Question

我有一张表<index> <letter> <title>A</title> <mainTerm> <title>Some description</title> <code>Q87.1</code> </mainTerm> <mainTerm> <title>Another description<nemod>(-some) (detail)</nemod></title> <code>F44.4</code> </mainTerm> <mainTerm> <title>A more detailed term</title> <seeAlso>similar terms</seeAlso> <term level="1"> <title>acute</title> <code>R10.0</code> </term> <term level="1"> <title>angina</title> <code>K55.1</code> </term> </mainTerm> </letter> ..... </index>，其中包含这样的数据

  $(document).find("letter").each(function(){
 // Define letters variable
 $letters =  $(this).find('> title').text();

//Find descendants of each letter            
$codes = ($(this).find('code').text());
$desc = ($(this)).find('mainTerm title').text();

//Build table
$("#content").append(
  '<table border="1"><tr><th>Code</th><th>Description</th></tr><tr><td> '+$codes+' </td><td> '+$desc+' </td></tr></table>'
    );            
(end brackets for the above)

我想用查询

创建一个这样的表

delay

我搜索了给定的问题。但是有些答案与表中类似值的计数有关，但没有一个对数据进行分类。

请帮忙

Answer 1

<强> SQL DEMO

with ranges as (
    SELECT 'On time' r UNION ALL
    SELECT 'Late < 1 min' UNION ALL
    SELECT 'Late by 1-2 min' UNION ALL
    SELECT 'Late by 2-3 min' UNION ALL
    SELECT 'Late by 3-4 min' UNION ALL
    SELECT 'Late by 4-5 min' UNION ALL
    SELECT 'Late by 5-6 min' UNION ALL
    SELECT 'Late by 6-7 min' UNION ALL
    SELECT 'Late by 7-8 min' UNION ALL
    SELECT 'Late >8 minutes'
), setup as (    
    SELECT CASE WHEN [Delay] = 0 THEN 'On time'
                WHEN [Delay] < 1 THEN 'Late < 1 min'
                WHEN [Delay] < 2 THEN 'Late by 1-2 min'
                WHEN [Delay] < 3 THEN 'Late by 2-3 min'
                WHEN [Delay] < 4 THEN 'Late by 3-4 min'
                WHEN [Delay] < 5 THEN 'Late by 4-5 min'
                WHEN [Delay] < 6 THEN 'Late by 5-6 min'
                WHEN [Delay] < 7 THEN 'Late by 6-7 min'            
                WHEN [Delay] < 8 THEN 'Late by 7-8 min'         
                ELSE 'Late >8 minutes'
           END as [Range],
           [Delay]
    FROM Table1       
)
SELECT t.r as Range, COALESCE(SUM(s.[Delay]),0) as Value, COUNT(s.[Delay]) as Freq
FROM ranges t
LEFT JOIN setup  s
  ON t.r = s.[Range]
GROUP BY t.r

<强>输出

Answer 2

使用CASE或IIF可以解决此问题，但我建议使用函数使代码更具可读性和可扩展性。

用户功能 - ufRange

CREATE FUNCTION dbo.ufRange (@Delay INT)
RETURNS INT
AS
BEGIN
    DECLARE @Ret INT
    IF @Delay <= 0 SET @Ret = 0         -- not late/ early
    ELSE IF @Delay > 320 SET @Ret = 9  -- more than 8 minutes late
    ELSE SET @Ret = @Delay / 60         -- late <=1 to <=8 minutes
    RETURN @Ret
END

然后将range的详细信息放在支持表LateRange中，就像在excel中一样，如果您熟悉vlookup用于此

支持表 - LateRange

idrange          Range                               Values
  0               On time                               0
  1               Late < 1 min                          60
  2               Late by 1-2 min                       120
  3               Late by 2-3 min                       180 
  4               Late by 3-4 min                       240
  5               Late by 4-5 min                       300
  6               Late by 5-6 min                       360
  7               Late by 6-7 min                       420
  8               Late by 7-8 min                       480
  9               Late >8 minutes

行。最后假设delay已保存的表名为LateList，查询为

SELECT Range, Values, 
      (SELECT COUNT(*) FROM LateList WHERE dbo.ufRange([delay]) = idrange) Frequency
FROM LateRange

或者，如果您愿意，可以使用INNER JOIN和子查询进行修改。

使用T-SQL查询创建频率分布

2 个答案: