Question

我想用gulp写一个监视任务来观察我的服务器php文件。但我不会看整个供应商的目录。现在我的代码看起来像这样：

var paths = {
    src: [
        'src/**/*.html',
        'src/**/*.js',
        'src/**/*.css',
        'src/server/**/*',
        'src/server/**/.*'

    ],
    dist: 'dist',
    base: 'src',
    watchSrc: [
        'src/**/*.html',
        'src/**/*.css',
        '!src/server/vendor/**/*',
        'src/server/**/*.php',
        'src/server/**/.*'

    ]
};

gulp.task('copy:src', function() {
    gulp.src('node_modules').pipe(symlink(paths.dist + '/node_modules', {force: true}));
    gulp.src('src/server/vendor').pipe(symlink(paths.dist + '/server/vendor', {force: true}));
    return gulp.src(paths.src, {base: paths.base}).pipe(gulp.dest(paths.dist));
});

gulp.task('clean:dist', function(cb) {
    rimraf(paths.dist, cb);
});

gulp.task('watch', function() {
    runSequence('clean:dist', 'copy:src');
    return watch(paths.src, {verbose: true, base: paths.base}).pipe(gulp.dest(paths.dist))
});

但这对我不起作用。也许你知道如何处理这个？

Answer 1

好的，我已经知道有一个复制和粘贴错误，它使用错误的路径调用监视任务：

时：

gulp.task('watch', function() {
    runSequence('clean:dist', 'copy:src');
    return watch(paths.src, {verbose: true, base: paths.base}).pipe(gulp.dest(paths.dist))
});

已改为：

gulp.task('watch', function() {
        runSequence('clean:dist', 'copy:src');
        return watch(paths.watchSrc, {verbose: true, base: paths.base}).pipe(gulp.dest(paths.dist))
    });

现在似乎有效了

在gulp手表中忽略forlder

1 个答案: