MS Access create table from query c# - c#

Unable to create table via
CREATE TABLE AAA (
[Percentage] Number(15),
[ID] AutoIncrement CONSTRAINT PRIMARY KEY ,
[CaseRecNo] Number(15),
[CaseName] Text,
[Amount] Currency
)
Gives Syntax Error,
How to set Number filed type double ?

The field size is a double by default. Create this table in access and you will see the result. You also need to name your constraint
CREATE TABLE AAAB (
[Percentage] NUMBER,
[ID] AutoIncrement CONSTRAINT MyFieldConstraint PRIMARY KEY ,
[CaseRecNo] Number,
[CaseName] Text,
[Amount] Currency
)

Related

Relationship between Identity tables and Other tables in ASP.NET MVC 5

In ASP.NET MVC 5 project, the database and identity tables are created by code first, and i created other tables by SQL (not by code first) in this database, and i want to join User table with some table by user Id.
Say database called Qwerty and the identity tables are:
dbo.Users
dbo.Roles
dbo.UserClaims
... ect
I want to create tables by SQL like this:
Create Table Topic.Topic
(
TopicID int Primary key identity(1,1) not null,
TopicAddress nvarchar(255) not null
)
Create Table dbo.Bookmark
(
BookmarkID int Primary key identity(1,1) not null,
BookmarkDate datetime default getdate() not null,
UserID int constraint FK_Favorites_Users_UserID foreign key (UserID) references Users(UserID) not null
)
Topic table is created successfully, but when i run SQL code for Bookmark table, it give me error and mark Users (table name) word with red line
By default the primary key of Users is nvarchar, so, your foreign key should be defined with that type.
Create Table dbo.Bookmark
(
BookmarkID int Primary key identity(1,1) not null,
BookmarkDate datetime default getdate() not null,
UserID [nvarchar](128) constraint FK_Favorites_Users_UserID foreign key (UserID) references Users(UserID) not null
)

The INSERT statement conflicted with the FOREIGN KEY constraint "FK_

I have two tables , one to one relationship i makes DetailsID of projectDetails FK in ID of projects table:
projects:
ID, //has FK With DetailsID in Details table & auto identity (1,1)
ProjectName,
Areas,
PaymentSystem,
ReceivedDate,
PropertyClassification,
ProjectImage
ProjectDetails:
DetailsID ,auto identity ( 1,1)
ProjectDetailName,
ProjectDetailImage
I am trying to insert new record in projects table , gives me this error at this line of code :
con.Open();
comm.ExecuteNonQuery(); // when execute
System.Data.SqlClient.SqlException: 'The INSERT statement conflicted with the FOREIGN KEY constraint "FK_Projects_ProjectDetails". The conflict occurred in database "AlamaarRealEstate", table "dbo.ProjectDetails", column 'DetailsID'.
and this is my stored to insert :
ALTER proc [Insert_Project]
#Projectname NVARCHAR(MAX) ,
#areas NVARCHAR(MAX) ,
#Paymentsystem NVARCHAR(MAX) ,
#Receiveddate date ,
#Classification NVARCHAR(MAX) ,
#Projectimage Nvarchar(MAX)
as
INSERT INTO dbo.Projects
(
ProjectName,
Areas,
PaymentSystem,
ReceivedDate,
PropertyClassification,
ProjectImage
)
VALUES
(
#Projectname ,
#areas,
#Paymentsystem ,
#Receiveddate ,
#Classification,
#Projectimage
)
The question explains the answer. Referential Integrity is not maintained properly, you are trying to insert into a child table for which the master value does not exist. Please insert values to Project_details first. This will resolve your issue. If you did not what this to throw an error, just check the existence of the DetailID in Projects table before inserting.
Without more detail your question is hard to answer. For instance in your procedure you do explicit column naming for inserts and LEAVE OUT the column you are having a problem with. So if the proc was wrong it would be bombing with your example foreign key constraint error. But you are not even listing the 'ID' field to insert into with the procedure. So this is common if you are using an 'Identity' field to self seed, but you are claiming it is a foreign key. So like others have commented, without more code to show the exact way your tables are made it's hard to guess. Here is a self extracting example you could run that shows if a column is nullable and I had a key constraint it would work. Without the exact code of the tables as well as the proc it is hard to tell. What you gave is pseudo code.
USE Tester --just a test database I have, you can use whatever database you want
GO
IF OBJECT_ID('Projects') IS NOT NULL
DROP TABLE Projects
IF OBJECT_ID('ProjectDetails') IS NOT NULL
DROP TABLE ProjectDetails
create TABLE ProjectDetails
(
DetailsID INT CONSTRAINT PK_DetailsId PRIMARY KEY,
ProjectDetailName VARCHAR(32)
)
CREATE TABLE Projects
(
Id INT CONSTRAINT FK_Projects_ProjectDetails FOREIGN KEY (Id) REFERENCES ProjectDetails(DetailsId),
ProjectName varchar(32)
)
GO
IF OBJECT_ID('Insert_Project') IS NOT NULL
DROP PROC Insert_Project
GO
Create proc Insert_Project
#Projectname NVARCHAR(MAX)
as
INSERT INTO dbo.Projects ( ProjectName )
VALUES ( #Projectname )
GO
Select *
From dbo.Projects
EXEC dbo.Insert_Project #Projectname = N'Test' -- nvarchar(max)
Select *
From dbo.Projects

Asp.Net Core Entity Framework Migration Error: index needed in a foreign key constraint [duplicate]

I need to ALTER my existing database to add a column. Consequently I also want to update the UNIQUE field to encompass that new column. I'm trying to remove the current index but keep getting the error MySQL Cannot drop index needed in a foreign key constraint
CREATE TABLE mytable_a (
ID TINYINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
UNIQUE(Name)
) ENGINE=InnoDB;
CREATE TABLE mytable_b (
ID TINYINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
UNIQUE(Name)
) ENGINE=InnoDB;
CREATE TABLE mytable_c (
ID TINYINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
UNIQUE(Name)
) ENGINE=InnoDB;
CREATE TABLE `mytable` (
`ID` int(11) NOT NULL AUTO_INCREMENT,
`AID` tinyint(5) NOT NULL,
`BID` tinyint(5) NOT NULL,
`CID` tinyint(5) NOT NULL,
PRIMARY KEY (`ID`),
UNIQUE KEY `AID` (`AID`,`BID`,`CID`),
KEY `BID` (`BID`),
KEY `CID` (`CID`),
CONSTRAINT `mytable_ibfk_1` FOREIGN KEY (`AID`) REFERENCES `mytable_a` (`ID`) ON DELETE CASCADE,
CONSTRAINT `mytable_ibfk_2` FOREIGN KEY (`BID`) REFERENCES `mytable_b` (`ID`) ON DELETE CASCADE,
CONSTRAINT `mytable_ibfk_3` FOREIGN KEY (`CID`) REFERENCES `mytable_c` (`ID`) ON DELETE CASCADE
) ENGINE=InnoDB;
mysql> ALTER TABLE mytable DROP INDEX AID;
ERROR 1553 (HY000): Cannot drop index 'AID': needed in a foreign key constraint
You have to drop the foreign key. Foreign keys in MySQL automatically create an index on the table (There was a SO Question on the topic).
ALTER TABLE mytable DROP FOREIGN KEY mytable_ibfk_1 ;
Step 1
List foreign key ( NOTE that its different from index name )
SHOW CREATE TABLE <Table Name>
The result will show you the foreign key name.
Format:
CONSTRAINT `FOREIGN_KEY_NAME` FOREIGN KEY (`FOREIGN_KEY_COLUMN`) REFERENCES `FOREIGN_KEY_TABLE` (`id`),
Step 2
Drop (Foreign/primary/key) Key
ALTER TABLE <Table Name> DROP FOREIGN KEY <Foreign key name>
Step 3
Drop the index.
If you mean that you can do this:
CREATE TABLE mytable_d (
ID TINYINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
Name VARCHAR(255) NOT NULL,
UNIQUE(Name)
) ENGINE=InnoDB;
ALTER TABLE mytable
ADD COLUMN DID tinyint(5) NOT NULL,
ADD CONSTRAINT mytable_ibfk_4
FOREIGN KEY (DID)
REFERENCES mytable_d (ID) ON DELETE CASCADE;
> OK.
But then:
ALTER TABLE mytable
DROP KEY AID ;
gives error.
You can drop the index and create a new one in one ALTER TABLE statement:
ALTER TABLE mytable
DROP KEY AID ,
ADD UNIQUE KEY AID (AID, BID, CID, DID);
A foreign key always requires an index. Without an index enforcing the constraint would require a full table scan on the referenced table for every inserted or updated key in the referencing table. And that would have an unacceptable performance impact.
This has the following 2 consequences:
When creating a foreign key, the database checks if an index exists. If not an index will be created. By default, it will have the same name as the constraint.
When there is only one index that can be used for the foreign key, it can't be dropped. If you really wan't to drop it, you either have to drop the foreign key constraint or to create another index for it first.
Because you have to have an index on a foreign key field you can just create a simple index on the field 'AID'
CREATE INDEX aid_index ON mytable (AID);
and only then drop the unique index 'AID'
ALTER TABLE mytable DROP INDEX AID;
I think this is easy way to drop the index.
set FOREIGN_KEY_CHECKS=0; //disable checks
ALTER TABLE mytable DROP INDEX AID;
set FOREIGN_KEY_CHECKS=1; //enable checks
drop the index and the foreign_key in the same query like below
ALTER TABLE `your_table_name` DROP FOREIGN KEY `your_index`;
ALTER TABLE `your_table_name` DROP COLUMN `your_foreign_key_id`;
Dropping FK is tedious and risky. Simply create the new index with new columns and new index name, such as AID2. After the new Unique Index is created, you can drop the old one with no issue. Or you can use the solution given above to incorporate the "drop index, add unique index" in the same alter table command. Both solutions will work
In my case I dropped the foreign key and I still could not drop the index. That was because there was yet another table that had a foreign key to this table on the same fields. After I dropped the foreign key on the other table I could drop the indexes on this table.
If you are using PhpMyAdmin sometimes it don't show the foreign key to delete.
The error code gives us the name of the foreign key and the table where it was defined, so the code is:
ALTER TABLE your_table DROP FOREIGN KEY foreign_key_name;
You can show Relation view in phpMyAdmin and first delete foreign key. After this you can remove index.
You can easily check it with DBeaver. Example:
As you can see there are 3 FKs but only 2 FK indexes. There is no index for FK_benefCompanyNumber_beneficiaries_benefId as UK index provide uniqueness for that FK.
To drop that UK you need to:
DROP FK_benefCompanyNumber_beneficiaries_benefId
DROP UK
CREATE FK_benefCompanyNumber_beneficiaries_benefId
The current most upvoted answer is not complete.
One needs to remove all the foreign keys whose "source" column is also present in the UNIQUE KEY declaration.
So in this case, it is not enough to remove mytable_ibfk_1 for the error to go away, mytable_ibfk_2 and mytable_ibfk_3 must be deleted as well.
This is the complete answer:
ALTER TABLE mytable DROP FOREIGN KEY mytable_ibfk_1;
ALTER TABLE mytable DROP FOREIGN KEY mytable_ibfk_2;
ALTER TABLE mytable DROP FOREIGN KEY mytable_ibfk_3;
Its late now but I found a solution which might help somebody in future.
Just go to table's structure and drop foreign key from foreign keys list. Now you will be able to delete that column.

