如何使用gulp-if退出任务

Question

我有一个处理图像的任务4。首先，它将它们最小化，如果满足条件，则应将SVG转换为PHP局部函数。如果不满足，我该如何实现这种条件并停止任务。

function processImages() {
    return src(src)
        .pipe(
            imagemin([
                imagemin.gifsicle({
                    interlaced: true,
                }),
                imagemin.jpegtran({
                    progressive: true,
                }),
                imagemin.optipng({
                    optimizationLevel: 3,
                }),
                imagemin.svgo({
                    plugins: [
                        {
                            removeViewBox: false,
                        },
                        {
                            cleanupIDs: false,
                        },
                    ],
                }),
            ]),
        )
        .pipe(dest(dest))
        // STOP AFTER THIS STEP IF CONDITION IS NOT MET
        .pipe(gulpif(!shouldCreatePartials(), exit()))
        .pipe(filter(file => /svg$/.test(file.path)))
        .pipe(
            rename({
                extname: '.php',
            }),
        )
        .pipe(dest(partialsDest));
}

Answer 1

请参阅gulp-fail-看起来是与gulp-if一起使用的。

var fail   = require('gulp-fail');
var gulpIf = require('gulp-if');

var condition = true;


// Will fail with the given message shown as the reason.
gulp.task('mytask', function() {
  return gulp.src('**/*.js')
             .pipe(gulpIf(condition, fail("It failed cause I told it to! ...Hooray?")));
})