通過實體框架更新MySQL視圖=「每個派生表都必須具有自己的別名」

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我試圖通過實體框架更新MySQL視圖並獲取錯誤「每個派生表都必須有自己的別名」。通過實體框架更新MySQL視圖=「每個派生表都必須具有自己的別名」

有關如何改變我的觀點以防止這種情況的任何想法？這裏是我的MySQL視圖代碼：

select 
    a.compid AS compid, 
    a.loid AS loid, 
    a.purchase_rate AS purchase_rate, 
    a.purchase_min AS purchase_min, 
    a.purchase_max AS purchase_max, 
    a.refi_rate AS refi_rate, 
    a.refi_min AS refi_min, 
    a.refi_max AS refi_max, 
    a.quarter AS quarter, 
    a.year AS year, 
    a.changed_on AS changed_on, 
    a.user_changed AS user_changed, 
    a.is_active AS is_active, 
    a.notes_id AS notes_id, 
    b.branch AS branchID, 
    b.name AS lo_name, 
    c.branch_name_friendly AS branchname 
from ((encompassdata.comp a 
    inner join default.users_encompass b on 
    (
     (a.loid = b.un_enc) 
    )) 
    inner join default.branch_info c on 
    (
     (b.branch = c.branch_id) 
    ))

來源

2013-03-22 Chronos

什麼用'FROM'畢竟這些括號的起來？從我可以告訴他們都不需要。 – 2013-03-22 00:31:30

是的......當我在我使用的MySQL管理工具中創建視圖時，括號會自動生成。我只是在沒有它們的情況下創建了視圖，並且仍然出現相同的錯誤。 – Chronos 2013-03-22 19:37:04

嘗試溝所有的括號內的FROM條款

SELECT a.compid AS compid, 
     ... 
    FROM encompassdata.comp a INNER JOIN 
     default.users_encompass b on a.loid = b.un_enc INNER JOIN 
     default.branch_info c on b.branch = c.branch_id

來源

2013-03-22 01:05:13 peterm

通過實體框架更新MySQL視圖=「每個派生表都必須具有自己的別名」

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