Question

我有两个彼此相关的表（feature和features），我想让开发人员透明id（主键）一个表（{ {1}}）。因此，为了插入一个表（feature），您不必放置另一个features的{{1}}，您可以使用另一个表的辅助键具有。这是我的代码：

id

我希望能够执行以下类型的插入，因此不使用feature表的CREATE TABLE feature( id INTEGER NOT NULL, description VARCHAR(15) NOT NULL, value VARCHAR(15) NOT NULL, price MONEY NOT NULL, CONSTRAINT feature_pk PRIMARY KEY(id), CONSTRAINT feature_un UNIQUE(description, value), CONSTRAINT feature_ck_price CHECK(price >= 0) ); CREATE TABLE features( name VARCHAR(20) NOT NULL, year NUMERIC(4) NOT NULL, feature_id INTEGER NOT NULL, CONSTRAINT features_pk PRIMARY KEY(name,year,feature_id), CONSTRAINT features_fk_model FOREIGN KEY(name,year) REFERENCES model(name,year), CONSTRAINT features_pk_feature FOREIGN KEY(feature_id) REFERENCES feature(id) ); -- Trigger not working because the features table doesnt have the columns I'm trying to use. CREATE TRIGGER trigger_features_get_feature_id ON features INSTEAD OF INSERT AS DECLARE @tr_features_name VARCHAR(20), @tr_features_year NUMERIC(4), @tr_features_description VARCHAR(15), @tr_features_value VARCHAR(15), @tr_features_feature_id INT SELECT @tr_features_name=(SELECT name FROM INSERTED) SELECT @tr_features_year=(SELECT year FROM INSERTED) SELECT @tr_features_description=(SELECT description FROM INSERTED) SELECT @tr_features_value=(SELECT value FROM INSERTED) SELECT @tr_features_feature_id=(SELECT id FROM feature WHERE description = @tr_features_description AND value = @tr_features_value) BEGIN PRINT CONVERT(varchar(100),@tr_features_feature_id) + @tr_features_description + @tr_features_value + CONVERT(varchar(100),@tr_features_name) + @tr_features_year INSERT INTO features VALUES(@tr_features_name,@tr_features_year,@tr_features_id) END（自动生成），而是可以使用此表的辅助键（ id和feature）。

value

Answer 1

我按照@adam说的那样，这是结果

CREATE PROCEDURE [dbo].procedure_insert_features 
@pr_car_name VARCHAR(20) = NULL,
@pr_car_year NUMERIC(4) = NULL,
@pr_feature_description VARCHAR(15),
@pr_feature_value VARCHAR(15)
AS
SET NOCOUNT ON
BEGIN
DECLARE @pr_feature_id AS INTEGER
    SET @pr_feature_id = (SELECT ID FROM [automobile].[dbo].[feature] WHERE description = @pr_feature_description AND value = @pr_feature_value);
    -- PRINT @pr_feature_id
    IF @pr_feature_id IS NOT NULL
        BEGIN
            INSERT INTO [automobile].[dbo].available_features VALUES(@pr_car_name,@pr_car_year,@pr_feature_id);
    END
ELSE
    BEGIN
        RAISERROR ('ID NOT FOUND, PLEASE CHECK SPELLING', 10,1);
    END
END

在表中插入另一个表的列

1 个答案: