Question

我有以下数据库结构：

tbl_record（_id，_id_user，...）
tbl_photo（_id，_id_record，...）
tbl_note（_id，_id_record，...）

在计算记录所拥有的照片数量时列出特定用户的记录时，我使用以下查询，该工作正常：

SELECT tbl_record._id, COUNT(tbl_photo._id_record) AS photo_count FROM tbl_record
   LEFT OUTER JOIN tbl_photo ON tbl_record._id=tbl_photo._id_record
WHERE tbl_record._id_user=? GROUP BY tbl_record._id;

现在，我想像上面那样做，但也要计算记录的笔记数量：

SELECT tbl_record._id, COUNT(tbl_photo._id_record) AS photo_count, COUNT(tbl_note._id_record) AS note_count FROM tbl_record
   LEFT OUTER JOIN tbl_photo ON tbl_record._id=tbl_photo._id_record
   LEFT OUTER JOIN tbl_note ON tbl_record._id=tbl_note._id_record
WHERE tbl_record._id_user=? GROUP BY tbl_record._id;

当记录包含＆gt; 0张照片时，第二个查询的计数无法正常工作＆gt; 0个音符，例如3张照片＆amp; 5张照片，每张照片的计数为15（3 * 5）。

知道如何让第二个查询返回正确的计数吗？

谢谢！

Answer 1

您可以使用COUNT(DISTINCT some_id)过滤掉重复项，但这样效率很低。更好地使用相关子查询：

SELECT _id,
       (SELECT COUNT(*)
        FROM tbl_photo
        WHERE _id_record = tbl_record._id
       ) AS photo_count,
       (SELECT COUNT(*)
        FROM tbl_note
        WHERE _id_record = tbl_record._id
       ) AS note_count
FROM tbl_record
WHERE _id_user = ?

Sqlite双重左外连接与计数

1 个答案: