2017-03-16 62 views
1

我有這個疑問,運作良好,這一起加入了幾個表,這都沒有問題,運作良好MYSQL加入所有表並獲得2值之間的字段記錄?

"SELECT post_info.post_id,username,profile_image_path,status_body,image_path,post_created_at FROM post_info 
LEFT OUTER JOIN users_info ON post_info.user_id = users_info.user_id 
LEFT OUTER JOIN status_of_post ON post_info.status_id = status_of_post.status_id 
LEFT OUTER JOIN image_of_post ON post_info.image_id =image_of_post.image_id" 

我想要得到的記錄(POST_ID + 1)和(POST_ID + 20) ,所以我修改它成爲這個

SELECT post_info.post_id,username,profile_image_path,status_body,image_path,post_created_at FROM post_info 
LEFT OUTER JOIN users_info ON post_info.user_id = users_info.user_id 
LEFT OUTER JOIN status_of_post ON post_info.status_id = status_of_post.status_id 
LEFT OUTER JOIN image_of_post ON post_info.image_id =image_of_post.image_id 
WHERE post_id BETWEEN (post_id + 1) AND (post_id + 20) 

它陳述了這個結果。

1046 - 無數據庫中選擇

我想每次20後查詢運行的基礎上,post_id的範圍內,是什麼做的正確方法?

+0

你在哪裏運行你的查詢? –

+0

我運行在phpMyAdmin – ken

+0

確保你想運行查詢的數據庫被選中 –

回答

0

用於在phpMyAdmin

enter image description here

OR

選擇確保您正在運行的查詢數據庫

您可以更改您的查詢,包括數據庫名是這樣的:

SELECT ..... FROM DATABASE-Name.post_info 
+1

好的,謝謝..我在這裏犯了一個愚蠢的錯誤..感謝您的幫助 – ken

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