Question

在按摩系统中，我想选择用户的所有对话。

我的msg表格如下：

id  | mfrom | mto  | date
=========================
1   | 10    | 12   | 0000
2   | 14    | 10   | 0000
3   | 10    | 14   | 0000
4   | 16    | 10   | 0000
5   | 10    | 16   | 0000

这里我的用户ID说：10。如上所述，我有3个用户ID为12,14,16的对话

我使用这个sql让用户ID与我交谈：

$results = mysqli_query($db,"SELECT mfrom FROM msg WHERE `mto`='10' GROUP BY `mfrom` ORDER BY `date` DESC") or die(mysqli_error($db));
    while($rows = mysqli_fetch_assoc($results)) {
    $mfroms[] = $rows['mfrom'];
}

$results = mysqli_query($db,"SELECT mto FROM msg WHERE `mfrom`='10' GROUP BY `mto` ORDER BY `date` DESC") or die(mysqli_error($db));
    while($rows = mysqli_fetch_assoc($results)) {
    $mtos[] = $rows['mto'];
}

$result = array_unique(array_merge($mfroms,$mtos), SORT_REGULAR);

上面的查询给了我一个数组结果12,14,16

现在我的问题是，我无法通过小组获取与我的所有对话。这意味着，我想从上表中得到如下结果：

user 12 conversation with user 10(me) total 1 display recent massage only
-------------------------------------------------------------------------
user 14 conversation with user 10(me) total 2 display recent massage only
-------------------------------------------------------------------------
user 16 conversation with user 10(me) total 2 display recent massage only

Answer 1

将您的表名替换为AccountTable，将10替换为您的用户ID并对其进行测试。

{{1}}

Answer 2

如果我理解正确的话：

SELECT s.user1,s.user2,count(*) as recent_messages
FROM (
    SELECT CASE WHEN t.mfrom > t.mto THEH t.mfrom ELSE t.mto END as user1,
           CASE WHEN t.mfrom > t.mto THEH t.mfrom ELSE t.mto END as user2
    FROM YourTable t) s
GROUP BY s.user1,s.user2

Answer 3

这是一种方法：

select (case when m.mfrom = 10 then m.mto else m.mfrom end) as other_user,
       10 as me,
       count(*),
       min(id) as message_id
from messages m
where 10 in (m.mfrom, m.mto)
group by other_user;

Sql按组选择所有会话

3 个答案: