pg-promise任务和映射具有多个相同级别的嵌套查询

Question

我正在使用节点和pg-promise创建一个基本的rest API，并且遇到某个问题需要查询特定用户的所有数据。以下是返回的数据应该是什么样子。地址，电话号码和技能都在不同的表格中。我没有任何问题检索地址或电话号码，它只是我看来无法得到的技能。不太确定如何在主要查询后获取多个查询以获取所有这些其他字段，请参阅附加的代码以供参考，我将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

我的pg-promise作者。

---

简化你的函数的版本是：

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 
    }); 

如果您正在使用bluebird的承诺库，那么你可以将这段代码：这一个

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

：

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

而不要使用像var deferred = Q.defer();这样的东西，那里不需要它。该图书馆已经基于诺言。

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