gulp-useref与几个html文件，以及如何获取包含在资产中的文件列表

Question

If I use gulp-useref with couple of html files, it process each time with same bundles.

1. Is there a way to process first incoming bundle and just replace string in other files? For example, if you look to code below - gulp-useref create vendor.js twice (at second time it rewrite previous file). If you have a lot of files and you use minification of css and js then it can take much time depending on how many files with the same bundles you use.

2. How I can copy other files, which not included to bundles?

Structure:

html-file-1.html

<!-- build:js js/vendor.js -->
<script scr="js/script-1.js"></script>
<script scr="js/script-2.js"></script>
<!-- endbuild -->
<script scr="js/file-1-script.js"></script>

html-file-2.html

<!-- build:js js/vendor.js -->
<script scr="js/script-1.js"></script>
<script scr="js/script-2.js"></script>
<!-- endbuild -->
<script scr="js/file-2-script.js"></script>

Task:

var files = [],
    getExcludedFiles = function() {
       var temp = files;
       for(var i = 0; i < temp.length; i++) {
          temp[i] = "!" + temp[i]
       }

       return temp;
    };

gulp.task("useref", function() {
   var assets = useref.assets();

   return gulp.src(path_to_htmls)
              .pipe(assests)
              .pipe(some_function_to_store_bundle_files_in_some_var)
              .pipe(gulpif('*.js', uglify()))
              .pipe(assets.restore())
              .pipe(useref())
              .pipe(gulp.dest(dets_path))
});


gulp.task("js", ["useref"], function() {
   return gulp.src([path_to_JS, getExcludedFiles()])
              .pipe(gulp.dest(JS_dets_path))
});

Answer 1

I think gulp-cached does what you need.

https://github.com/contra/gulp-cached

This keeps an in-memory cache of files (and their contents) that have passed >through it. If a file has already passed through on the last run it will not be >passed downstream. This means you only process what you need and save time + resources.