我从MySQL切换到PostgreSQL,我想知道我如何能有一个INT列与自动递增。我在PostgreSQL文档中看到了一个名为SERIAL的数据类型,但我在使用它时遇到了语法错误。


当前回答

如果你想在已经存在的表中添加sequence到id,你可以使用:

CREATE SEQUENCE user_id_seq;
ALTER TABLE user ALTER user_id SET DEFAULT NEXTVAL('user_id_seq');

其他回答

如果你想在已经存在的表中添加sequence到id,你可以使用:

CREATE SEQUENCE user_id_seq;
ALTER TABLE user ALTER user_id SET DEFAULT NEXTVAL('user_id_seq');

你必须小心不要直接插入到你的SERIAL或sequence字段中,否则当序列达到插入值时,你的写入将失败:

-- Table: "test"

-- DROP TABLE test;

CREATE TABLE test
(
  "ID" SERIAL,
  "Rank" integer NOT NULL,
  "GermanHeadword" "text" [] NOT NULL,
  "PartOfSpeech" "text" NOT NULL,
  "ExampleSentence" "text" NOT NULL,
  "EnglishGloss" "text"[] NOT NULL,
  CONSTRAINT "PKey" PRIMARY KEY ("ID", "Rank")
)
WITH (
  OIDS=FALSE
);
-- ALTER TABLE test OWNER TO postgres;
 INSERT INTO test("Rank", "GermanHeadword", "PartOfSpeech", "ExampleSentence", "EnglishGloss")
           VALUES (1, '{"der", "die", "das", "den", "dem", "des"}', 'art', 'Der Mann küsst die Frau und das Kind schaut zu', '{"the", "of the" }');


 INSERT INTO test("ID", "Rank", "GermanHeadword", "PartOfSpeech", "ExampleSentence", "EnglishGloss")
           VALUES (2, 1, '{"der", "die", "das"}', 'pron', 'Das ist mein Fahrrad', '{"that", "those"}');

 INSERT INTO test("Rank", "GermanHeadword", "PartOfSpeech", "ExampleSentence", "EnglishGloss")
           VALUES (1, '{"der", "die", "das"}', 'pron', 'Die Frau, die nebenen wohnt, heißt Renate', '{"that", "who"}');

SELECT * from test; 

自从PostgreSQL 10

CREATE TABLE test_new (
    id int GENERATED BY DEFAULT AS IDENTITY PRIMARY KEY,
    payload text
);

这种方法肯定有效,我希望它能有所帮助:

CREATE TABLE fruits(
   id SERIAL PRIMARY KEY,
   name VARCHAR NOT NULL
);

INSERT INTO fruits(id,name) VALUES(DEFAULT,'apple');

or

INSERT INTO fruits VALUES(DEFAULT,'apple');

你可以在下一个链接中查看详细信息: http://www.postgresqltutorial.com/postgresql-serial/

您可以使用任何其他整数数据类型,例如smallint。

例子:

CREATE SEQUENCE user_id_seq;
CREATE TABLE user (
    user_id smallint NOT NULL DEFAULT nextval('user_id_seq')
);
ALTER SEQUENCE user_id_seq OWNED BY user.user_id;

最好使用您自己的数据类型,而不是用户串行数据类型。