检查MySQL中的Null嵌套如果

我正在尝试编写一个MySQL函数，从格式化为JSON dict的文本字段中提取一个值。但是，由于某些原因，我无法使嵌套的IF语句在没有语法错误的情况下工作

以下是函数：

DROP FUNCTION IF EXISTS json_key;
DELIMITER $$

CREATE FUNCTION `json_key`(`col` TEXT, `key` VARCHAR(255)) RETURNS VARCHAR(255)
DETERMINISTIC
BEGIN
    DECLARE ret VARCHAR(255);
    DECLARE pos INT;
    IF (col IS NULL) THEN SET ret = "";
    ELSEIF (col = "") THEN SET ret =  "";
    ELSEIF (LOCATE(key, col) = 0) THEN SET ret =  "";
    ELSE 
        SET pos := LOCATE(key, col);
        SET len := LENGTH(key);
        SET bg  := pos + len;
        SET ret = SUBSTR(col, bg, LOCATE('"', col, bg) - bg));
    END IF;
    RETURN ret;
END $$
DELIMITER ;

以下是我得到的错误：

ERROR 1064 (42000): You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to 
use near 'key, col) = 0) THEN SET ret =  "";

ELSE 
SET pos := LOCATE(key, col);
SET len :=' at line 7

知道我遗漏了什么吗？

试试这个：

SET len := (coalesce(LENGTH(key),0));

---

移除条件周围的括号并检查；像

ELSEIF LOCATE(key, col) = 0 THEN SET ret =  "";

你也可以像检查一样改变你的状况

IF (col IS NULL OR  col = "" OR LOCATE(key, col) = 0) THEN SET ret = "";
ELSE 
    SET pos := LOCATE(key, col);
    SET len := LENGTH(key);
    SET bg  := pos + len;
    SET ret = SUBSTR(col, bg, LOCATE('"', col, bg) - bg));
END IF;

---

键

是保留字

或者使用不同的名称，比如说

jKey

，或者使用反勾（`）使其被接受

更改这些行：

ELSEIF (LOCATE(key, col) = 0) THEN SET ret =  "";
ELSE 
    SET pos := LOCATE(key, col);
    SET len := LENGTH(key);

ELSEIF (LOCATE(`key`, col) = 0) THEN SET ret =  "";
ELSE 
    SET pos := LOCATE(`key`, col);
    SET len := LENGTH(`key`);

至：

ELSEIF (LOCATE(key, col) = 0) THEN SET ret =  "";
ELSE 
    SET pos := LOCATE(key, col);
    SET len := LENGTH(key);

ELSEIF (LOCATE(`key`, col) = 0) THEN SET ret =  "";
ELSE 
    SET pos := LOCATE(`key`, col);
    SET len := LENGTH(`key`);

---

参考：

Ravinder发现了这个错误，但我也投了赞成票，因为它清理了整个结构。