javascript - 如何返回 promise ？

Question

我有一个函数

parseJobs(userId: string) {
    this.getLikedJobs(userId).subscribe(result => {
        result.map(key =>{
            let rows = {
                name : (key as any).jobsUser.firstName,
                jobType: 'Liked'
            }
            let job = {...rows,...(key as any).jobPosting};
            this.result.push(job);
        });
    });

    this.getSavedJobs(userId).subscribe(result => {
        result.map(key =>{
            let rows = {
                name : (key as any).jobsUser.firstName,
                jobType: 'Saved'
            }
            let job = {...rows,...(key as any).jobPosting};
            this.result.push(job);
        });
    });
    return this.result;
}

如何将结果返回给promise，我尽力了，但我不知道该怎么做，也许是因为我里面有两个可观察的，

Answer 1

您将 Promisify 两个可观察量，然后使用 Promise.all 来获得在所有操作完成后实现的 Promise:

parseJobs(userId: string) {
    // Create a promise
    const p1 = new Promise(resolve => {
        this.getLikedJobs(userId).subscribe(result => {
            // Resolve with the modified array
            resolve(result.map(key =>{
                let rows = {
                    name : (key as any).jobsUser.firstName,
                    jobType: 'Liked'
                }
                let job = {...rows,...(key as any).jobPosting};
                // In a map, you want to return:
                return job;
            }));
        });
    });
    // Same here:
    const p2 = new Promise(resolve => {
        this.getSavedJobs(userId).subscribe(result => {
            resolve(result.map(key =>{
                let rows = {
                    name : (key as any).jobsUser.firstName,
                    jobType: 'Saved'
                }
                let job = {...rows,...(key as any).jobPosting};
                return job;
            }));
        });
    });
    // Return a promise that will fulfill when both promises fulfill
    //    and concatenate the results
    return Promise.all([p1, p2]).then(result => [].concat(...result));
}

现在，您不再将结果存储在 this.result 中，而是将其设为 promise 值，如下所示:

parseJobs(1).then(result =>
    console.log(result);
});

您当然仍然可以将结果存储在 this.result 中，但这不是最佳实践，因为它表明一段代码可能会在它可用之前尝试访问它:您会始终使用 then 方法获取结果。