2017-08-04 70 views
0

這是我工作的功能。聲明不funtion使用PostgreSQL

這是輸出:

ERROR: invalid type name "CONCAT(translate,'=true') INTO tempo FROM seo_content WHERE name"

即使當我刪除的concat和我只是不停地翻譯,我得到

ERROR: invalid type name "translate INTO tempo FROM seo_content WHERE name"

感謝

+0

嘗試在開始之前聲明..是'翻譯'一些column?..你需要返回函數返回不是void –

+0

translate是一個列並且翻譯是OUT var –

+0

啊,IC和'SET translation ='A';'應該怎麼做? –

回答

0

試試這個:

CREATE OR REPLACE FUNCTION public.func_leonard_test2(
    IN url text, 
    OUT translation text) 
    RETURNS text AS 
$BODY$ 

DECLARE tempo text; 
BEGIN 

SELECT CONCAT(translate,'=true') INTO tempo FROM seo_content WHERE "name"=url;  

    CASE 

     WHEN tempo = 'location_town' THEN 

      translation := 'A'; 

     ELSE 

      translation := 'B'; 

    END CASE; 

END;$BODY$ language plpgsql;