Combining two task in gulp version 4





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var gulp = require('gulp'); 

var concat = require('gulp-concat');



var autoprefix = require('gulp-autoprefixer');

var minifyCSS = require('gulp-minify-css');

var uglify = require('gulp-uglify');



gulp.task('minifyCSS', function() {

   gulp.src(['src/styles/*.css'])

   .pipe(concat('styles.css'))

   .pipe(autoprefix('last 2 versions'))

   .pipe(minifyCSS())

   .pipe(gulp.dest('build/styles/'));

});



gulp.task('compressJS', function() {

  gulp.src('src/scripts/*.js')

    .pipe(concat('all.js'))

    .pipe(uglify())

    .pipe(gulp.dest('build/scripts/'))

});



gulp.task('default', gulp.series('compressJS','minifyCSS'));


We have defined two tasks in the gulpfile.js - compressJS and minifyCSS.
Below command gives the desired output:



gulp minifyCSS compressJS


But if we give the following command:



gulp default


We get an output of the first task passed in the series, that is only the compressJS is executed in this case



Could anyone help me to combine the above 2 tasks "minifyCSS and compressJS" when i run command "gulp default"






gulp







share|improve this question


















  • 1





    Add a return statement before gulp.src in both tasks.

    – TheDancingCode
    Jan 3 at 8:06











  • thank you so much! adding return statement before gulp.src worked!

    – anuja
    Jan 3 at 9:18


















0















var gulp = require('gulp'); 

var concat = require('gulp-concat');



var autoprefix = require('gulp-autoprefixer');

var minifyCSS = require('gulp-minify-css');

var uglify = require('gulp-uglify');



gulp.task('minifyCSS', function() {

   gulp.src(['src/styles/*.css'])

   .pipe(concat('styles.css'))

   .pipe(autoprefix('last 2 versions'))

   .pipe(minifyCSS())

   .pipe(gulp.dest('build/styles/'));

});



gulp.task('compressJS', function() {

  gulp.src('src/scripts/*.js')

    .pipe(concat('all.js'))

    .pipe(uglify())

    .pipe(gulp.dest('build/scripts/'))

});



gulp.task('default', gulp.series('compressJS','minifyCSS'));


We have defined two tasks in the gulpfile.js - compressJS and minifyCSS.
Below command gives the desired output:



gulp minifyCSS compressJS


But if we give the following command:



gulp default


We get an output of the first task passed in the series, that is only the compressJS is executed in this case



Could anyone help me to combine the above 2 tasks "minifyCSS and compressJS" when i run command "gulp default"






gulp







share|improve this question


















  • 1





    Add a return statement before gulp.src in both tasks.

    – TheDancingCode
    Jan 3 at 8:06











  • thank you so much! adding return statement before gulp.src worked!

    – anuja
    Jan 3 at 9:18














0












0








0








var gulp = require('gulp'); 

var concat = require('gulp-concat');



var autoprefix = require('gulp-autoprefixer');

var minifyCSS = require('gulp-minify-css');

var uglify = require('gulp-uglify');



gulp.task('minifyCSS', function() {

   gulp.src(['src/styles/*.css'])

   .pipe(concat('styles.css'))

   .pipe(autoprefix('last 2 versions'))

   .pipe(minifyCSS())

   .pipe(gulp.dest('build/styles/'));

});



gulp.task('compressJS', function() {

  gulp.src('src/scripts/*.js')

    .pipe(concat('all.js'))

    .pipe(uglify())

    .pipe(gulp.dest('build/scripts/'))

});



gulp.task('default', gulp.series('compressJS','minifyCSS'));


We have defined two tasks in the gulpfile.js - compressJS and minifyCSS.
Below command gives the desired output:



gulp minifyCSS compressJS


But if we give the following command:



gulp default


We get an output of the first task passed in the series, that is only the compressJS is executed in this case



Could anyone help me to combine the above 2 tasks "minifyCSS and compressJS" when i run command "gulp default"






gulp







share|improve this question














var gulp = require('gulp'); 

var concat = require('gulp-concat');



var autoprefix = require('gulp-autoprefixer');

var minifyCSS = require('gulp-minify-css');

var uglify = require('gulp-uglify');



gulp.task('minifyCSS', function() {

   gulp.src(['src/styles/*.css'])

   .pipe(concat('styles.css'))

   .pipe(autoprefix('last 2 versions'))

   .pipe(minifyCSS())

   .pipe(gulp.dest('build/styles/'));

});



gulp.task('compressJS', function() {

  gulp.src('src/scripts/*.js')

    .pipe(concat('all.js'))

    .pipe(uglify())

    .pipe(gulp.dest('build/scripts/'))

});



gulp.task('default', gulp.series('compressJS','minifyCSS'));


We have defined two tasks in the gulpfile.js - compressJS and minifyCSS.
Below command gives the desired output:



gulp minifyCSS compressJS


But if we give the following command:



gulp default


We get an output of the first task passed in the series, that is only the compressJS is executed in this case



Could anyone help me to combine the above 2 tasks "minifyCSS and compressJS" when i run command "gulp default"






gulp




gulp






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Jan 3 at 7:25









anujaanuja

588




588








  • 1





    Add a return statement before gulp.src in both tasks.

    – TheDancingCode
    Jan 3 at 8:06











  • thank you so much! adding return statement before gulp.src worked!

    – anuja
    Jan 3 at 9:18














  • 1





    Add a return statement before gulp.src in both tasks.

    – TheDancingCode
    Jan 3 at 8:06











  • thank you so much! adding return statement before gulp.src worked!

    – anuja
    Jan 3 at 9:18








1




1





Add a return statement before gulp.src in both tasks.

– TheDancingCode
Jan 3 at 8:06





Add a return statement before gulp.src in both tasks.

– TheDancingCode
Jan 3 at 8:06













thank you so much! adding return statement before gulp.src worked!

– anuja
Jan 3 at 9:18





thank you so much! adding return statement before gulp.src worked!

– anuja
Jan 3 at 9:18












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