Question

我正在尝试在 MySQL 中SET 3个变量，并获取其中两个变量的和。

第一个变量@cFollow和@cComment，应该各自返回一个整数值（返回多少行的计数）；第三个是这两个整数的和。

这是我的 SQL ：

SET @cFollow = (SELECT COUNT(*) FROM followers WHERE unix > :unix AND following = :user);
SET @cComment = (SELECT COUNT(*) FROM comments WHERE comment_unix > :unix AND comment_track IN (SELECT upload_id FROM uploads WHERE upload_artist = :user));

SET @total = @cFollow + @cComment;

SELECT @total;

当我在PHPMyAdmin上对其进行测试时，它返回了正确的值，并且运行良好。但是，当我在 PHP 中对其进行测试时，它返回了一个空数组。

这是我的 PHP ：

$holdPoint = (int)Input::get("hold_point");

$_SQL = "
SET @cFollow = (SELECT COUNT(*) FROM followers WHERE unix > :unix AND following = :user);
SET @cComment = (SELECT COUNT(*) FROM comments WHERE comment_unix > :unix AND comment_track IN (SELECT upload_id FROM uploads WHERE upload_artist = :user));

SET @total = @cFollow + @cComment;

SELECT @total;";
$_PARAMS = [":unix" => $holdPoint, ":user" => $user_id];

$check = DB::getInstance()->queryPro($_SQL, $_PARAMS);  
var_dump($check);

这是该var_dump的结果：

array(0){} // not very impressive...
           // should be something like int(1) instead

我整夜都在搜索如何通过 MySQL 查询在 PHP 中返回变量，这是我所了解的。 / p>

感谢所有帮助，
干杯。

Answer 1

此答案并不是真正的答案，而是更多的评论。
另请注意，您的查询

SET @cFollow = (SELECT COUNT(*) FROM followers WHERE unix > :unix AND following = :user);
SET @cComment = (SELECT COUNT(*) FROM comments WHERE comment_unix > :unix AND comment_track IN (SELECT upload_id FROM uploads WHERE upload_artist = :user));

SET @total = @cFollow + @cComment;

SELECT @total;

最有可能被重写为一个查询

SELECT 
 SUM(alias.c) AS total
FROM (
  SELECT COUNT(*) AS c FROM followers WHERE unix > :unix AND following = :user
  UNION ALL
  SELECT COUNT(*) AS c FROM comments WHERE comment_unix > :unix AND comment_track IN (SELECT upload_id FROM uploads WHERE upload_artist = :user)
) AS alias

选择SQL变量返回一个空数组

1 个答案: