Sqlite 返回在列中具有相同项的不同名称对

我想在表中找到不同的名称对，它们在items列中具有相同的精确项。例如：

CREATE TABLE t
(
    name    VARCHAR(255),
    item    VARCHAR(255)
);

INSERT INTO t VALUES("Alice", "Orange");
INSERT INTO t VALUES("Alice", "Pear");
INSERT INTO t VALUES("Alice", "Lemon");
INSERT INTO t VALUES("Bob", "Orange");
INSERT INTO t VALUES("Bob", "Pear");
INSERT INTO t VALUES("Bob", "Lemon");
INSERT INTO t VALUES("Charlie", "Pear");
INSERT INTO t VALUES("Charlie", "Lemon");

这里的答案是爱丽丝，鲍勃，因为他们拿的是完全相同的东西

我想用NOT EXISTS/NOT进行双重否定，我认为这更适合这个问题，但我想不出任何接近功能的东西

---

这有点类似于问题，但我使用的是SQLite，所以我不能使用GROUP_CONCAT，但我想知道如何使用关系划分（使用NOT EXISTS/NOT IN）来完成。我可能已经找到了解决您问题的方法。我的是使用MySQL测试的，但它没有使用GROUP_CONCAT。它可能适用于您的SQLite数据库。我的查询用于查找购买相同物品的人

尝试使用以下语句： 从t e1、t e2中选择不同的e1.name、e2.name，其中e1.item=e2.item和e1.name！=e2.按e1命名组。计数*大于1的项目；

---

这似乎适用于SQLLite

    select t1.name
    from t t1
        join t t2 on t1.name <> t2.name and t1.item = t2.item 
        join (select name, count(*) as cnt from t group by name) t3 on t3.name = t1.name
        join (select name, count(*) as cnt from t group by name) t4 on t4.name = t2.name
    group by t1.name, t3.cnt, t4.cnt
    having count(*) = max(t3.cnt, t4.cnt)

---

要获取所有名称对之间的公共项目数，可以使用以下查询：

SELECT t1.name AS name1, t2.name AS name2, COUNT(*) AS cnt
FROM t AS t1
INNER JOIN t AS t2 ON t1.item = t2.item AND t1.name < t2.name
GROUP BY t1.name, t2.name

现在，您只需要过滤掉计数不等于name1和name2项数的name1和name2对。可以使用具有相关子查询的HAVING子句执行此操作：

SELECT t1.name AS name1, t2.name AS name2
FROM t AS t1
INNER JOIN t AS t2 ON t1.item = t2.item AND t1.name < t2.name
GROUP BY t1.name, t2.name
HAVING COUNT(*) = (SELECT COUNT(*) FROM t WHERE name = t1.name) AND 
       COUNT(*) = (SELECT COUNT(*) FROM t WHERE name = t2.name)

与：

使用NOT IN是可能的，bit以更复杂的方式表达完全相同的机制：

SELECT t1.name, t2.name
FROM t AS t1, t AS t2
GROUP BY t1.name, t2.name
HAVING t1.name < t2.name
   AND NOT EXISTS (SELECT item
                   FROM t
                   WHERE name = t1.name
                     AND item NOT IN (SELECT item
                                      FROM t
                                      WHERE name = t2.name))
   AND NOT EXISTS (SELECT item
                   FROM t
                   WHERE name = t2.name
                     AND item NOT IN (SELECT item
                                      FROM t
                                      WHERE name = t1.name));

---

我给你做了一把SQLFiddle在这里玩~你的桌子上可以有多少不同的东西？@TimBiegeleisen，你想要多少就有多少。我认为它不会影响任何东西，只要它仍然可以返回包含完全相同的项集的对@队长，好吧，我的错，但我真的很想得到一些提示，在不使用集体协商的情况下解决这个问题。。。

SELECT t1.name, t2.name
FROM t AS t1, t AS t2
GROUP BY t1.name, t2.name
HAVING t1.name < t2.name
   AND NOT EXISTS (SELECT item FROM t WHERE name = t1.name
                   EXCEPT
                   SELECT item FROM t WHERE name = t2.name)
   AND NOT EXISTS (SELECT item FROM t WHERE name = t2.name
                   EXCEPT
                   SELECT item FROM t WHERE name = t1.name);

SELECT t1.name, t2.name
FROM t AS t1, t AS t2
GROUP BY t1.name, t2.name
HAVING t1.name < t2.name
   AND NOT EXISTS (SELECT item
                   FROM t
                   WHERE name = t1.name
                     AND item NOT IN (SELECT item
                                      FROM t
                                      WHERE name = t2.name))
   AND NOT EXISTS (SELECT item
                   FROM t
                   WHERE name = t2.name
                     AND item NOT IN (SELECT item
                                      FROM t
                                      WHERE name = t1.name));