如何获取基本文件名并将其作为变量传递给gulp中的管道？

Question

I'm trying to grab the base name of each html file that passes through a stream. The code here obviously doesn't work because _base is no longer in scope. It seems like there should be a really easy way to do this, or maybe there's something built into gulp-ejs that I don't know about. Thoughts?

gulp.task('html', function () {
    return gulp.src('app/*.html')
    .pipe(tap(function(file, t){
        var _base = path.basename(file.path, '.html');
    })
    .pipe($.ejs(require('./app/'+_base+'.json')))
});

Answer 1

Can't you just declare _base above the return statement?

gulp.task('html', function () {
    var _base;
    return gulp.src('app/*.html')
    .pipe(tap(function(file, t){
        _base = path.basename(file.path, '.html');
    })
    .pipe($.ejs(require('./app/'+_base+'.json')))
});

Answer 2

You cannot access the filename like that because you are out of the scope of the running stream. What you need to do is use the filename in your tap or use through2. What are you trying to achieve specifically?