在 gulp 文件中只调用最后一个回调

Question

在 gulpfile 中，我有 3 个任务，当我运行以下代码时，它只执行最后一个任务的回调。我希望如果我运行gulp 命令，在完成 clean 任务后，它应该执行 copy:db & default 任务的回调.

Gulp.task('clean', function () {
    console.log('Clean');
    return Gulp.src("./dbSchema/*")
        .pipe(VinylPaths(Del));
})

Gulp.task('copy:db', Gulp.series("clean"), function () {
    console.log('Copy DB');
    return Gulp.src("./db/*")
        .pipe(Gulp.dest("./dbSchema"));
})

Gulp.task('default', Gulp.series("copy:db"), function () {
    console.log('defaulp');
    return TypeScriptProject.src()
        .pipe(TypeScriptProject())
        .js.pipe(Gulp.dest('dist'));
});

当我运行命令gulp 时，它会显示以下日志。

[12:46:37] Starting 'default'...
[12:46:37] Starting 'copy:db'...
[12:46:37] Starting 'clean'...
Clean
[12:46:37] Finished 'clean' after 26 ms
[12:46:37] Finished 'copy:db' after 28 ms
[12:46:37] Finished 'default' after 31 ms

谁能告诉我哪里出错了？

Answer 1

要让您的代码按照您描述的方式运行，需要将回调函数作为参数传递给.series()。例如：

Gulp.task('clean', function () {
    console.log('Clean');
    return Gulp.src("./dbSchema/*")
        .pipe(VinylPaths(Del));
})

Gulp.task('copy:db', Gulp.series(clean, function () {
    console.log('Copy DB');
    return Gulp.src("./db/*")
        .pipe(Gulp.dest("./dbSchema"));
}))

Gulp.task('default', Gulp.series(copy:db, function () {
    console.log('defaulp');
    return TypeScriptProject.src()
        .pipe(TypeScriptProject())
        .js.pipe(Gulp.dest('dist'));
}));

---

恕我直言，拥有三个完全独立的任务会更简单：

Gulp.task('clean', function () {
    console.log('Clean');
    return Gulp.src("./dbSchema/*")
        .pipe(VinylPaths(Del));
});

Gulp.task('copy:db', function () {
    console.log('Copy DB');
    return Gulp.src("./db/*")
        .pipe(Gulp.dest("./dbSchema"));
});

Gulp.task('default', function () {
    console.log('defaulp');
    return TypeScriptProject.src()
        .pipe(TypeScriptProject())
        .js.pipe(Gulp.dest('dist'));
});

然后调用它们：

Gulp.task('default', gulp.series(clean, copy:db, js));

或

Gulp.task('default', gulp.series(clean, gulp.parallel(copy:db, js)));

希望有帮助:)

---

补充说明：

1. gulp/vars 的命名约定通常是驼峰式，例如：gulp 和 typeScriptProject 而不是 Gulp 或 TypeScriptProject。

2. 您可以通过使用：var {gulp, task, src, dest, watch, series, parallel} = require('gulp');

完全消除编写 gulp. 的需要

3. 您可以使用 CommonJS exports 模块符号来声明任务，而不是直接定义您的任务，从而使您的代码更易于阅读。

4. 如果您在使用引号时保持一致，而不是混合单打和双打，那么生活会更轻松一些。两者都允许通配符

---

按照 Gulp 自己的文档可能是开始的地方，their sample code on github 有一些很好的例子来设置基本的 gulpfile。

如果你把这一切都总结起来，你会得到这个：

/*
 * Example of requires with gulp methods also requiring gulp.
 */
var { 
    gulp,
    dest,
    series,
    parallel,
    src,
    task,
    watch 
}           = require('gulp'),
vinylPaths  = require('vinyl-paths'), // may not be required, see note in clean func.
del         = require('del'),
ts          = require('gulp-typescript');

/*
 * Added a basic TypeScript Project so the example is complete and could run.
 */
var typeScriptProject = ts.createProject({
    declaration: true
});

/*
 * Your tasks converted to plain/standard js functions.
 */
function clean () {
    return src('dbSchema/*')
        .pipe(vinylPaths(del));

    // Looking at your example code the vinylPaths is redundant,
    // as long as you’re using del ^2.0 it will return its promise,
    // so you could replace the above with:
    return del([ 'dbSchema' ]);
}

function copyDb () {
    return src('db/*')
        .pipe(dest('dbSchema'));
}

function scripts () {
    // example src path
    return src('lib/*.ts')
        .pipe(typeScriptProject())
        .pipe(dest('dist'));
}

/*
 * By defining all the tasks separately it makes it really clear how the tasks will run.
 */
var build = gulp.series (
    clean, 
    gulp.parallel (
        copyDb,
        scripts
    )
);

/*
 * Example of using `exports` module notation to declare tasks.
 */
exports.clean = clean;
exports.copyDb = copyDb;
exports.scripts = scripts;
exports.build = build;

/*
 * Good practise to define the default task as a reference to another task.
 */
exports.default = build;