具有多个相同级别嵌套查询的 pg-promise 任务和映射

Question

我正在使用 node 和 pg-promise 创建一个基本的 rest API，并且在查询特定用户的所有数据时遇到了一些问题。下面是返回的数据应该是什么样子。地址、电话号码和技能都位于不同的表格中。我在检索地址或电话号码时没有问题，这只是我似乎无法获得的技能。不太确定如何在让用户获取所有其他字段的主查询之后进行多个查询，请参阅随附的代码以供参考，我很乐意回答任何问题。

---

{
"user_id": 1,
"first_name": "Eugene",
"last_name": "Hanson",
"display_name": "Eugene Hanson",
"email": "ehanson0@typepad.com",
"hash": "88a6aa27235d2e39dd9cb854cc246487147050f265578a3e1aee35be5db218ef",
"privilege_id": 14,
"seniority": 1,
"birthday": "19-11-1940 00:00:00.0",
"shift_count_total": 587,
"shift_count_year": 62,
"address_id": 1,
"street": "92 Schmedeman Lane",
"city": "Fort Smith",
"state": "AR",
"zip": 72905,
"phone_numbers": [
  {
    "phone_number": "62-(705)636-2916",
    "name": "PRIMARY"
  }
],
"skills": [
    "Head Audio",
    "Head Video",
    "Head Electrician",
    "Carpenter",
    "rigger"
    ]
}

---

    function getAllUsers() {
    // console.time("answer time")
    var deferred = Q.defer();
    db.task(t => {
        return t.map('SELECT * \
                        FROM users \
                        JOIN addresses \
                        ON users.address_id = addresses.address_id',[], user => {
            var user_id = user.user_id;
            // console.log(user_id)
            console.time("answer time")
            return t.manyOrNone('SELECT phone_numbers.phone_number, phone_types.name \
                                    FROM users \
                                    JOIN users_phone_numbers \
                                    ON users.user_id = users_phone_numbers.user_id \
                                    JOIN phone_numbers \
                                    ON users_phone_numbers.phone_id = phone_numbers.phone_id \
                                    JOIN phone_types \
                                    ON phone_numbers.phone_type_id = phone_types.phone_type_id \
                                    WHERE users.user_id = $1', user.user_id)
                                    .then(phone_numbers=> {
                                        // logger.log('info', phone_numbers)
                                        user.phone_numbers = phone_numbers;
                                        return user;
                                    })
        }).then(t.batch);
    })
    .then(data => {
        // console.log(data)
        console.timeEnd("answer time");
        var response = {code: "200", 
                        message: "", 
                        payload: data};
        deferred.resolve(response);
    })
    .catch(error => {
    var response = {code: error.code, 
                    message: error.message, 
                    payload: ""};
    logger.log('error', error)
    deferred.reject(response)
});

Answer 1

我是pg-promise的作者。

---

您的函数的简化版本是：

function getAllUsers() {
    return db.task(t => {
        return t.map('SELECT * FROM users', [], user => {
            return t.batch([
                t.any('SELECT * FROM phones'), // plus formatting params
                t.any('SELECT * FROM skills'), // plus formatting params
            ])
                .then(data => {
                    user.phones = data[0];
                    user.skills = data[1];
                    return user;
                });
        }).then(t.batch);
    });
}

getAllUsers()
    .then(data => {
        // data tree
    })
    .catch(error => {
        // error
    });

如果你使用bluebird作为promise库，那么你可以替换这个代码：

.then(data => {
    user.phones = data[0];
    user.skills = data[1];
    return user;
});

用这个：

.spread((phones, skills) => {
    user.phones = phones;
    user.skills = skills;
    return user;
});

不要使用var deferred = Q.defer(); 之类的东西，那里不需要它。该库已经基于 Promise。

有关高性能替代方案，请参阅：get JOIN table as array of results with PostgreSQL/NodeJS。