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。以下是返回的数据应该是什么样子。地址，电话号码和技能都存在于单独的表中。我没有问题检索地址或电话号码，这是我似乎无法获得的技能。在主要查询获得用户获取所有其他字段后，不太确定如何进行多次查询，请参阅附带的代码以供参考，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"
    ]
}

＆＃13;

    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)
});

＆＃13;

Answer 1

我是pg-promise的作者。

您的功能的简化版本将是：

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
    });

如果您使用bluebird作为promise库，那么您可以替换此代码：

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

这一个：

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

不要使用像var deferred = Q.defer();这样的东西，那里不需要它。图书馆已经以承诺为基础。

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

pg-promise任务和具有多个相同级别嵌套查询的映射

1 个答案: