Question

我一直在尝试编写此查询，但由于某种原因似乎无法正确使用..

我需要做的是：

如果在过去24小时内没有与此问题相关联的更新插入qUpdateTable，请将问题的状态更改为“已关闭”。

如果工作人员至少回复过一次，我只希望关闭它。

您可以通过检查qUpdateTable并查看StaffID字段是否为空或具有特定故障单更新的值来确定工作人员或用户是否已回复该问题。如果有一个staffID，那么它已由工作人员更新，但如果没有，那么qUpdate是由用户完成的。

实际上，用户可以通过插入问题表来发布问题，并通过插入qUpdate表并使用外键链接到原始问题 - “QuestionID”来回复。

表格：

CREATE TABLE Staff
(
    ID      INTEGER NOT NULL PRIMARY KEY,
    Name        VARCHAR(40) NOT NULL
);

CREATE TABLE Customer
(
    ID      INTEGER NOT NULL PRIMARY KEY,
    Name        VARCHAR(40) NOT NULL,
    Email       VARCHAR(40) NOT NULL
);

CREATE TABLE Product
(
    ID      INTEGER NOT NULL PRIMARY KEY,
    Name    TEXT NOT NULL
);

CREATE TABLE Question
(
    ID      INTEGER NOT NULL PRIMARY KEY,
    Problem VARCHAR(1000),
    Status  VARCHAR(20) NOT NULL DEFAULT 'open', 
    Priority    INTEGER NOT NULL,
    LoggedTime  TIMESTAMP NOT NULL,
    CustomerID  INTEGER NOT NULL,
    ProductID   INTEGER NOT NULL,
    FOREIGN KEY (ProductID) REFERENCES Product(ID),
    FOREIGN KEY (CustomerID) REFERENCES Customer(ID),
    CHECK (Status IN ('open','closed') AND Priority IN (1,2,3))

);

CREATE TABLE qUpdate
(
    ID      INTEGER NOT NULL PRIMARY KEY,
    Message VARCHAR(1000) NOT NULL,
    UpdateTime  TIMESTAMP NOT NULL,
    QuestionID  INTEGER NOT NULL,
    StaffID INTEGER,
    FOREIGN KEY (StaffID) REFERENCES Staff(ID),
    FOREIGN KEY (QuestionID) REFERENCES Question(ID)
);

一些示例插页：

INSERT INTO Customer (ID, Name, Email) VALUES (1, 'testname1', 'testemail1');

INSERT INTO Customer (ID, Name, Email) VALUES (2, 'testname2', 'testemail2');

INSERT INTO Staff (ID, Name) VALUES (1, 'Don Keigh');

INSERT INTO Product (ID, Name) VALUES (1, 'Xbox');

INSERT INTO Question (ID, Problem, Status, Priority, LoggedTime, CustomerID, ProductID) 
VALUES  (1, 'testproblem1', 'open', 3, '2012-04-14 09:30', 2, 1);

INSERT INTO Question (ID, Problem, Status, Priority, LoggedTime, CustomerID, ProductID) 
VALUES  (2, 'testproblem2', 'open', 3, '2012-04-14 09:30', 2, 1);

INSERT INTO qUpdate (ID, Message, UpdateTime, StaffID, QuestionID) VALUES (2, 'testmessage1','2012-07-12 14:27', 1, 1);

INSERT INTO qUpdate (ID, Message, UpdateTime, QuestionID) VALUES (3, 'testmessage1','2012-06-18 19:42', 2);

到目前为止我做了什么（显然不起作用）

UPDATE Question
SET Status = 'closed' 
WHERE EXISTS
(SELECT qUpdate.QuestionID  
MAX(qUpdate.UpdateTime - Now() = INTERVAL '1 day') FROM qUpdate
LEFT JOIN Question ON qUpdate.QuestionID = Question.ID
WHERE qUpdate.StaffID IS NOT NULL);

我意识到我的解释可能有点令人困惑，所以如果您需要更多信息，我会尽快回复

Answer 1

UPDATE Question
SET Status = 'closed' 
where
-- this clause asserts there's at least one staff answer
exists (
  select null from qUpdate
  where qUpdate.QuestionID = Question.ID
  and StaffID is not null
)
-- this clause asserts there's been no update in the last 24 hours
and not exists (
  select null from qUpdate
  where qUpdate.QuestionID = Question.ID
  and qUpdate.UpdateTime > (now() - interval '24 hours') 
)
and Status = 'open';

您几乎肯定希望在qUpdate（QuestionID）或qUpdate（QuestionID，UpdateTime）或qUpdate（QuestionID，StaffID）上使用索引来获得exists（）测试中子选择的良好性能。

需要帮助在Postgresql中编写SQL查询

1 个答案: