我想要的是找到与user2相关联的每个朋友（id = 2），然后找到每个user2朋友和user1之间的共同朋友的数量。
function whereIsAlice(names) {
for (var i = 0; i < names.length; i++) {
if (names[i].firstName == "Alice") {
return i;
}
}
// When not found in above loop then return -1, not found!
return -1;
}
var friends1 = [
{ firstName: 'John', lastName: 'Gaudet' },
{ firstName: 'Lisa', lastName: 'Mcclanahan' },
{ firstName: 'Alice', lastName: 'Vore' }, // Alice is here, at index 2
{ firstName: 'Marine', lastName: 'Salsbury' }
];
var friends2 = [
{ firstName: 'Tim' },
{ firstName: 'Arthur' },
{ firstName: 'Juan' }
];
console.log(whereIsAlice(friends1)); //Should be 2
console.log(whereIsAlice(friends2)); // Should be -1
表用户
mysql to find user2 Friends and get mutual_count of Friends with user1 ( with every user2 friend)表朋友
user_id | username
------------------
1 | user1
2 | user2
3 | user3
4 | user4
5 | user5
6 | user6
7 | user7
预期输出：用户2的朋友//（朋友姓名，朋友ID，当前朋友ID和用户1之间的朋友相互计数（id = 1））
user_one_id | user_two_id
------------------------
1 | 4
1 | 5
1 | 6
2 | 3
2 | 4
3 | 1
3 | 4
5 | 2
5 | 3
5 | 4
6 | 2
6 | 3
7 | 2
mysql语句
username | user_id | mutual_count
------------------------
user3 | 3 | 3 // user3 and user1 mutual -> (user4,user5,user6)
user4 | 4 | 2 // user4 and user1 mutual -> (user3 ,user5)
user5 | 5 | 2 // user5 and user1 mutual -> (user3 ,user4)
user6 | 6 | 1 // user6 and user1 mutual -> ( user3)
user7 | 7 | 0 // user7 and user1 mutual -> (0)
我得到的错误 未知栏＆＃39; users.user_id＆＃39;在＆＃39; where子句＆＃39; 获取mutual_count canot的问题子查询知道users.user_id
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答案 0 :(得分：2)
我修改了你获得所需输出的尝试（但可能有更好的方法来获得你想要的东西）。
似乎主要问题是所有子查询都无法访问顶部外部查询中的users表。所以我修改了需要该表的子查询，以便它可以通过其直接外部查询中的users子句访问WHERE表。
SELECT users.username,users.user_id,
(SELECT count(a.friendID) FROM
( SELECT user_two_id friendID, user_one_id where_id FROM friends
UNION
SELECT user_one_id friendID, user_two_id where_id FROM friends
) AS a
JOIN
( SELECT user_two_id friendID FROM friends WHERE user_one_id = 1
UNION
SELECT user_one_id friendID FROM friends WHERE user_two_id = 1
) AS b
ON a.friendID = b.friendID
WHERE a.where_id = users.user_id
) as mutual_count
FROM friends LEFT JOIN users
ON friends.user_one_id = users.user_id or friends.user_two_id = users.user_id
WHERE (friends.user_one_id = 2 OR friends.user_two_id = 2) AND users.user_id !=2
order by user_id
<强>输出：
+----------+---------+--------------+
| username | user_id | mutual_count |
+----------+---------+--------------+
| user3 | 3 | 3 |
+----------+---------+--------------+
| user4 | 4 | 2 |
+----------+---------+--------------+
| user5 | 5 | 2 |
+----------+---------+--------------+
| user6 | 6 | 1 |
+----------+---------+--------------+
| user7 | 7 | 0 |
+----------+---------+--------------+
修改后的行：
第4行： SELECT user_two_id friendID, user_one_id where_id FROM friends
第6行： SELECT user_one_id friendID, user_two_id where_id FROM friends
第14行： WHERE a.where_id = users.user_id