在phpmyadmin中顯示錯誤的Mysql查詢

Question

我想找出具有短代碼的頁面ID。這是我想在MySQL phpmyadmin

SELECT ID FROM wp_posts WHERE post_type = "page" AND post_status="publish" AND post_content LIKE "%[registration id="17"]%" 

運行查詢，但它顯示以下錯誤

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '17"]%" 

LIMIT 0，25' 列提前1個

感謝。

Answer 1

你在你的SQL

SELECT ID FROM wp_posts WHERE post_type = "page" AND 
post_status="publish" AND post_content LIKE "%[registration id="17"]%" 
                   ^^^^^^ 

使用'，而不是"像部分地方有問題。

SELECT ID FROM wp_posts WHERE post_type = "page" AND 
post_status="publish" AND post_content LIKE "%[registration id='17']%" 

如果你真的想匹配 registration id="17"，試圖逃脫"。

SELECT ID FROM wp_posts WHERE post_type = "page" AND 
post_status="publish" AND post_content LIKE "%[registration id=\"17\"]%" 

OR

SELECT ID FROM wp_posts WHERE post_type = "page" AND 
post_status="publish" AND post_content LIKE '%[registration id="17"]%' 

Answer 2

嘗試這個

SELECT ID FROM wp_posts WHERE post_type = "page" AND post_status="publish" AND post_content LIKE "%[registration id='17']%"; 

或

SELECT ID FROM wp_posts WHERE post_type = "page" AND post_status="publish" AND post_content LIKE '%[registration id="17"]%';