表因外键而发生冲突;

本文介绍了表因外键而发生冲突;的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

在我的第一个tbl_uni Uni_id，University_Name是列 in tbl_course C_id，University_name，Course，Major，Uni_id（FK refer tbl_uni） in tbl_major Mid，University_name，Course，Major，Uni_id（FK refer tbl_uni），C_id（FK refer tbl_course） textfield = University_name valuefield = Uni_id 错误将是 INSERT语句与FOREIGN KEY约束FK__tbl_cours__Uni_i__36B12243冲突。冲突发生在数据库大学，表dbo.tbl_university，列'Uni_id'。

in my first tbl_uni Uni_id,University_Name are the columns in tbl_course C_id,University_name,Course,Major,Uni_id(FK refer tbl_uni) in tbl_major Mid,University_name,Course,Major,Uni_id(FK refer tbl_uni),C_id(FK refer tbl_course) while inserting the vale by dropdownlist in tbl_course textfield=University_name valuefield=Uni_id the error will be The INSERT statement conflicted with the FOREIGN KEY constraint "FK__tbl_cours__Uni_i__36B12243". The conflict occurred in database "University", table "dbo.tbl_university", column 'Uni_id'.

推荐答案

首先，数据库设计是完全不可接受的： 1.为什么university_name出现在tbl_course和tbl_major？ 2.为什么课程出现在tbl_major？ 3.为什么major出现在tbl_course？ 您正在复制数据并显示您的表格未正确规范化。 数据库规范化 [ ^ ] 接下来，错误说明您尝试将新的uni_id插入tbl_course（child）并且在tbl_uni表（父）中不存在uni_id。阅读更多 SQL FOREIGN KEY Constraint [ ^ ] 了解详情： www.datanamic/support/lt-dez005-introduction-db-modeling.html [ ^ ] First thing first, the database design is totally unacceptable: 1. Why is university_name appears in tbl_course and tbl_major? 2. Why is course appears in tbl_major? 3. Why is major appears in tbl_course? You are duplicating data and that shows your tables are not properly normalized. Normalization of Database[^] Next, the error stated that you attempted to insert a new uni_id into the tbl_course (child) and that uni_id did not exist in the tbl_uni table (parent). Read more SQL FOREIGN KEY Constraint[^] Learn more: www.datanamic/support/lt-dez005-introduction-db-modeling.html[^]

下面查询将有助于找到数据库的PK和FK的链接。 为了更容易，你也可以使用数据库图。 Below query will be helpful in finding the linkage of PK and FK of your Database. For more easier you can use Database diagram also. SELECT f.name AS ForeignKey, SCHEMA_NAME(f.SCHEMA_ID) SchemaName, OBJECT_NAME(f.parent_object_id) AS TableName, COL_NAME(fc.parent_object_id,fc.parent_column_id) AS ColumnName, SCHEMA_NAME(o.SCHEMA_ID) ReferenceSchemaName, OBJECT_NAME (f.referenced_object_id) AS ReferenceTableName, COL_NAME(fc.referenced_object_id,fc.referenced_column_id) AS ReferenceColumnName FROM sys.foreign_keys AS f INNER JOIN sys.foreign_key_columns AS fc ON f.OBJECT_ID = fc.constraint_object_id INNER JOIN sys.objects AS o ON o.OBJECT_ID = fc.referenced_object_id