Question

我想用我的数据库中的数据返回带有mysql的10个json字符串，但是我得到了以下错误：SQL错误（1172）：结果包含多行！是的我知道结果包含更多行，它们是10，问题是如何在不同的行中打印它们？

CREATE DEFINER=`root`@`localhost` FUNCTION `search_for`(`value` TEXT)
    RETURNS text CHARSET latin1
    LANGUAGE SQL
    NOT DETERMINISTIC
    CONTAINS SQL
    SQL SECURITY DEFINER
    COMMENT ''
BEGIN
    DECLARE sid TEXT;
    DECLARE semail TEXT;
    DECLARE sname TEXT;
    DECLARE slastname TEXT;
    DECLARE response TEXT;

    SELECT id,email,name,lastname INTO sid,semail,sname,slastname 
    FROM user WHERE email LIKE CONCAT('%',value,'%') 
    OR lastname LIKE CONCAT('%',value,'%') 
    OR name LIKE CONCAT('%',value,'%') LIMIT 10;

    IF (semail IS NOT NULL) THEN             
        SET response = CONCAT('{"response":1,"id":',sid,',"email":"',semail,'","name":"',sname,'","lastname":"',slastname,'"}');
    ELSE
        SET response = CONCAT('{"response":0}');
    END IF;

    RETURN response;
END

在我使用游标并循环后，该方法在多行中获取搜索结果如下所示：

BEGIN
    DECLARE finished INTEGER DEFAULT 0;
    DECLARE ids TEXT;
    DECLARE response TEXT;
    DECLARE ids_cursor CURSOR FOR
        SELECT id,email,name,lastname  -- INTO ids,semail,sname,slastname 
        FROM user WHERE email LIKE CONCAT('%',value,'%') 
        OR lastname LIKE CONCAT('%',value,'%') 
        OR name LIKE CONCAT('%',value,'%');
    DECLARE CONTINUE HANDLER FOR NOT FOUND SET finished = 2;

    OPEN ids_cursor;
    get_ids: LOOP

        FETCH ids_cursor INTO ids;

        IF finished = 2 THEN
            LEAVE get_ids;
        END IF;

        SET response = CONCAT(ids,";",response);

    END LOOP get_ids;
    CLOSE ids_cursor;

    RETURN response;
END

我仍然得到一点错误（1328）：FETCH变量数量不正确！为什么呢？

更新

编辑下一行：

SELECT id --,email,name,lastname -- INTO ids,semail,sname,slastname

获得NULL回复

Answer 1

不得使用功能。也许从这个查询定义视图？像这样：

create or replace view userquery select if(semail is not null, CONCAT('{"response":1,"id":',sid,',"email":"',semail,'","name":"',sname,'","lastname":"',slastname,'"}'),"0") response, email,lastname,name FROM user ;

并查询

select response from userquery  where ...

或者只是在查询中使用“if”，没有功能。

Answer 2

是的，最后我不会使用循环和光标来响应此查询的响应，我将使用这样的json字符串：

BEGIN
    DECLARE response TEXT;

    SELECT CONCAT('{',GROUP_CONCAT(id SEPARATOR ','),'}') INTO response
    FROM user 
    WHERE email LIKE CONCAT('%',value,'%') 
       OR lastname LIKE CONCAT('%',value,'%') 
       OR name LIKE CONCAT('%',value,'%');

    IF( response IS NOT NULL ) THEN
        RETURN response;
    ELSE
        RETURN CONCAT('{"response":0}');
    END IF;

END

我认为sql游标和循环用于获得更高级的结果，无论如何工作就像我在这里做的那样，tcx！

SQL Result由多行组成

2 个答案: