PostgreSQL中的Base58編碼器函數

Question

我發現這個MySQL函數Base58編碼器在Github Gist中。

DELIMITER $$ 

CREATE FUNCTION base58_encode (num int) RETURNS varchar(255) 
    DETERMINISTIC 

BEGIN 
    DECLARE alphabet varchar(255); 
    DECLARE base_count int DEFAULT 0; 
    DECLARE encoded varchar(255); 
    DECLARE divisor DECIMAL(10,4); 
    DECLARE mode int DEFAULT 0; 

    SET alphabet = '123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ'; 
    SET base_count = CHAR_LENGTH(alphabet); 
    SET encoded = ""; 

    WHILE num >= base_count DO 
    SET divisor = num/base_count; 
    SET mode = (num - (base_count* TRUNCATE(divisor,0))); 
    SET encoded = CONCAT(SUBSTRING(alphabet FROM mode+1 FOR 1), encoded); 
    SET num = TRUNCATE(divisor,0); 
    END WHILE; 

    SET encoded = CONCAT(SUBSTRING(alphabet FROM num+1 FOR 1), encoded); 

    RETURN (encoded); 

END 

我是的新手，難以將上面的函數轉換爲PostgreSQL函數。

對於Base58編碼器，上面的SQL代碼段的等效PostgreSQL函數將如何工作？

Answer 1

我想出來的Pos​​tgreSQL中的等價函數如下。

CREATE FUNCTION base58_encode(num INT) 
    RETURNS VARCHAR(255) AS $encoded$ 

DECLARE 
    alphabet VARCHAR(255); 
    base_count INT DEFAULT 0; 
    encoded VARCHAR(255); 
    divisor DECIMAL(10, 4); 
    mod  INT DEFAULT 0; 

BEGIN 
    alphabet := '123456789abcdefghijkmnopqrstuvwxyzABCDEFGHJKLMNPQRSTUVWXYZ'; 
    base_count := char_length(alphabet); 
    encoded := ''; 

    WHILE num >= base_count LOOP 
    divisor := num/base_count; 
    mod := (num - (base_count * trunc(divisor, 0))); 
    encoded := concat(substring(alphabet FROM mod + 1 FOR 1), encoded); 
    num := trunc(divisor, 0); 
    END LOOP; 

    encoded = concat(substring(alphabet FROM num + 1 FOR 1), encoded); 

    RETURN (encoded); 

END; $encoded$ 
LANGUAGE PLPGSQL;