gulp 文件中只调用了最后一个回调

Question

在gulpfile中，我有3个任务，当我运行下面的代码时，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[=24 的回调=] & 默认任务。

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

谁能告诉我哪里错了？

Answer 1

为了让您的代码按照您描述的方式工作，回调函数需要作为参数传递给 .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)));

希望对您有所帮助:)

补充说明：

gulp/vars 的命名约定通常是驼峰式命名，例如：gulp 和 typeScriptProject 而不是 Gulp 或 TypeScriptProject。
您可以完全消除使用 gulp. 的需要：var {gulp, task, src, dest, watch, series, parallel} = require('gulp');
与其直接定义任务，不如使用 CommonJS exports 模块符号来声明任务，从而使代码更易于阅读。
如果您在引用时保持一致，而不是混合使用单打和双打，那么生活会更轻松一些。两者都允许通配

遵循 Gulp 自己的文档也许是开始的地方，their sample code on github 有一些设置基本 gulpfile 的很好的例子。

如果你把所有这些都包起来，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;

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