Migrating to identity with [Id] replacement

I have an existing database with my users. But the [Id] column is just a number
[Id] BIGINT IDENTITY (1, 1) NOT NULL
CONSTRAINT [Users] PRIMARY KEY CLUSTERED ([Id] ASC)
I want to transfer all of my users to a new AspNetUsers table where
[Id] NVARCHAR (128) NOT NULL
CONSTRAINT [PK_dbo.AspNetUsers] PRIMARY KEY CLUSTERED ([Id] ASC)
I saw all the explanations on migrating, but I couldnt find how can I generate a new Id in SQL Insert statment. All examples show only how to transfer data from the old database to a new one.
Can someone please give me an example of SQL code that copies all the data from one table and inserts it to another table with new generated unique value.
The final script that worked:
INSERT INTO AspNetUsers
(
Id,
Email,
UserName,
.
.
.
)
SELECT
CAST(NEWID() AS NVARCHAR (128)),
Email,
UserName,
.
.
.
FROM Users

Access sql for create table with autonumber

I am trying to create a table with a column whose AutoIncrement is Yes.
Here is my query not sure what's wrong in it
CREATE TABLE testallcols(SOCycle Text(3), AutoKey integer AUTOINCREMENT NOT NULL, SOData LongBinary NOT NULL)
How do I get my AutoKey column to be an auto increment integer?
AUTOINCREMENT and integer are two different datatypes as far as Access DDL is concerned. Use only AUTOINCREMENT. And to make it function correctly as an autonumber, include the PRIMARY KEY constraint.
This one works without error when tested with ADO/OleDb in Access 2010:
CREATE TABLE testallcols(SOCycle Text(3), AutoKey AUTOINCREMENT PRIMARY KEY, SOData LongBinary NOT NULL)
CREATE TABLE Tblcontacts
(
contactid AUTOINCREMENT(101,1) PRIMARY KEY ,
firstname CHAR (60),
lastname CHAR (60),
email VARCHAR (75)
);

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