SQL Server：結合兩個表

Question

我有2個大表，每個大約25列。它們都包含相同類型的數據，但列的順序不同。

如何組合這些表而無需重新排列我的查詢代碼？我寧願不要爲〜600行腳本做這件事。

如果需要，我可以舉例說明它的樣子，但我並不真正看到需要。

我到目前爲止嘗試過;

SELECT * FROM [guest].table1 UNION ALL SELECT * FROM [guest].table2; 
SELECT * FROM [guest].table1, [guest].table2; 

我也嘗試插入數據像這樣;

SET IDENTITY_INSERT [guest].table1 ON; 
SET IDENTITY_INSERT [guest].table2 ON; 

INSERT INTO [guest].table1 id, short_name, name, invention_title, reference, client_id, client_ref, date_case_opened, date_case_closed, case_type, notes, fee_earner, created, last_updated, file_location, foreign_attorney_id, foreign_attorney_ref, country_code, application_number, filing_date, publication_number, invoice_currency, publication_date, status, pct_application_number, pct_case_id, national_phase_entry, base_number, base_country, base_date, base_case_id, divisional_date_lodged, illustrative_image, parent_case_id, parent_application_number) 
select id, short_name, name, invention_title, reference, client_id, client_ref, date_case_opened, date_case_closed, case_type, notes, fee_earner, created, last_updated, file_location, foreign_attorney_id, foreign_attorney_ref, country_code, application_number, filing_date, publication_number, invoice_currency, publication_date, status, pct_application_number, pct_case_id, national_phase_entry, base_number, base_country, base_date, base_case_id, divisional_date_lodged, illustrative_image, parent_case_id, parent_application_number 
FROM [guest].table2; 

但是，這給了我這個錯誤：

SQL Error: Cannot insert explicit value for identity column in table 'exported_cases' when IDENTITY_INSERT is set to OFF.

任何幫助將不勝感激！

爲了澄清，我導出了一個當前存在於2個部分中的數據庫（表1和表2）。它已經生成了PK，並且在它完全導出之前我沒有插入它，此時PK將被設置爲比任何當前PK更高的數字。

（我需要保持保留ID字段。）

Answer 1

由於大家都在專注於插入，讓我回答選擇部分:)

爲聯盟工作，確保以相同的順序明確選擇所有列：

SELECT id, short_name, name, invention_title, reference, client_id, client_ref, date_case_opened, date_case_closed, case_type, notes, fee_earner, created, last_updated, file_location, foreign_attorney_id, foreign_attorney_ref, country_code, application_number, filing_date, publication_number, invoice_currency, publication_date, status, pct_application_number, pct_case_id, national_phase_entry, base_number, base_country, base_date, base_case_id, divisional_date_lodged, illustrative_image, parent_case_id, parent_application_number 
FROM [guest].table1 
UNION ALL 
SELECT id, short_name, name, invention_title, reference, client_id, client_ref, date_case_opened, date_case_closed, case_type, notes, fee_earner, created, last_updated, file_location, foreign_attorney_id, foreign_attorney_ref, country_code, application_number, filing_date, publication_number, invoice_currency, publication_date, status, pct_application_number, pct_case_id, national_phase_entry, base_number, base_country, base_date, base_case_id, divisional_date_lodged, illustrative_image, parent_case_id, parent_application_number 
FROM [guest].table2 

這個結果集可以用來查詢，插入到一個新的表格或者其他東西:)

Answer 2

you should not insert identity element its generated auto: 

just try inserting all except id(which is your identity) 

請試試這個：

SET IDENTITY_INSERT [guest].table1 ON; 
    SET IDENTITY_INSERT [guest].table2 ON; 

    INSERT INTO [guest].table1 id, short_name, name, invention_title, reference, client_id, client_ref, date_case_opened, date_case_closed, case_type, notes, fee_earner, created, last_updated, file_location, foreign_attorney_id, foreign_attorney_ref, country_code, application_number, filing_date, publication_number, invoice_currency, publication_date, status, pct_application_number, pct_case_id, national_phase_entry, base_number, base_country, base_date, base_case_id, divisional_date_lodged, illustrative_image, parent_case_id, parent_application_number) 
    select id, short_name, name, invention_title, reference, client_id, client_ref, date_case_opened, date_case_closed, case_type, notes, fee_earner, created, last_updated, file_location, foreign_attorney_id, foreign_attorney_ref, country_code, application_number, filing_date, publication_number, invoice_currency, publication_date, status, pct_application_number, pct_case_id, national_phase_entry, base_number, base_country, base_date, base_case_id, divisional_date_lodged, illustrative_image, parent_case_id, parent_application_number 
    FROM [guest].table2; 

SET IDENTITY_INSERT [guest].table1 OFF; 
SET IDENTITY_INSERT [guest].table2 OFF; 

Answer 3

您已設置Identity Specification-> Is Identity to True

所以你不能插入自動手動生成主鍵！

Answer 4

INSERT INTO short_name, name, invention_title, reference, client_id, client_ref, date_case_opened, date_case_closed, case_type, notes, fee_earner, created, last_updated, file_location, foreign_attorney_id, foreign_attorney_ref, country_code, application_number, filing_date, publication_number, invoice_currency, publication_date, status, pct_application_number, pct_case_id, national_phase_entry, base_number, base_country, base_date, base_case_id, divisional_date_lodged, illustrative_image, parent_case_id, parent_application_number) 
select short_name, name, invention_title, reference, client_id, client_ref, date_case_opened, date_case_closed, case_type, notes, fee_earner, created, last_updated, file_location, foreign_attorney_id, foreign_attorney_ref, country_code, application_number, filing_date, publication_number, invoice_currency, publication_date, status, pct_application_number, pct_case_id, national_phase_entry, base_number, base_country, base_date, base_case_id, divisional_date_lodged, illustrative_image, parent_case_id, parent_application_number 
FROM [guest].table2; 

試試這個沒有ID。