Question

再次使用count和sql

在sqlite上，我有桌子

论文：paper_id，doi，年份
作者：paper_id，author_id，inst_id
作者：author_id，姓名，名字
inst：inst_id，名称，see_id

inst是院校列表：大学等。书面形式的每一行都给出了一篇论文，一位作者，该作者当时所隶属的机构。可能有一个以上的机构，并且每个机构都重复一对夫妇paper_id，author_id。对于给定的作者，我想要一个列表和一份papers.doi，papers.year以及与他一起撰写论文的同事的数量。我尝试过

class Foo {val bar = "a"}
class Bar {val foo = "b"}

我遇到的问题可能是，如果我正在寻找的作者针对某篇论文出现两次（因为有两个机构）数量增加了一倍。对于给定的论文，2890的预期结果由

的行数给出

 SELECT  papers.doi, papers.year, count(*) as c
 FROM authors 
 INNER JOIN writtenby ON authors.author_id =  writtenby.author_id
 INNER JOIN writtenby AS writtenby_1 ON writtenby.paper_id =  
 writtenby_1.paper_id
 INNER JOIN papers on  writtenby_1.paper_id = papers.paper_id 
 WHERE authors.name ='Beck' AND authors.firstname= 'H P' 
 GROUP BY papers.doi, papers.year
 ORDER BY c DESC

（使用我的数据：2890行）如果没有区别，我将有3023行，上面的第一个查询给出6046的计数。我尝试在上面的Count子句中使用DISTINCT，但这仍然无法正常工作。

我可以在子查询中使用count吗？感谢您的帮助...

样本数据：

SELECT DISTINCT author_id
FROM writtenby 
WHERE paper_id = 4593

检查查询：

-- Make the tables

CREATE TABLE 'authors' (name collate nocase, firstname collate nocase, see_id integer, 'author_id' INTEGER PRIMARY KEY NOT NULL );
CREATE TABLE 'inst' ('name' TEXT NOT NULL, 'country' TEXT NOT NULL , 'see_id' INTEGER, 'inst_id' INTEGER PRIMARY KEY NOT NULL );
CREATE TABLE 'papers' ('doi' TEXT NOT NULL,'year' TEXT NOT NULL, 'paper_id' INTEGER PRIMARY KEY NOT NULL );
CREATE TABLE 'writtenby' ('paper_id' INTEGER NOT NULL, 'author_id' INTEGER NOT NULL, 'inst_id' INTEGER NOT NULL, PRIMARY KEY ('paper_id', 'author_id', 'inst_id'));

-- Insert the data 

-- authors : 5 names, one with 2 variants

INSERT INTO 'authors' (name, firstname, see_id, author_id) VALUES ('Doe', 'J', 1, 1);
INSERT INTO 'authors' (name, firstname, see_id, author_id) VALUES ('Klein', 'K', 2, 2);
INSERT INTO 'authors' (name, firstname, see_id, author_id) VALUES ('Lang', 'F', 3, 3);
INSERT INTO 'authors' (name, firstname, see_id, author_id) VALUES ('Rue', 'A De La', 6, 4);
INSERT INTO 'authors' (name, firstname, see_id, author_id) VALUES ('La Rue', 'A De', 6, 5);
INSERT INTO 'authors' (name, firstname, see_id, author_id) VALUES ('De La Rue', 'A', 6, 6);
INSERT INTO 'authors' (name, firstname, see_id, author_id) VALUES ('Smith', 'S', 7, 7);

-- inst 4 name, 2 variants

INSERT INTO 'inst' (name, country, see_id, inst_id) VALUES ('Universite de Paris', 'France', 1, 1);
INSERT INTO 'inst' (name, country, see_id, inst_id) VALUES ('Paris University', 'France', 1, 2);
INSERT INTO 'inst' (name, country, see_id, inst_id) VALUES ('Universite de Lyon', 'France', 3, 3);
INSERT INTO 'inst' (name, country, see_id, inst_id) VALUES ('Univ Freiburg', 'Germany', 4, 4);
INSERT INTO 'inst' (name, country, see_id, inst_id) VALUES ('EPFZ', 'Switzerland', 5, 5);
INSERT INTO 'inst' (name, country, see_id, inst_id) VALUES ('Eidg Techn Hochschule', 'Switzerland', 5, 6);

-- papers: 3 papers

INSERT INTO 'papers' (doi, year, paper_id) VALUES ('doi1', '2017', 1);
INSERT INTO 'papers' (doi, year, paper_id) VALUES ('doi2', '2018', 2);
INSERT INTO 'papers' (doi, year, paper_id) VALUES ('doi3', '2018', 3);

-- paper 1: 4 authors

INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (1, 6, 1);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (1, 6, 3);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (1, 1, 5);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (1, 2, 4);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (1, 7, 1);

-- paper 2: 3 authors

INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (2, 6, 1);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (2, 6, 3);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (2, 1, 5);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (2, 2, 5);

-- paper 3: 3 authors

INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (3, 6, 1);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (3, 2, 4);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (3, 6, 3);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (3, 2, 1);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (3, 3, 4);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (3, 3, 5);
INSERT INTO 'writtenby' (paper_id, author_id,  inst_id) VALUES (3, 3, 1);

两个查询均给出错误的结果第一个：

 SELECT  papers.doi, papers.year, count(*) as c
 FROM authors 
 INNER JOIN writtenby ON authors.author_id =  writtenby.author_id
 INNER JOIN writtenby AS writtenby_1 ON writtenby.paper_id =  
 writtenby_1.paper_id
 INNER JOIN papers on  writtenby_1.paper_id = papers.paper_id 
 WHERE authors.name ='De La Rue' AND authors.firstname= 'A' 
 GROUP BY papers.doi, papers.year
 ORDER BY c DESC


SELECT p.doi, p.year, COUNT(w2.author_id) AS cnt
FROM authors a
INNER JOIN writtenby w1
    ON a.author_id =  w1.author_id
INNER JOIN writtenby w2
    ON w1.paper_id = w2.paper_id AND w1.author_id <> w2.author_id
INNER JOIN papers p
    ON w2.paper_id = p.paper_id 
WHERE
    a.name = 'De La Rue' AND a.firstname = 'A'
GROUP BY
    p.doi, p.year
ORDER BY
    cnt DESC;

第二个查询

doi3|2018|14
doi1|2017|10
doi2|2018|8

François

Answer 1

我发现正在发生的一个计数问题是您在writtenby表的自联接中。此处，您没有检查匹配的行是否具有不同 author_id。如果author_id是相同的，那么您不应该计算它。另外，您要计算的共享作者数量是第二个writtenby表。这样，如果给定作者没有任何共同作者，则计数将显示为零。

SELECT p.doi, p.year, COUNT(w2.author_id) AS cnt
FROM authors a
INNER JOIN writtenby w1
    ON a.author_id =  w1.author_id
INNER JOIN writtenby w2
    ON w1.paper_id = w2.paper_id AND w1.author_id <> w2.author_id
INNER JOIN papers p
    ON w2.paper_id = p.paper_id 
WHERE
    a.name = 'Beck' AND a.firstname = 'H P'
GROUP BY
    p.doi, p.year
ORDER BY
    cnt DESC;

Answer 2

借助Tim Biegeleisen和示例数据，我发现计数中缺少子句DISTINCT

SELECT p.doi, p.year, COUNT(DISTINCT w2.author_id) AS cnt
FROM authors a
INNER JOIN writtenby w1
ON a.author_id =  w1.author_id
INNER JOIN writtenby w2
ON w1.paper_id = w2.paper_id 
INNER JOIN papers p
ON w2.paper_id = p.paper_id 
WHERE
a.name = 'De La Rue' AND a.firstname = 'A'
GROUP BY
p.doi, p.year
ORDER BY
cnt DESC;

提供作者总数。

doi1    2017    4
doi2    2018    3
doi3    2018    3

在子句w1.author_id <> w2.author_id中，计数减少了一个。

F。

SQL：计数和子查询

2 个答案: