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综合框架说明中的协同错误
原标题:Syntax error in nested IF statement

我需要一种程序,以核实某一数字是否属首要。 The syntax on the syntax似乎是罚款的,但有以下错误:

Query : CREATE PROCEDURE prime_ number (IN n INT) BEGIN SET @prime = “Prime number”; SET @n_prime = “Not prime number”; SET @divider = ......

《刑法》:1064

您在座椅上有一处错误;检查了与MySQL服务器版本相对应的手册,供右yn在28行附近使用。

Here s my code so far:

DELIMITER $
CREATE PROCEDURE prime_number(IN n INT)
BEGIN
    SET @prime = "Prime number";
    SET @n_prime = "Not prime number";
    SET @divider = 0;
    SET @i = 5;
    IF n % 2 = 0 THEN
        SELECT @n_prime;
        
    ELSEIF (n < 2) OR ((n % 3) = 0) THEN /* Was using ELSE IF, separatedly*/
        SELECT @n_prime;
        
    ELSE
        SET @root = SQRT(n);
        
        WHILE @i <= @root && @divider = 0 DO
            IF (n % @i) = 0 THEN
                SET @divider = 1;
            END IF;
            SET @i = @i + 1;
        END WHILE;
        
        IF @divider = 1 THEN SELECT @n_prime;
        ELSE SELECT @prime;
        END IF;
    
    END IF; /*Line 28*/
END $
DELIMITER ;

它为什么会发生?

最佳回答

您的问题很可能是ELSE FI。 这就产生了另一个<代码>IF说明,要求另加<代码>END FI。 http://dev.mysql.com/doc/refman/5.0/en/if-statement.html”

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