根据您想要的输出，这里使用内部联接。但若代码（b01）不存在于stock_in和stock_out表中，则使用左联接而不是内部联接。在金额字段中使用合并函数来忽略null。

-- MySQL
SELECT s.code, s.name, s.stock, si.stock_in, so.stock_out
     , (s.stock + si.stock_in - so.stock_out) last_stock
FROM Stock s
INNER JOIN (SELECT code, MAX(name) name
                 , SUM(stock_in) stock_in
            FROM Stock_in
            GROUP BY code) si
        ON s.code = si.code
INNER JOIN (SELECT code, MAX(name) name
                 , SUM(stock_out) stock_out
            FROM Stock_out             
            GROUP BY code) so
            ON s.code = so.code;

请检查url的结果http://sqlfiddle.com/#!9/6dc765/4
请检查使用左联接的结果http://sqlfiddle.com/#!9/9e97ef3/2

这里有一种方法（使用rahulbiswas深思熟虑地提供的小提琴）。。。

SELECT s.code   
     , s.name
     , s.stock
     , SUM(CASE WHEN direction = 'in' THEN qty END) stock_in
     , SUM(CASE WHEN direction = 'out' THEN qty END) stock_out
     , s.stock+SUM(CASE WHEN direction = 'in' THEN qty END)-SUM(CASE WHEN direction = 'out' THEN qty END) balance
  FROM stock s
  JOIN 
     ( SELECT code
            , name
            , stock_in qty
            , 'in' direction
         FROM stock_in
        UNION ALL
       SELECT code
            , name
            , stock_out 
            , 'out' 
         FROM stock_out
    ) x
   ON x.code = s.code;

http://sqlfiddle.com/#!9/6dc765/10