Question

我有两张这样的表：

帖子表：

post_id    subject    
   1         abc
   2         def

和内容表，其中post_id列是外键：

content_id    post_id     content              date
     1           1        main content       2014-07-31 03:02:00
     2           1        reply one          2014-07-31 03:03:20
     3           1        reply two          2014-07-31 03:06:32

如您所见，内容表存储了帖子的内容及其可能的复制内容。在此示例中，post_id＃1有其主要内容和两个回复。

我想要做的是回显主题及其内容，但不是回复。所以我编写了这样的查询：

$query = "SELECT subject, content
          FROM posts As p
          INNER JOIN contents AS c USING (post_id)
          ORDER BY post_id DESC";
$result = mysqli_query ($dbc, $query)

while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
    echo "{$row['content']}n\";
}

然后你知道吗？它回应了帖子的最新内容，即content_id＃3。但是，我想要做的是回应帖子的内容，这是最早的content_id post_id＃1。

请帮帮我吗？

Answer 1

MIN(DATE)必须满足要求，因此您需要将id与DATE

分组

在post_id的基础上加入两个表格，然后根据min(date)

进行分组

$query = "SELECT p.post_id,  p.subject, c.content, c.date
          FROM posts p
          INNER JOIN contents c on p.post_id = c.post_id
          WHERE date = (select min(date) 
                        from contents ct 
                        where ct.post_id = p.post_id 
                        )
          GROUP BY post_id,subject,content, date
          ORDER BY post_id DESC";

$result = mysqli_query ($dbc, $query)

while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
    echo "{$row['content']}n\";
}

Answer 2

<强>已更新

它基本上选择子查询中每个帖子的最小日期，并将其用作“正常”查询的关键字。

SELECT distinct subject, content FROM content c
JOIN post p 
on c.post_id = p.post_id
WHERE cast(c.post_id as char)+cast(c.contentdate as char)in               
        (
        SELECT 
        cast(p.post_id as char)+cast(min(c.contentdate) as char) as KeyValue
        FROM post p
        INNER JOIN content c on p.post_id = c.post_id
        GROUP BY p.post_id
        )

这让我回报：

subject | content
----------------------
asd     | main content
def     | main content

Answer 3

您还需要按content_id（或日期）订购结果。所以sql查询可能如下所示：

$query = "SELECT subject, content
      FROM posts As p
      INNER JOIN contents AS c USING (post_id)
      ORDER BY post_id DESC, content_id ASC";

当然，我假设主要内容已在回复之前添加到数据库中，并且您不会更改行索引，这可能导致第一行具有某些post_id的情况不是实际内容。

因此，如果您想打印出代码中需要执行某些操作的所有内容。您只需要打印具有相同post_id的第一行：

$last_post_id = 0;
while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)) {
    if ($last_post_id != $row['post_id']) {
        echo "{$row['content']}n\";
    }
    $last_post_id = $row['post_id'];
}

Answer 4

试试这个

$query = "SELECT subject, content
          FROM posts As p
          INNER JOIN contents AS c USING (post_id)
          ORDER BY date ASC, post_id DESC";

Answer 5

在测试了所有答案之后，我选择了 @NoobEditor 的答案。但是，我必须更清楚地说明如下：

$query = "SELECT p.post_id,  p.subject, c.content, c.date
          FROM posts p
          INNER JOIN contents c on p.post_id = c.post_id
          WHERE date = (select min(date) 
                        from contents c 
                        where c.post_id = p.post_id 
                       )
         GROUP BY c.post_id
         ORDER BY p.post_id DESC";

$result = mysqli_query ($dbc, $query)

我在GROUP和ORDER BY函数中更清楚了。并使用 c 作为内容表的别名，但不需要使用 c 和 ct

while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {
     echo "{$row['content']}n\";
}

如何回应帖子的内容

5 个答案: