Gulp from command line with error handling

First step is installing the gulp.
Install gulp
Install gulp modules

There will be node-modules directory in your present working directory (pwd).

Create a gulpfile.js in your current working directory which would include the JS functions and gulp tasks.
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Sample gulpfile.js
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function handleError(err) {
    console.log("Error: ", err.toString());
    process.exit(-1);
}

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

gulp.task('minify-css', function() {
  return gulp.src('./*.css')
    .pipe(minifyCss({compatibility: 'ie8'}))
    // handle error
    .on("error", handleError)
    .pipe(gulp.dest('CSS'));
});

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

gulp.task('minify-js', function() {
  return gulp.src('./*.js')
    .pipe(uglify())
    // handle error
    .on("error", handleError)
    .pipe(gulp.dest('JS'))
    ;
});


Open terminal in the directory where gulpfile.js is located.
Run the following command.
gulp <task-name>
Ex. gulp minify-js

Error handling

Error code returned from the gulp process will be handled in the shell script that invoked the gulp command.

On error, there will be an exit code of -1 from gulp handleError function which can be used to take corresponding action in the shell script.

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Sample shell script (powershell)
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gulp minify-js
if ($LASTEXITCODE -ne 0) {
    throw "Gulp Failed. Refer gulp log files."
    exit
}
echo ":) Gulp tasks finished successfully."