Question

尝试进行生成此类结果的查询

EmpId,  EmpSickLeave,EmpCasualLeave, EmpAnnualLeave, TotalLeave
1,       1,           0,                0            1
2,       0,           0,                2            2
3,       0,           1,                0            1
4,       0,           1,                0            1  
5,       1,           0,                0            1

虽然我有两张桌子

表EmpLeave ::包含一列EmpId，Date，EmpLeaveTypeID

表EmpLeaveType :: include EmpLeaveTypeID和LeaveName

DATA表格

EmpLeaveType Table
EmpLeaveTypeID , LeaveName
1,              Sick Leave
2,              Annual
3,              Casual

EmpLeave Table
column EmpId, Date, EmpLeaveTypeID
            1, 2015-07-01, 1
            3, 2015-07-02, 2
            5, 2015-07-04, 1
            4, 2015-07-04, 2
            2, 2015-07-05, 2
            2, 2015-07-07, 2

我正在拉我的头发，无法提出这种结果。这可能吗？

  EEmpId,  EmpSickLeave,EmpCasualLeave, EmpAnnualLeave, TotalLeave
1,       1,           0,                0            1
2,       0,           0,                2            2
3,       0,           1,                0            1
4,       0,           1,                0            1  
5,       1,           0,                0            1

我尝试过这个错误的事情！

    SELECT * count(EmpLeaveTypeID) FROM `EmpLeaveType`
WHERE SwitchDate Between '2015-07-01' AND '2015-07-28'
group by EmpLeaveType, EmpId

Answer 1

请尝试下面的SELECT查询。它没有考虑到有一天可能会改变休假类型。 SQL小提琴链接http://sqlfiddle.com/#!9/0fd13/2 查询只是执行条件聚合来计算每个员工的叶子。

SELECT 
EmpId, 
SUM(CASE WHEN EmpLeaveTypeID=1 THEN 1 ELSE 0 END) as EmpSickLeave,
SUM(CASE WHEN EmpLeaveTypeID=2 THEN 1 ELSE 0 END) as EmpCasualLeave,
SUM(CASE WHEN EmpLeaveTypeID=3 THEN 1 ELSE 0 END) as EmpAnnualLeave,
COUNT(1) as TotalLeave
FROM `EmpLeave`
WHERE Date Between '2015-07-01' AND '2015-07-28'
group by EmpId

Answer 2

这看起来是一个条件计数，您只想计算某些行。为此，您可以使用带有select EmpId as EEmpId, sum(case when EmpLeaveTypeID = 1 then 1 else 0 end) as EmpSickLeave, sum(case when EmpLeaveTypeID = 2 then 1 else 0 end) as EmpCasualLeave, sum(case when EmpLeaveTypeID = 3 then 1 else 0 end) as EmpAnnualLeave, count(*) as TotalLeave from EmpLeave where ... group by EmpId语句的0：

这样做会根据EmpLeaveTypeID是否匹配为每行添加COUNT或TotalLeave。然后，您可以使用{{1}}获取{{1}}列的总行数，如上所示。

SQL Fiddle demo

Answer 3

我省略了package com.you.take.me.to.funky; import java.io.FileNotFoundException; import java.io.IOException; import java.io.PrintWriter; import java.nio.charset.StandardCharsets; import java.nio.path.Files; import java.nio.path.Path; import java.nio.path.Paths; import java.util.ArrayList; import java.util.List; import java.util.UUID; import java.util.regex.Matcher; import java.util.regex.Pattern; public class town { private static List<String> list = new ArrayList<>(); private static List<String> misclist = new ArrayList<>(); private static Pattern p0 = Pattern.compile("(Bench:)(.*)"); private static Pattern p1 = Pattern.compile("(Author):(.*)"); private static Pattern p2 = Pattern.compile("JUDGMENT(.*?)J[.]"); private static Pattern p3 = Pattern.compile("(([H|h]on)|(HON)).*((ble)|BLE)(.*)"); private static Pattern p4 = Pattern.compile(",\\s*([^,]+),[^,]*\\b(J|JJ)\\.");//CAKE private static Pattern p5 = Pattern.compile("\\s\\w*(?=\\w*[,]\\sJ[.]*\\b)"); private static Pattern p4 = Pattern.compile("\\w*(?=\\w*[,]*\\s*((JJ)|(L.J)|(C.J)|(J))[.]\\s\\b)"); private static Pattern p6 = Pattern.compile("BENCH:.*?BENCH.*?CITATION", Pattern.DOTALL); public static void main(String[] args) { Path path = Paths.get(args[0]); String str; try { str = new String(Files.readAllBytes(path), StandardCharsets.UTF_8); } catch (IOException e) { e.printStackTrace(); } Matcher match = p0.matcher(str); if (match.find()) { Matcher match1 = p1.matcher(str); if (match1.find()) { // pathname judgename authorname list.add(path.toString() + "\t" + match.group(2) + "\t" + match1.group(2)); System.out.println(match1.group(2)); } else { list.add(path.toString() + "\t" + match.group(2) + "\t "); System.out.println(match.group(2)); } } else { match = p1.matcher(str); if (match.find()) { list.add(path.toString() + "\t \t" + match.group(2)); } else { match = p2.matcher(str); if (match.find()) { list.add(path.toString() + "\t" + match.group() + "\t "); } else { match = p3.matcher(str); if (match.find()) { list.add(path.toString() + "\t" + match.group() + "\t "); } else { match = p4.matcher(str); if (match.find()) { //do nothing } else { match = p5.matcher(str); if (match.find()) { list.add(path.toString() + "\t" + match.group() + "\t "); System.out.println(path.toString()); } else { match = p6.matcher(str); if (match.find()) { System.out.println("DEBUG MARKER"); } else { // list of documents which have no Judgenames misclist.add(path.toString()); String name = UUID.randomUUID().toString(); try { PrintWriter pw = new PrintWriter("/h/g/d/d/m/" + name); pw.write(str); pw.flush(); } catch (FileNotFoundException e) { e.printStackTrace(); } } } } } } } } } }，因为它是一个简单的date between添加。

尝试下面的SQL查询：

where

检查sqlfiddle

Answer 4

请检查以下内容：

 SELECT 
 temp.EmpId , 
 temp.EmpSickLeave, 
 temp.EmpCasualLeave, 
 temp.EmpAnnualLeave,
 (temp.EmpSickLeave + temp.EmpCasualLeave + temp.EmpAnnualLeave ) 
 as  TotalLeave 
 FROM 
( SELECT 
 EmpId, 
 SUM( CASE WHEN EmpLeaveTypeID=1 THEN 1 ELSE 0 END ) as EmpSickLeave,
 SUM( CASE WHEN EmpLeaveTypeID=2 THEN 1 ELSE 0 END ) as EmpCasualLeave,
 SUM( CASE WHEN EmpLeaveTypeID=3 THEN 1 ELSE 0 END ) as EmpAnnualLeave
 FROM EmpLeave 
 WHERE Date Between '2015-07-01' AND '2015-07-28'
 group by EmpId ) temp

SQL Query将列的组结果带入多个列

4 个答案: