Question

我需要编写一个运行相同命令但需要更改LIKE字段的过程。它将从'aa％'开始，然后一直到'ab％'到'az％'。那么它将以'ba％'，'bb％'开始，依此类推，直到'zz％'。我尝试在循环中运行它，但不知道从哪里获取字母。

SELECT t1.colA, t2.colA, count(*)
FROM (SELECT colA FROM table WHERE colA LIKE "a[a-z]%") t1
JOIN table t2
ON t1.date = t2.date
GROUP BY t1.colA, t2.colA
HAVING count(*) > 10
INSERT INTO /tmp/[a[a-z]].csv
FIELDS TERMINATED BY ',' ENCLOSED BY '"'
LINES TERMINATED BY '\r\n'

经过26次迭代后，下一次将是

 SELECT t1.colA, t2.colA, count(*)
    FROM (SELECT colA FROM table WHERE colA LIKE "b[a-z]%") t1
    JOIN table t2
    ON t1.date = t2.date
    GROUP BY t1.colA, t2.colA
    HAVING count(*) > 10
    INSERT INTO /tmp/[b[a-z]].csv
    FIELDS TERMINATED BY ',' ENCLOSED BY '"'
    LINES TERMINATED BY '\r\n'

接下来将是

SELECT t1.colA, t2.colA, count(*)
FROM (SELECT colA FROM table WHERE colA LIKE "c[a-z]%") t1
JOIN table t2
ON t1.date = t2.date
GROUP BY t1.colA, t2.colA
HAVING count(*) > 10
INSERT INTO /tmp/[c[a-z]].csv
FIELDS TERMINATED BY ',' ENCLOSED BY '"'
LINES TERMINATED BY '\r\n'

等等，直到

SELECT t1.colA, t2.colA, count(*)
FROM (SELECT colA FROM table WHERE colA LIKE "z[a-z]%") t1
JOIN table t2
ON t1.date = t2.date
GROUP BY t1.colA, t2.colA
HAVING count(*) > 10
INSERT INTO /tmp/[z[a-z]].csv
FIELDS TERMINATED BY ',' ENCLOSED BY '"'
LINES TERMINATED BY '\r\n'

Answer 1

一种选择是使用REPEAT。字符的ASCII码值＆＃39; a＆＃39;通过＆＃39; z＆＃39;在97到122之间，我们可以使用CHAR()函数返回给定ASCII代码值的字符。

例如：

BEGIN
    DECLARE c VARCHAR(1);
    DECLARE i INT;
    SET i = 97;
    REPEAT
        SET c = CHAR(i);

        -- SELECT c;

        SET i = i + 1;
    UNTIL i > 122 END REPEAT;
END

写程序来完成所有字母

1 个答案: