sum_distinct.result

来自「这是linux下运行的mysql软件包,可用于linux 下安装 php + m」· RESULT 代码 · 共 98 行

DROP TABLE IF EXISTS t1, t2;
CREATE TABLE t1 (
id INTEGER NOT NULL PRIMARY KEY AUTO_INCREMENT, 
gender CHAR(1),
name VARCHAR(20)
);
SELECT SUM(DISTINCT LENGTH(name)) s1 FROM t1;
s1
NULL
INSERT INTO t1 (gender, name) VALUES (NULL, NULL);
INSERT INTO t1 (gender, name) VALUES (NULL, NULL);
INSERT INTO t1 (gender, name) VALUES (NULL, NULL);
SELECT SUM(DISTINCT LENGTH(name)) s1 FROM t1;
s1
NULL
INSERT INTO t1 (gender, name) VALUES ('F', 'Helen'), ('F', 'Anastasia'),
('F', 'Katherine'), ('F', 'Margo'), ('F', 'Magdalene'), ('F', 'Mary');
CREATE TABLE t2 SELECT name FROM t1;
SELECT (SELECT SUM(DISTINCT LENGTH(name)) FROM t1) FROM t2;
(SELECT SUM(DISTINCT LENGTH(name)) FROM t1)
18
18
18
18
18
18
18
18
18
DROP TABLE t2;
INSERT INTO t1 (gender, name) VALUES ('F', 'Eva'), ('F', 'Sofia'),
('F', 'Sara'), ('F', 'Golda'), ('F', 'Toba'), ('F', 'Victory'),
('F', 'Faina'), ('F', 'Miriam'), ('F', 'Beki'), ('F', 'America'),
('F', 'Susan'), ('F', 'Glory'), ('F', 'Priscilla'), ('F', 'Rosmary'),
('F', 'Rose'), ('F', 'Margareth'), ('F', 'Elizabeth'), ('F', 'Meredith'),
('F', 'Julie'), ('F', 'Xenia'), ('F', 'Zena'), ('F', 'Olga'),
('F', 'Brunhilda'), ('F', 'Nataly'), ('F', 'Lara'), ('F', 'Svetlana'),
('F', 'Grethem'), ('F', 'Irene');
SELECT
SUM(DISTINCT LENGTH(name)) s1,
SUM(DISTINCT SUBSTRING(NAME, 1, 3)) s2,
SUM(DISTINCT LENGTH(SUBSTRING(name, 1, 4))) s3
FROM t1;
s1	s2	s3
42	0	7
SELECT
SUM(DISTINCT LENGTH(g1.name)) s1,
SUM(DISTINCT SUBSTRING(g2.name, 1, 3)) s2,
SUM(DISTINCT LENGTH(SUBSTRING(g3.name, 1, 4))) s3
FROM t1 g1, t1 g2, t1 g3;
s1	s2	s3
42	0	7
SELECT
SUM(DISTINCT LENGTH(g1.name)) s1,
SUM(DISTINCT SUBSTRING(g2.name, 1, 3)) s2,
SUM(DISTINCT LENGTH(SUBSTRING(g3.name, 1, 4))) s3
FROM t1 g1, t1 g2, t1 g3 GROUP BY LENGTH(SUBSTRING(g3.name, 5, 10));
s1	s2	s3
42	0	NULL
42	0	7
42	0	4
42	0	4
42	0	4
42	0	4
42	0	4
SELECT SQL_BUFFER_RESULT
SUM(DISTINCT LENGTH(name)) s1,
SUM(DISTINCT SUBSTRING(NAME, 1, 3)) s2,
SUM(DISTINCT LENGTH(SUBSTRING(name, 1, 4))) s3
FROM t1;
s1	s2	s3
42	0	7
SELECT SQL_BUFFER_RESULT
SUM(DISTINCT LENGTH(g1.name)) s1,
SUM(DISTINCT SUBSTRING(g2.name, 1, 3)) s2,
SUM(DISTINCT LENGTH(SUBSTRING(g3.name, 1, 4))) s3
FROM t1 g1, t1 g2, t1 g3 GROUP BY LENGTH(SUBSTRING(g3.name, 5, 10));
s1	s2	s3
42	0	NULL
42	0	7
42	0	4
42	0	4
42	0	4
42	0	4
42	0	4
SET @l=1;
UPDATE t1 SET name=CONCAT(name, @l:=@l+1);
SELECT SUM(DISTINCT RIGHT(name, 1)) FROM t1;
SUM(DISTINCT RIGHT(name, 1))
45
SELECT SUM(DISTINCT id) FROM t1;
SUM(DISTINCT id)
703
SELECT SUM(DISTINCT id % 11) FROM t1;
SUM(DISTINCT id % 11)
55
DROP TABLE t1;