如何连接表并显示？

Question

我尝试合并两张桌子。示例，表 A 和 B。表 A 有 2 列（id 和 value）表 A 有 3 行 (1, a)、(1, b) 和 (1, c)。表 B 有 1 列和 1 行（id）为（1）我尝试使用join table执行此操作，结果：3 行 id 和值为 (1, a), (1, b), (1, c)这不是我想要的，我需要创建如下：结果：只有 1 行，id 为 (1)，value 为 (a, b, c)但是，我只想从“value”中获取“c”数据。那么，我想要的结果是(1, c)，怎么样？编辑：检查，有工作：http://sqlfiddle.com/#! 9/b0fd3b/2它不适用于我的代码，未找到结果。检查一下：<?phpif (isset($_GET['page_no']) && $_GET['page_no']!="") {  $page_no = $_GET['page_no'];}else {  $page_no = 1;}$total_records_per_page = 3;$offset = ($page_no-1) * $total_records_per_page;$previous_page = $page_no - 1;$next_page = $page_no + 1;$adjacents = "2"; $result_count = mysqli_query($conn, "SELECT COUNT(*) As total_records FROM b");$total_records = mysqli_fetch_array($result_count);$total_records = $total_records['total_records'];  $total_no_of_pages = ceil($total_records / $total_records_per_page);$second_last = $total_no_of_pages - 1; // total page minus 1$result = mysqli_query($conn, "SELECT id, a.meta_key, a.meta_value FROM a JOIN b USING (id) WHERE a.meta_value = 'value2' LIMIT $offset, $total_records_per_page");if (mysqli_num_rows($result) > 0) {  while ($row = mysqli_fetch_array($result)) {    echo $row['id'];    echo $row['key'];    echo $row['value'];  }}else {  echo "no found";}?>// pagination....

Answer 1

尝试下面的查询。我希望它能得到你想要的结果。

SELECT a.id, a.value, 

FROM a

INNER JOIN b ON a.id=b.id; 

ORDER BY

    a.id DESC;

Answer 2

只有 1 行，id 为 (1)，value 为 (a, b, c)

SELECT id, GROUP_CONCAT(value ORDER BY value) values

FROM a

JOIN b USING (id)

GROUP BY id

但是，我只想从“value”中获取“c”数据。

SELECT id, a.value

FROM a

JOIN b USING (id)

WHERE a.value = 'c'

Answer 3

你可以这样做：

select a.*

from a join

     b

     on a.id = b.id

order by a.value desc

limit 1;

我对为什么join需要 a 有点困惑：

select a.*

from a

order by value desc

limit 1;

但你似乎想要join.