は... XML RAW

Question

へのクエリ

私はのTransact SQLを使用していると私は、これらのテーブルを持って

main.locality - id,locality 
main.job - id,jobs 
main.person - name,surname,jobid,localityid,salary,gender 

を作る私はこの出力（XML RAW）PSを取得する必要があります。 第1行はすべてのジョブの平均給与 第2行はすべての女性の平均給与 第3行はすべての男性の平均給与 第4行はその仕事と女性の平均給与 第5行その仕事と男性の平均給与です。

<row Average="19827" /> 
<row gender="F" Average="19835" /> 
<row gender="M" Average="19819" /> 
<row job="C# Developer" Average="20941" /> 
<row job="C# Developer" gender="F" Average="20845" /> 
<row job="C# Developer" gender="M" Average="21027" /> 

誰かが質問をしてくれますか？

Answer 1

私は考えている：

-- 1. SAMPLE DATA 
DECLARE @locality TABLE (localityid int, locality int); 
DECLARE @job TABLE (id int, Job varchar(100)); 
DECLARE @person TABLE 
(
    name varchar(100), 
    surname varchar(100), 
    jobid int, 
    localityid int, 
    salary money, 
    gender char(1) 
); 

INSERT @locality VALUES (1, 10), (2, 20), (3, 35), (4, 1000); 
INSERT @job VALUES (1, 'C# Developer'), (2, 'DBA'); 
INSERT @person VALUES 
('joe', 'blow', 1, 1, 100000, 'M'), 
('Sally', 'Doe', 1, 2, 135000, 'F'), 
('Sue', 'Smith', 2, 2, 115000, 'F'), 
('Kyle', 'Jones', 2, 3, 107500, 'M') 
; 

-- 2. SOLUTION: 
SELECT gender = NULL, job = NULL, average = AVG(p.salary) 
FROM @person p 
UNION ALL 
SELECT p.gender, job = NULL, average = AVG(p.salary) 
FROM @person p 
JOIN @job j ON p.jobid = j.id 
GROUP BY p.gender 
UNION ALL 
SELECT gender = NULL, job = j.job, average = AVG(p.salary) 
FROM @person p 
JOIN @job j ON p.jobid = j.id 
GROUP BY j.Job 
UNION ALL 
SELECT p.gender, j.Job, average = AVG(p.salary) 
FROM @person p 
JOIN @job j ON p.jobid = j.id 
GROUP BY p.gender, j.Job 
FOR XML RAW; 

戻り値：

<row average="114375.0000" /> 
<row gender="F" average="125000.0000" /> 
<row gender="M" average="103750.0000" /> 
<row job="C# Developer" average="117500.0000" /> 
<row job="DBA" average="111250.0000" /> 
<row gender="F" job="C# Developer" average="135000.0000" /> 
<row gender="M" job="C# Developer" average="100000.0000" /> 
<row gender="F" job="DBA" average="115000.0000" /> 
<row gender="M" job="DBA" average="107500.0000" /> 

Answer 2

これはa job for ROLLUP

（私はアラン・バースタインのサンプルデータを借り - ありがとう！）である

-- 1. SAMPLE DATA 
DECLARE @locality TABLE (localityid int, locality int); 
DECLARE @job TABLE (id int, Job varchar(100)); 
DECLARE @person TABLE 
(
    name varchar(100), 
    surname varchar(100), 
    jobid int, 
    localityid int, 
    salary money, 
    gender char(1) 
); 

INSERT @locality VALUES (1, 10), (2, 20), (3, 35), (4, 1000); 
INSERT @job VALUES (1, 'C# Developer'), (2, 'DBA'); 
INSERT @person VALUES 
('joe', 'blow', 1, 1, 100000, 'M'), 
('Sally', 'Doe', 1, 2, 135000, 'F'), 
('Sue', 'Smith', 2, 2, 115000, 'F'), 
('Kyle', 'Jones', 2, 3, 107500, 'M'); 


SELECT j.Job 
     ,p.gender 
     ,AVG(p.salary) AS Average 
FROM @person AS p 
INNER JOIN @locality AS l ON p.localityid=l.localityid 
INNER JOIN @job AS j ON p.jobid=j.id 
GROUP BY ROLLUP (gender,job) 
ORDER BY job,gender 
FOR XML RAW 

を

Th eの結果

<row Average="114375.0000" /> 
<row gender="F" Average="125000.0000" /> 
<row gender="M" Average="103750.0000" /> 
<row Job="C# Developer" gender="F" Average="135000.0000" /> 
<row Job="C# Developer" gender="M" Average="100000.0000" /> 
<row Job="DBA" gender="F" Average="115000.0000" /> 
<row Job="DBA" gender="M" Average="107500.0000" />