分组+行差异？ php mysql

Question

我该怎么做？我有这个表

╔════════╦══════════╦══════════╗ 
║ ITEM ║ PRICE ║ DATE  ║ 
╠════════╬══════════╬══════════╣ 
║ Dollar ║  60 ║ 1.3.2016 ║ 
║ Dollar ║  50 ║ 2.3.2016 ║ 
║ Bound ║  100 ║ 1.3.2016 ║ 
║ Bound ║  110 ║ 2.3.2016 ║ 
║ Euro ║  600 ║ 1.3.2016 ║ 
║ Euro ║  580 ║ 3.3.2016 ║ 
╚════════╩══════════╩══════════╝ 

输出应该从每个类型与最后日期和前一行差价最后价格dedepnding显示一个项目，像这样

╔════════╦══════════╦════════════════════════╗ 
║ ITEM ║ PRICE ║ DATE  ║ Differnece ║ 
╠════════╬══════════╬════════════════════════╣ 
║ Dollar ║  50 ║ 2.3.2016 ║ -10  ║ 
║ Bound ║  110 ║ 2.3.2016 ║ 10  ║ 
║ Euro ║  580 ║ 3.3.2016 ║ -20  ║ 
╚════════╩══════════╩════════════════════════╝ 

这是我做的，但它是抛出错误

SELECT a.item, a.price, a.date, a.price - b.price AS difference 
FROM (SELECT t1.item, t1.price, t1.date 
     FROM yourTable AS t1 
     JOIN (SELECT item, MAX(date) AS maxdate 
      FROM yourTable 
      GROUP BY item) AS t2 
     ON t1.item = t2.item, t1.date = t2.maxdate) AS a 
JOIN (SELECT t1.item, t1.price, t1.date 
     FROM yourTable AS t1 
     JOIN (SELECT item, MIN(date) AS mindate 
      FROM yourTable 
      GROUP BY item) AS t2 
     ON t1.item = t2.item, t1.date = t2.mindate) AS b 

这是错误

1064 - 您的SQL语法错误;检查对应于你的MySQL服务器版本使用附近的正确语法手册“.maxdate）为连接（SELECT t1.item_type，t1.purchasing_price，t1.date_price”在第7行

Answer 1

写一个子查询拿到第一排的每个项目，而另一个子查询，以获得最后一行对每个项目，加入他们的行列中减去。

SELECT a.item, a.price, a.date, a.price - b.price AS difference 
FROM (SELECT t1.item, t1.price, t1.date 
     FROM yourTable AS t1 
     JOIN (SELECT item, MAX(date) AS maxdate 
      FROM yourTable 
      GROUP BY item) AS t2 
     ON t1.item = t2.item, t1.date = t2.maxdate) AS a 
JOIN (SELECT t1.item, t1.price, t1.date 
     FROM yourTable AS t1 
     JOIN (SELECT item, MIN(date) AS mindate 
      FROM yourTable 
      GROUP BY item) AS t2 
     ON t1.item = t2.item, t1.date = t2.mindate) AS b