});
It's not clear how I would do this.
使用gulp-tap，它可以让你获得文件名，甚至可以改变下游管道。
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