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How do I group tasks?


How to combine gulp-watch and gulp-inject?How to pass a parameter to gulp-watch invoked taskGulp SASS, autoprefixer, minify and sourcemaps generate huge fileStart hexo to generate a watch in a gulpfile.js?gulp - chaining several gulp.src-gulp.dest expressions in one task?Gulp-sass won't compile scss files to css instead copy all files from scss folder to css folderPlumber, notify and a custom error fucntion give me an Error in plugin 'plumber': Message: Can't pipe to undefinedGulp watch with browserSyncHave to run gulp more than once to get Style changesGulp v.4 CSS Inject Using BrowserSync not Working






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;








0















I'm experimenting with gulp 4 and I have a couple of tasks(buildIndex, buildContent, buildStyling and buildScripts) that insert files in the build folder.



I'd like to group those tasks into one task, so I can execute the group at the top-level.



This is how I got it to work for now. It might not work in the future as this comment states that gulp 4 intents to execute all tasks at top-level.



const src, dest, parallel, series, watch = require('gulp');
const browserSync = require('browser-sync').create();
const clean = require('gulp-clean');
const uglify = require('gulp-uglify');
const sass = require('gulp-sass');
var rename = require('gulp-rename');

exports.default = series(
rebuild,
parallel(
host,
watchSource));

function rebuild(cb)
series(
deleteBuild,
build)();
cb();


function host(cb)
browserSync.init(
server:
baseDir: 'build'

);
watch('build/**').on('change', browserSync.reload);
cb();


function watchSource()
watch('src/**', rebuild);


function deleteBuild()
return src('build', allowEmpty: true, read: false )
.pipe(clean());


function build(cb)
parallel(
buildIndex,
buildContent,
buildStyling,
buildScripts)();
cb();


function buildIndex()
return src('src/index.html')
.pipe(dest('build'))
.pipe(browserSync.stream());


function buildContent()
return src('src/content/**/*.html')
.pipe(dest('build/content'))
.pipe(browserSync.stream());


function buildStyling()
return src('src/styling/**/*.scss')
.pipe(sass.sync().on('error', sass.logError))
.pipe(dest('build/styling'))
.pipe(browserSync.stream());


function buildScripts()
return src('src/scripts/**/*.js')
.pipe(uglify())
.pipe(rename( extname: '.min.js' ))
.pipe(dest('build/scripts', sourcemaps: true ))
.pipe(browserSync.stream());



What is the correct way to group tasks?










share|improve this question




























    0















    I'm experimenting with gulp 4 and I have a couple of tasks(buildIndex, buildContent, buildStyling and buildScripts) that insert files in the build folder.



    I'd like to group those tasks into one task, so I can execute the group at the top-level.



    This is how I got it to work for now. It might not work in the future as this comment states that gulp 4 intents to execute all tasks at top-level.



    const src, dest, parallel, series, watch = require('gulp');
    const browserSync = require('browser-sync').create();
    const clean = require('gulp-clean');
    const uglify = require('gulp-uglify');
    const sass = require('gulp-sass');
    var rename = require('gulp-rename');

    exports.default = series(
    rebuild,
    parallel(
    host,
    watchSource));

    function rebuild(cb)
    series(
    deleteBuild,
    build)();
    cb();


    function host(cb)
    browserSync.init(
    server:
    baseDir: 'build'

    );
    watch('build/**').on('change', browserSync.reload);
    cb();


    function watchSource()
    watch('src/**', rebuild);


    function deleteBuild()
    return src('build', allowEmpty: true, read: false )
    .pipe(clean());


    function build(cb)
    parallel(
    buildIndex,
    buildContent,
    buildStyling,
    buildScripts)();
    cb();


    function buildIndex()
    return src('src/index.html')
    .pipe(dest('build'))
    .pipe(browserSync.stream());


    function buildContent()
    return src('src/content/**/*.html')
    .pipe(dest('build/content'))
    .pipe(browserSync.stream());


    function buildStyling()
    return src('src/styling/**/*.scss')
    .pipe(sass.sync().on('error', sass.logError))
    .pipe(dest('build/styling'))
    .pipe(browserSync.stream());


    function buildScripts()
    return src('src/scripts/**/*.js')
    .pipe(uglify())
    .pipe(rename( extname: '.min.js' ))
    .pipe(dest('build/scripts', sourcemaps: true ))
    .pipe(browserSync.stream());



