如何在执行Gulp的过程中运行功能,以便Gulp将功能视为任务(将其打印到日志中)?
这是我的gulpfile.js的示例:
async function doStaff(todo) {
// ...
}
exports.default = async () => {
// Some asynchronous logic which gets the subtasks
const todos = await myAPI.getTodos();
await Promise.all(todos.map(async todo => {
const subTask = doStuff.bind(todo);
subTask.displayName = `todo «${todo.name}»`;
await subTask(); // I want it to run as a Gulp task
}));
};
预期的Gulp输出:
Starting 'default'...
Starting 'todo «foo»'...
Starting 'todo «bar»'...
Finished 'todo «foo»' after 1.2s
Finished 'todo «bar»' after 1.5s
Finished 'default' after 2s
实际Gulp输出:
Starting 'default'...
Finished 'default' after 2s
为什么:我的子任务列表无法同步获取,并且Gulp doesn't support asynchronous configuration也将不支持它。但是我仍然希望在Gulp输出中看到子任务的进度和计时。
答案 0 :(得分:0)
将该函数包装到gulp.series
或gulp.parallel
中,以使Gulp显示其状态。看起来这不是一个优雅的解决方案,但这是我发现的最好的解决方案。
这是一个通用示例:
function runGulpTask(func) {
return new Promise((resolve, reject) => gulp.series(func)(error => {
if (error) {
reject(error);
} else {
resolve();
}
}));
// The same solution using built-in Node.js functions:
// const {promisify} = require('util');
// return promisify(gulp.series(func))();
}
async function doStaff(todo) {
// ...
}
exports.default = async () => {
const todos = await myAPI.getTodos();
await Promise.all(todos.map(async todo => {
const subTask = doStuff.bind(todo);
subTask.displayName = `todo «${todo.name}»`;
await runGulpTask(subTask);
}));
};
Gulp输出与预期输出相同。
这是一个更具体的示例:
async function doStaff(todo) {
// ...
}
exports.default = async () => {
const todos = await myAPI.getTodos();
const subTasks = todos.map(todo => {
const subTask = doStuff.bind(todo);
subTask.displayName = `todo «${todo.name}»`;
return subTask;
});
await new Promise((resolve, reject) => gulp.parallel(...subTasks)(error => {
if (error) {
reject(error);
} else {
resolve();
}
}));
// Or using built-in Node.js functions:
// const {promisify} = require('util');
// await promisify(gulp.parallel(...subTasks))();
};