Question

我在MySQL中使用连接相对较新。

我有两张桌子：

sh_subscriptions
    => id
    => user_id
    => feed_id

sh_feeds
    => id
    => shop_name
    => feed_id

我正在尝试从sh_subscriptions获取所有字段feed_id对应于feed_id对应sh_feed的{{1}}以及$row['shop_name'] {1}}对应user_id。

这是我的尝试：

$id['id]

更新

我现在有以下内容：

SELECT * FROM sh_subscriptions s
INNER JOIN sh_feeds f ON s.shop_id = f.feed_id
WHERE s.id = '" . $id['id'] . "'
AND f.shop_name = '" . $row['shop_name'] . "'

我可以确认数据库字段全部存在并且变量正确打印出来，但是尽管已经包含了错误处理程序，但while($row = mysqli_fetch_array($query)) { echo "<div class='col-md-4'>"; echo $row['shop_name'] . " "; $query = mysqli_query($con, "SELECT * FROM sh_subscriptions s INNER JOIN sh_feeds f ON s.feed_id = f.feed_id WHERE s.user_id = '" . $id['id'] . "' AND f.shop_name = '" . $row['shop_name'] . "'") or die(mysql_error($con)); echo "</div>"; }内的任何内容都没有打印出来。

Answer 1

试试这个：

SELECT * FROM sh_subscriptions s
INNER JOIN sh_feeds f ON s.id = f.feed_id
WHERE s.id = '" . $id['id'] . "'
AND f.shop_name = '" . $row['shop_name'] . "'") or die(mysql_error($con));

Answer 2

我认为应该是

SELECT * FROM sh_subscriptions s
INNER JOIN sh_feeds f ON s.feed_id = f.feed_id
WHERE s.id = '" . $id['id'] . "'
AND f.shop_name = '" . $row['shop_name'] . "';

MySQL连接查询似乎有效但未按预期执行

2 个答案: