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我有一個表插入從選擇返回另一個值
|| *content_id* || *content_value* || *content_const* || *language_language_id* ||
|| 1 || title_1 || heading_1 || 1 ||
|| 2 || title_2 || heading_2 || 1 ||
|| 23 || subtitle_1 || sub_1 || 1 ||
|| 24 || subtitle_2 || sub_2 || 1 ||
|| 50 || title_1 || heading_1 || 2 ||
|| 51 || title_2 || heading_2 || 2 ||
|| 48 || title_2 || heading_2 || 3 ||
|| 53 || title_1 || heading_1 || 3 ||
我需要插回表LANGUAGE_ID究竟存在於表中的值(在這種情況下,id爲2和3)什麼什麼我添加任何語言ID。這是做什麼的正確方法?
我需要的結果:
|| *content_id* || *content_value* || *content_const* || *language_language_id* ||
|| 1 || title_1 || heading_1 || 1 ||
|| 2 || title_2 || heading_2 || 1 ||
|| 23 || subtitle_1 || sub_1 || 1 ||
|| 24 || subtitle_2 || sub_2 || 3 ||
|| 48 || title_2 || heading_2 || 3 ||
|| 50 || title_1 || heading_1 || 2 ||
|| 51 || title_2 || heading_2 || 2 ||
|| 53 || title_1 || heading_1 || 3 ||
|| 62 || subtitle_1 || sub_1 || 1 ||
|| 63 || subtitle_1 || sub_1 || 2 ||
|| 64 || subtitle_1 || sub_1 || 2 ||
|| 65 || subtitle_1 || sub_1 || 3 ||
我有一個查詢什麼給我很好的結果沒有語言ID。 SELECT * FROM .content GROUP BY .content.content_const HAVING COUNT(.content.content_const) = 1
,但我需要它插回與其他語言ID ... :( 謝謝你的幫助