javascript这个顺序异步任务问题怎么解决？

Question

In the below code snippet ths tasks are an object of tasks and the function defination should be such a way that the expected output should match.

    Let tasks = {
        ‘a’: {
            job: function(finish){
                setTimeout(() => {
                    console.log(‘A done’)
                    finish();
                }, 5000)
            },
            dependencies: []
        },
        ‘b’: {
            job: function(finish){
                setTimeout(() => {
                    console.log(‘B done’)
                    finish();
                }, 2000)
            },
            dependencies: []
        },
        ‘c’: {
            job: function(finish){
                setTimeout(() => {
                    console.log(‘C done’)
                    finish();
                }, 2000)
            },
            dependencies: [‘a’, ‘b’]
        },
        ‘d’: {
            job: function(finish){
                setTimeout(() => {
                    console.log(‘D done’)
                    finish();
                }, 1000)
            },
            dependencies: []
        },
        ‘e’: {
            job: function(finish){
                setTimeout(() => {
                    console.log(‘E  done’)
                    finish();
                }, 2000)
            },
            dependencies: [‘c’, ‘b’]
        },
    };

// implement a executeTasks function which can be invoked like below

executeTasks(tasks, function(){
    console.log(‘all done’);
})


function executeTasks(taskList, callback) {
    // write your code here
}

// Expected output

D done
B done
A done
C done
E done
All done

the function executeTasks will take a tasklist as shown example on the top and also a callback. The nature of the function is that the tasks will be executed with considering all the dependencies and once all tasks are executed the passed callback should invoke. The expected output is an example how the result will be for that given task list.

Answer 1

Use rxjs , forkJoin might help you do it gracefully. https://rxjs.dev/api/index/function/forkJoin