從Wordpress數據庫中的單個列中選擇多個值

Question

我試圖輸出帖子的詳細信息以及wordpress數據庫中的一些自定義字段。 我已經自定義字段稱爲「縮略圖」和「視頻」

SELECT ID, post_title, post_content, wp_postmeta.meta_value 
FROM wp_posts 
INNER JOIN wp_postmeta ON wp_posts.id=wp_postmeta.post_id 
WHERE wp_postmeta.meta_key = "thumnail" 
OR wp_postmeta.meta_key = "video" 

到目前爲止，我越來越喜歡這個人行：

| ID | Title  | Content  | Custom Fields      | 
|--------------------------------------------------------------------| 
| 234 | INCEPTION | Content etc | wp-content/thumbnail/inception.jpg | 
| 234 | INCEPTION | Content etc | wp-content/video/inception.flv  | 

但我想是（即平均每個職位一行）

|ID |Title  |Content  |Thumbnail     |Video      | 
|-----------------------------------------------------------------------------------| 
|234|INCEPTION |Content etc |wp-content/..inception.jpg |wp-content/..inception.flv | 

有人可能會建議如何將不同的值分隔到SELECT語句中的列中嗎？

Answer 1

這個過程被稱爲「數據透視表」或「交叉表報告」。 MySQL中沒有PIVOT命令，但可以手動完成，這裏有兩個例子。第一個例子使用IF statement，第二個例子使用CASE statement。 aggregate function用於擺脫重複的行。

SELECT ID, post_title, post_content, 
MAX(IF (wp_postmeta.meta_key = "thumnail", wp_postmeta.meta_value, '')) AS Thumbnail, 
MAX(IF (wp_postmeta.meta_key = "video", wp_postmeta.meta_value, '')) AS Video 
FROM wp_posts 
INNER JOIN wp_postmeta ON wp_posts.id=wp_postmeta.post_id 
WHERE wp_postmeta.meta_key IN ("thumnail", "video") 
GROUP BY ID 

SELECT ID, post_title, post_content, 
MAX(CASE WHEN wp_postmeta.meta_key = "thumnail" THEN wp_postmeta.meta_value ELSE '' END) AS Thumbnail, 
MAX(CASE WHEN wp_postmeta.meta_key = "video" THEN wp_postmeta.meta_value ELSE '' END) AS Video 
FROM wp_posts 
INNER JOIN wp_postmeta ON wp_posts.id=wp_postmeta.post_id 
WHERE wp_postmeta.meta_key IN("thumnail", "video") 
GROUP BY ID