    What is the correct way to group tasks?










    share|improve this question
























      0












      0








      0








      I'm experimenting with gulp 4 and I have a couple of tasks(buildIndex, buildContent, buildStyling and buildScripts) that insert files in the build folder.



      I'd like to group those tasks into one task, so I can execute the group at the top-level.



      This is how I got it to work for now. It might not work in the future as this comment states that gulp 4 intents to execute all tasks at top-level.



      const src, dest, parallel, series, watch = require('gulp');
      const browserSync = require('browser-sync').create();
      const clean = require('gulp-clean');
      const uglify = require('gulp-uglify');
      const sass = require('gulp-sass');
      var rename = require('gulp-rename');

      exports.default = series(
      rebuild,
      parallel(
      host,
      watchSource));

      function rebuild(cb)
      series(
      deleteBuild,
      build)();
      cb();


      function host(cb)
      browserSync.init(
      server:
      baseDir: 'build'

      );
      watch('build/**').on('change', browserSync.reload);
      cb();


      function watchSource()
      watch('src/**', rebuild);


      function deleteBuild()
      return src('build', allowEmpty: true, read: false )
      .pipe(clean());


      function build(cb)
      parallel(
      buildIndex,
      buildContent,
      buildStyling,
      buildScripts)();
      cb();


      function buildIndex()
      return src('src/index.html')
      .pipe(dest('build'))
      .pipe(browserSync.stream());


      function buildContent()
      return src('src/content/**/*.html')
      .pipe(dest('build/content'))
      .pipe(browserSync.stream());


      function buildStyling()
      return src('src/styling/**/*.scss')
      .pipe(sass.sync().on('error', sass.logError))
      .pipe(dest('build/styling'))
      .pipe(browserSync.stream());


      function buildScripts()
      return src('src/scripts/**/*.js')
      .pipe(uglify())
      .pipe(rename( extname: '.min.js' ))
      .pipe(dest('build/scripts', sourcemaps: true ))
      .pipe(browserSync.stream());



      What is the correct way to group tasks?










      share|improve this question














      I'm experimenting with gulp 4 and I have a couple of tasks(buildIndex, buildContent, buildStyling and buildScripts) that insert files in the build folder.



      I'd like to group those tasks into one task, so I can execute the group at the top-level.



      This is how I got it to work for now. It might not work in the future as this comment states that gulp 4 intents to execute all tasks at top-level.



      const src, dest, parallel, series, watch = require('gulp');
      const browserSync = require('browser-sync').create();
      const clean = require('gulp-clean');
      const uglify = require('gulp-uglify');
      const sass = require('gulp-sass');
      var rename = require('gulp-rename');

      exports.default = series(
      rebuild,
      parallel(
      host,
      watchSource));

      function rebuild(cb)
      series(
      deleteBuild,
      build)();
      cb();


      function host(cb)
      browserSync.init(
      server:
      baseDir: 'build'

      );
      watch('build/**').on('change', browserSync.reload);
      cb();


      function watchSource()
      watch('src/**', rebuild);


      function deleteBuild()
      return src('build', allowEmpty: true, read: false )
      .pipe(clean());


      function build(cb)
      parallel(
      buildIndex,
      buildContent,
      buildStyling,
      buildScripts)();
      cb();


      function buildIndex()
      return src('src/index.html')
      .pipe(dest('build'))
      .pipe(browserSync.stream());


      function buildContent()
      return src('src/content/**/*.html')
      .pipe(dest('build/content'))
      .pipe(browserSync.stream());


      function buildStyling()
      return src('src/styling/**/*.scss')
      .pipe(sass.sync().on('error', sass.logError))
      .pipe(dest('build/styling'))
      .pipe(browserSync.stream());


      function buildScripts()
      return src('src/scripts/**/*.js')
      .pipe(uglify())
      .pipe(rename( extname: '.min.js' ))
      .pipe(dest('build/scripts', sourcemaps: true ))
      .pipe(browserSync.stream());



      What is the correct way to group tasks?







      gulp gulp-4






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Mar 24 at 12:19









      Janneman96Janneman96

      1191316




      1191316






















          1 Answer
          1






          active

          oldest

          votes


















          0














          Edit:



          Your already decalring the default task in export.default. If you want to add another task that runs the tasks you mentioned, you can do it like this:



          Run the tasks in parallel:



          exports.mynewtask = parallel(buildIndex, buildContent, buildStyling, buildScripts);


          Run the tasks one after another:



          exports.mynewtask = series(buildIndex, buildContent, buildStyling, buildScripts);


          This seems to be the correct way for grouping tasks as you can see in this Gulp.js documentation for parallel().



          .



          The same with old Gulp syntax would be like this:



          const gulp = require("gulp");
          ...
          gulp.task('default', gulp.parallel('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));
          // or
          gulp.task('default', gulp.series('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));





          share|improve this answer

























          • I'm doing that in the first line: const src, dest, parallel, series, watch = require('gulp');. I'm importing all the functions I need. The syntax I'm using is new in gulp 4. Here's information: gulpjs.com

            – Janneman96
            Mar 26 at 14:16











          • Sure. I edited the answer to comply with the new syntax.

            – lofihelsinki
            Mar 26 at 14:27











          • @janneman96 Does the new syntax answer your question?

            – lofihelsinki
            Mar 27 at 12:44












          Your Answer






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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0














          Edit:



          Your already decalring the default task in export.default. If you want to add another task that runs the tasks you mentioned, you can do it like this:



          Run the tasks in parallel:



          exports.mynewtask = parallel(buildIndex, buildContent, buildStyling, buildScripts);


          Run the tasks one after another:



          exports.mynewtask = series(buildIndex, buildContent, buildStyling, buildScripts);


          This seems to be the correct way for grouping tasks as you can see in this Gulp.js documentation for parallel().



          .



          The same with old Gulp syntax would be like this:



          const gulp = require("gulp");
          ...
          gulp.task('default', gulp.parallel('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));
          // or
          gulp.task('default', gulp.series('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));





          share|improve this answer

























          • I'm doing that in the first line: const src, dest, parallel, series, watch = require('gulp');. I'm importing all the functions I need. The syntax I'm using is new in gulp 4. Here's information: gulpjs.com

            – Janneman96
            Mar 26 at 14:16











          • Sure. I edited the answer to comply with the new syntax.

            – lofihelsinki
            Mar 26 at 14:27











          • @janneman96 Does the new syntax answer your question?

            – lofihelsinki
            Mar 27 at 12:44
















          0














          Edit:



          Your already decalring the default task in export.default. If you want to add another task that runs the tasks you mentioned, you can do it like this:



          Run the tasks in parallel:



          exports.mynewtask = parallel(buildIndex, buildContent, buildStyling, buildScripts);


          Run the tasks one after another:



          exports.mynewtask = series(buildIndex, buildContent, buildStyling, buildScripts);


          This seems to be the correct way for grouping tasks as you can see in this Gulp.js documentation for parallel().



          .



          The same with old Gulp syntax would be like this:



          const gulp = require("gulp");
          ...
          gulp.task('default', gulp.parallel('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));
          // or
          gulp.task('default', gulp.series('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));





          share|improve this answer

























          • I'm doing that in the first line: const src, dest, parallel, series, watch = require('gulp');. I'm importing all the functions I need. The syntax I'm using is new in gulp 4. Here's information: gulpjs.com

            – Janneman96
            Mar 26 at 14:16











          • Sure. I edited the answer to comply with the new syntax.

            – lofihelsinki
            Mar 26 at 14:27











          • @janneman96 Does the new syntax answer your question?

            – lofihelsinki
            Mar 27 at 12:44














          0












          0








          0







          Edit:



          Your already decalring the default task in export.default. If you want to add another task that runs the tasks you mentioned, you can do it like this:



          Run the tasks in parallel:



          exports.mynewtask = parallel(buildIndex, buildContent, buildStyling, buildScripts);


          Run the tasks one after another:



          exports.mynewtask = series(buildIndex, buildContent, buildStyling, buildScripts);


          This seems to be the correct way for grouping tasks as you can see in this Gulp.js documentation for parallel().



          .



          The same with old Gulp syntax would be like this:



          const gulp = require("gulp");
          ...
          gulp.task('default', gulp.parallel('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));
          // or
          gulp.task('default', gulp.series('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));





          share|improve this answer















          Edit:



          Your already decalring the default task in export.default. If you want to add another task that runs the tasks you mentioned, you can do it like this:



          Run the tasks in parallel:



          exports.mynewtask = parallel(buildIndex, buildContent, buildStyling, buildScripts);


          Run the tasks one after another:



          exports.mynewtask = series(buildIndex, buildContent, buildStyling, buildScripts);


          This seems to be the correct way for grouping tasks as you can see in this Gulp.js documentation for parallel().



          .



          The same with old Gulp syntax would be like this:



          const gulp = require("gulp");
          ...
          gulp.task('default', gulp.parallel('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));
          // or
          gulp.task('default', gulp.series('buildIndex', 'buildContent', 'buildStyling', 'buildScripts'));






          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Mar 27 at 12:49

























          answered Mar 26 at 7:43









          lofihelsinkilofihelsinki

          1,29111020




          1,29111020












          • I'm doing that in the first line: const src, dest, parallel, series, watch = require('gulp');. I'm importing all the functions I need. The syntax I'm using is new in gulp 4. Here's information: gulpjs.com

            – Janneman96
            Mar 26 at 14:16











          • Sure. I edited the answer to comply with the new syntax.

            – lofihelsinki
            Mar 26 at 14:27











          • @janneman96 Does the new syntax answer your question?

            – lofihelsinki
            Mar 27 at 12:44


















          • I'm doing that in the first line: const src, dest, parallel, series, watch = require('gulp');. I'm importing all the functions I need. The syntax I'm using is new in gulp 4. Here's information: gulpjs.com

            – Janneman96
            Mar 26 at 14:16











          • Sure. I edited the answer to comply with the new syntax.

            – lofihelsinki
            Mar 26 at 14:27











          • @janneman96 Does the new syntax answer your question?

            – lofihelsinki
            Mar 27 at 12:44

















          I'm doing that in the first line: const src, dest, parallel, series, watch = require('gulp');. I'm importing all the functions I need. The syntax I'm using is new in gulp 4. Here's information: gulpjs.com

          – Janneman96
          Mar 26 at 14:16





          I'm doing that in the first line: const src, dest, parallel, series, watch = require('gulp');. I'm importing all the functions I need. The syntax I'm using is new in gulp 4. Here's information: gulpjs.com

          – Janneman96
          Mar 26 at 14:16













          Sure. I edited the answer to comply with the new syntax.

          – lofihelsinki
          Mar 26 at 14:27





          Sure. I edited the answer to comply with the new syntax.

          – lofihelsinki
          Mar 26 at 14:27













          @janneman96 Does the new syntax answer your question?

          – lofihelsinki
          Mar 27 at 12:44






          @janneman96 Does the new syntax answer your question?

          – lofihelsinki
          Mar 27 at 12:44




















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