gulp add folder in zip archive












0
















gulp.task( 'zip', function () {
return gulp.src( './dist/**' )
.pipe(zip( 'dist.zip' ))
.pipe(gulp.dest( './' ));
});


I have next structure



dist.zip
all-files


How do I make the following structure in the archive?



dist.zip
dist
all-files









share|improve this question





























    0
















    gulp.task( 'zip', function () {
    return gulp.src( './dist/**' )
    .pipe(zip( 'dist.zip' ))
    .pipe(gulp.dest( './' ));
    });


    I have next structure



    dist.zip
    all-files


    How do I make the following structure in the archive?



    dist.zip
    dist
    all-files









    share|improve this question



























      0












      0








      0









      gulp.task( 'zip', function () {
      return gulp.src( './dist/**' )
      .pipe(zip( 'dist.zip' ))
      .pipe(gulp.dest( './' ));
      });


      I have next structure



      dist.zip
      all-files


      How do I make the following structure in the archive?



      dist.zip
      dist
      all-files









      share|improve this question

















      gulp.task( 'zip', function () {
      return gulp.src( './dist/**' )
      .pipe(zip( 'dist.zip' ))
      .pipe(gulp.dest( './' ));
      });


      I have next structure



      dist.zip
      all-files


      How do I make the following structure in the archive?



      dist.zip
      dist
      all-files






      gulp zip






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 13 at 8:18

























      asked Nov 12 at 13:31









      Max

      12




      12
























          1 Answer
          1






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          0














          Your code is fine except for the gulp.dest which takes a string (or a function):



          gulp.task( 'zip', function () {
          return gulp.src( './dist/**' )
          .pipe(zip( 'dist.zip' ))

          // .pipe(gulp.dest( ./ )); your code, note the apostrophes below

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

          });





          share|improve this answer





















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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            0














            Your code is fine except for the gulp.dest which takes a string (or a function):



            gulp.task( 'zip', function () {
            return gulp.src( './dist/**' )
            .pipe(zip( 'dist.zip' ))

            // .pipe(gulp.dest( ./ )); your code, note the apostrophes below

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

            });





            share|improve this answer


























              0














              Your code is fine except for the gulp.dest which takes a string (or a function):



              gulp.task( 'zip', function () {
              return gulp.src( './dist/**' )
              .pipe(zip( 'dist.zip' ))

              // .pipe(gulp.dest( ./ )); your code, note the apostrophes below

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

              });





              share|improve this answer
























                0












                0








                0






                Your code is fine except for the gulp.dest which takes a string (or a function):



                gulp.task( 'zip', function () {
                return gulp.src( './dist/**' )
                .pipe(zip( 'dist.zip' ))

                // .pipe(gulp.dest( ./ )); your code, note the apostrophes below

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

                });





                share|improve this answer












                Your code is fine except for the gulp.dest which takes a string (or a function):



                gulp.task( 'zip', function () {
                return gulp.src( './dist/**' )
                .pipe(zip( 'dist.zip' ))

                // .pipe(gulp.dest( ./ )); your code, note the apostrophes below

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

                });






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 12 at 15:10









                Mark

                10.8k33148




                10.8k33148






























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