Question

我的手术有问题。我尝试从sales表中获取值并使用它们进行查询。程序如下所示：

DROP PROCEDURE IF EXISTS turnover;

DELIMITER $$

CREATE PROCEDURE turnover()
BEGIN
    DECLARE col INT;
    DECLARE q TEXT;
    DECLARE i INT DEFAULT 0;
    DECLARE m TEXT;
    SET col = (SELECT count(DISTINCT article) FROM sales);
    SET q = "SELECT article, ";
    WHILE i < co DO
        SET m = (SELECT DISTINCT month FROM sales LIMIT 1 OFFSET i);
        SET q = q + "SUM(IF(month=" + m + ",value,NULL)) AS " + m;
        IF i < (col - 1) THEN
            SET q = q + ", ";
        END IF;
        SET i = i + 1;
    END WHILE;
    SET q = q + " FROM sales GROUP BY article";
    EXECUTE q;
END$$

DELIMITER ;

CALL turnover();

我收到错误：

错误代码：1292。截断错误的DOUBLE值：＆＃39;，value，NULL））AS＆＃39;

我如何才能使它有效？

感谢。

Answer 1

col问题已修复或假设如下。

CREATE SCHEMA safe_Tuesday_01; -- safe sandbox
USE safe_Tuesday_01;    -- DO the work in this db to test it

-- a fake table, we need something
create table sales 
(   article varchar (100) not null,
    month int not null
);

第1步，找出字符串的样子：

DROP PROCEDURE IF EXISTS turnover;
DELIMITER $$
CREATE PROCEDURE turnover()
BEGIN
    DECLARE col INT;
    DECLARE q TEXT;
    DECLARE i INT DEFAULT 0;
    DECLARE m TEXT;
    SET col = (SELECT count(DISTINCT article) FROM sales);
    SET q = "SELECT article, ";
    WHILE i < col DO
        SET m = (SELECT DISTINCT month FROM sales LIMIT 1 OFFSET i);
        SET q = CONCAT(q,"SUM(IF(month=" + m + ",value,NULL)) AS ", m);
        IF i < (col - 1) THEN
            SET q = q + ", ";
        END IF;
        SET i = i + 1;
    END WHILE;
    SET q = CONCAT(q," FROM sales GROUP BY article");
    select q;
    -- EXECUTE q; -- No no no this is wrong anyway
END$$
DELIMITER ;

CALL turnover();

SELECT article，FROM sales GROUP BY article

以上SELECT看起来并不那么热。在第1步中重复修复逻辑以修复该字符串。

第2步，当您修复上面的代码时，请在下面进行操作。注意，目前，它没有修复。所以，再次，上面这样做。

但是在下面，请使用您不是的PREPARED STATEMENT。

DROP PROCEDURE IF EXISTS turnover;
DELIMITER $$
CREATE PROCEDURE turnover()
BEGIN
    DECLARE col INT;
    DECLARE q TEXT;
    DECLARE i INT DEFAULT 0;
    DECLARE m TEXT;
    SET col = (SELECT count(DISTINCT article) FROM sales);
    SET q = "SELECT article, ";
    WHILE i < col DO
        SET m = (SELECT DISTINCT month FROM sales LIMIT 1 OFFSET i);
        SET q = CONCAT(q,"SUM(IF(month=" + m + ",value,NULL)) AS ", m);
        IF i < (col - 1) THEN
            SET q = q + ", ";
        END IF;
        SET i = i + 1;
    END WHILE;
    SET q = CONCAT(q," FROM sales GROUP BY article");
    -- select q;
    SET @theSQL=q;
    PREPARE stmt1 FROM @theSQL;
    EXECUTE stmt1;
    DEALLOCATE PREPARE stmt1;
END$$
DELIMITER ;

完成后，

DROP SCHEMA safe_Tuesday_01; -- clean up, poof, sandbox is gone

CONCAT是你的朋友。你错过了这一步。重要的是PREPARE针对用户变量（带有@标志）而不是局部变量（来自DECLARE），否则它会爆炸。所以我使用@theSQL

修复了上述内容

再次，请参阅MySQL手册页PREPARE Syntax。让你的字符串正确是很重要的。这就是第1步的要点。只有这样，你才能继续进行第2步并使用它。

Answer 2

当SELECT DISTINCT month FROM sales没有返回任何内容时会发生这种情况。在下一行，查询片段生成为SUM(IF(month=,value,NULL)) AS，当然，那里有一个错误（可能MySQL没有产生正确的错误消息，但这是错误的地方）。

错误原因是将WHILE与未知变量i进行比较的co行。它可能应该是：

WHILE i < col DO

但它无法解决问题，因为col是article的不同值的数量，您在col的不同值上从1迭代到month 。很可能它们的数量不同，如果文章数量较大，那么错误将再次发生。

MySQL错误＆＃34;截断错误的DOUBLE值＆＃34;在程序中

2 个答案: