oracle中的timestamp是否有替代方法来显示时间？

Question

我想知道时间戳是否有替代方法来显示时间和日期？ 我真的不想使用时间戳，因为它会使我感到困惑，而且我不确定是否正确地使用了时间戳。

我的代码如下所示：

--CREATE SCRIPTS
/*put your create scripts here – your script should not commented out*/

-- this is creating a table called Project that contains 3 variables, the primary key being ProjectID
CREATE TABLE Project
(
    Proj_ID integer,
    Proj_Name varchar(10),
    Proj_Start_Date date,
    primary key (Proj_ID)
);
-- this is creating a table called Bug that has 4 variables, BugID being the primary key
CREATE TABLE Bug
(
    Bug_ID integer,
    Bug_Type varchar(20),
    Bug_Desc varchar(20),
    Bug_Time timestamp(4),
    primary key(Bug_ID)
);

-- this is creating a table called Bug_Project with 2 variables; BugID and ProjectID which combine and make a composite key
CREATE TABLE Bug_Project
(
    Bug_ID integer,
    Proj_ID integer,
    primary key(Bug_ID, Proj_ID),
    foreign key(Bug_ID) references Bug (Bug_ID),
    foreign key(Proj_ID) references  Project (Proj_ID)
);

CREATE TABLE Engineer
(
    Engineer_ID integer,
    Engineer_Name varchar(10),
    Engineer_Type varchar(20),
    primary key (Engineer_ID)
);

CREATE TABLE Fix_Allocation
(
    Engineer_ID integer,
    Bug_ID integer,
    primary key(Engineer_ID, Bug_ID),
    foreign key(Engineer_ID) references Engineer (Engineer_ID),
    foreign key(Bug_ID) references Bug (Bug_ID)
);

CREATE TABLE Test_Allocation
(
    Engineer_ID integer,
    Bug_ID integer,
    primary key(Engineer_ID, Bug_ID),
    foreign key(Engineer_ID) references Engineer (Engineer_ID),
    foreign key(Bug_ID) references Bug (Bug_ID)
);

CREATE TABLE Note
(
    Engineer_ID integer,
    Bug_ID integer,
    Note_author varchar(10),
    Note_contents varchar(20),
    primary key(Engineer_ID, Bug_ID),
    foreign key(Engineer_ID) references Engineer (Engineer_ID),
    foreign key(Bug_ID) references Bug (Bug_ID)
);

COMMIT;
--INSERT SCRIPTS
/*put your insert scripts here – your script should not commented out */

INSERT INTO Project(Proj_ID, Proj_Name, Proj_Start_Date) VALUES (00, 'Project 1', DATE '1900-02-14');
INSERT INTO Project(Proj_ID, Proj_Name, Proj_Start_Date) VALUES (01, 'Project 2', DATE '1950-12-11');
INSERT INTO Project(Proj_ID, Proj_Name, Proj_Start_Date) VALUES (02, 'Project 3', DATE '1974-07-01');
INSERT INTO Project(Proj_ID, Proj_Name, Proj_Start_Date) VALUES (03, 'Project 4', DATE '2000-07-22');
INSERT INTO Project(Proj_ID, Proj_Name, Proj_Start_Date) VALUES (04, 'Project 5', DATE '2012-03-19');

INSERT INTO Bug(Bug_ID, Bug_Type, Bug_Desc, Bug_Time) VALUES (00, 'BugType1', 'Bug Description', timestamp '1997-01-31 09:26:50.12' );
INSERT INTO Bug VALUES ();
INSERT INTO Bug VALUES ();

INSERT INTO Bug_Project VALUES ();
INSERT INTO Bug_Project VALUES ();
INSERT INTO Bug_Project VALUES ();

INSERT INTO Engineer VALUES (00, "James", "Tester");
INSERT INTO Engineer VALUES (01, "Jeff", "Fixer");
INSERT INTO Engineer VALUES (02, "Jacob", "Fixer");
INSERT INTO Engineer VALUES (03, "John", "Tester");

INSERT INTO Fix_Allocation VALUES ();
INSERT INTO Fix_Allocation VALUES ();
INSERT INTO Fix_Allocation VALUES ();

INSERT INTO Test_Allocation VALUES ();
INSERT INTO Test_Allocation VALUES ();
INSERT INTO Test_Allocation VALUES ();

INSERT INTO Note VALUES ();
INSERT INTO Note VALUES ();
INSERT INTO Note VALUES ();

COMMIT;
--SELECT SCRIPT
/*put your select scripts here (with indication of which query is answered) – your script should not commented out

-- Query 1:  List of all the bugs, and their details.
SELECT * FROM Bug;

-- Query 2: List of all bugs, and their notes.

-- Query 3: List of all bugs, with their notes, and the engineers who have written them; sorted by name of engineer.

-- Query 4: List the bugs and how much cumulative time (in hours) they have taken; ordered by time taken.

-- Query 5: The bug that has taken most time to fix, and the projects it is connected to.

COMMIT;
--DROP SCRIPT
/*put your drop scripts here (in the correct order)– your script should not commented out

DROP TABLE Note;
DROP TABLE Test_Allocation;
DROP TABLE Fix_Allocation;
DROP TABLE Engineer;
DROP TABLE Bug_Project;
DROP TABLE Bug;
DROP TABLE Project;

COMMIT;

这是第一个错误插入语句的结果，也试图摆脱所有0。

任何帮助将不胜感激！

Answer 1

DATE数据类型存储时间和日期。阅读Oracle Docs - DATE Data Type了解更多信息。