Question

我有点卡在列的订单上。我从堆栈溢出中得到了另一个query的好答案，但现在只需要正确订购它。

考虑我有这些结果（无序）

|LloydA|20|
|LloydC|5 |
|LloydB|0 |
|JonesA|10|
|JonesB|0 |
|ZuberB|10|
|ZuberA|0 |

然后想要先按第二列排序它们，但也要将它排在第1列，这样结果就是

(function(){

    function drawLine(points){
        context.beginPath();
        points.forEach(function(point){
            context.lineTo(point.x, point.y);   
        });
        context.stroke();
    }

    var canvas = document.createElement('canvas');
    canvas.setAttribute('width', 500);
    canvas.setAttribute('height', 300);
    document.body.appendChild(canvas);
    context = canvas.getContext("2d");

    // top
    drawLine([{x: 100, y: 40}, {x: 400, y: 10}]);
    // bottom
    drawLine([{x: 50, y: 220}, {x: 300, y: 290}]);
    // left
    drawLine([{x: 40, y: 20}, {x: 80, y: 260}]);
    // right
    drawLine([{x: 490, y: 60}, {x: 440, y: 290}]);

})();

因为20是hte最大值，我想在顶部显示这一行，然后显示所有具有相同名称开头的行（即所有Lloyd ..）

我正在考虑订单的子选择，但我正在努力。有没有人有任何提示。

Answer 1

这是一个产生您需要的结果的查询：

WITH 
  -- "t" contains the raw data
  t(name, score) AS (
    SELECT 'Lloyd0', 20 FROM DUAL UNION ALL
    SELECT 'Lloyd1', 0  FROM DUAL UNION ALL
    SELECT 'Lloyd2', 5  FROM DUAL UNION ALL
    SELECT 'Jones0', 10 FROM DUAL UNION ALL
    SELECT 'Jones1', 0  FROM DUAL UNION ALL
    SELECT 'Zuber0', 0  FROM DUAL UNION ALL
    SELECT 'Zuber1', 10 FROM DUAL
  ),

  -- "u" generates the "name prefix" by removing the numbers form the names
  u(name, score, name_prefix) AS (
    SELECT name, score, regexp_replace(name, '\d+', '')
    FROM t
  ),

  -- "v" generates the max score per name_prefix ("group_rank") for each "group"
  v(name, score, group_rank, name_prefix) AS (
    SELECT name, score, MAX(score) OVER (PARTITION BY name_prefix), name_prefix
    FROM u
  )
SELECT name, score
FROM v
ORDER BY 

  -- Order by the group's rank first
  group_rank DESC, 

  -- Order equally ranked groups by name
  name_prefix ASC, 

  -- Order entries within each group by score
  score DESC

当然，您不需要为此使用公用表表达式。这可以通过派生表或视图来完成，也可以通过重复某些表达式来完成。

SQLFiddle here

注意：您的原始问题是Lloyd0而不是LloydA。它并不十分清楚＆＃34;有趣的＆＃34;名称的一部分确实是（例如Lloyd），以及＆＃34;非有趣的＆＃34;部分是（例如A或0）。但我怀疑这对答案来说并不重要。

Answer 2

根据您的previous SO question，您无需提取名称，只需添加GROUP MAX：

-- copied from the accepted answer
select p.* from people p  
join (select p1.first_name, p1.last_name 
      from people p1 where p1.id = 17
     )un
  on un.first_name = p.first_name 
where p.last_name like un.last_name || '%'
-- ADDED
order by  -- find the MAX for each name
   max(col2) over (partition by un.last_name),
   last_name

在SQL中按顺序排序

2 个答案: