alter table statment to insert duplicate into another table - mysql

I have a table in which there is a column name with SP varchar(10) NOT NULL. I want that column always to be unique so i created unique index on that column . My table schema as follows :
CREATE TABLE IF NOT EXISTS `tblspmaster` (
`CSN` bigint(20) NOT NULL AUTO_INCREMENT,
`SP` varchar(10) NOT NULL,
`FileImportedDate` date NOT NULL,
`AMZFileName` varchar(50) NOT NULL,
`CasperBatch` varchar(50) NOT NULL,
`BatchProcessedDate` date NOT NULL,
`ExpiryDate` date NOT NULL,
`Region` varchar(50) NOT NULL,
`FCCity` varchar(50) NOT NULL,
`VendorID` int(11) NOT NULL,
`LocationID` int(11) NOT NULL,
PRIMARY KEY (`CSN`),
UNIQUE KEY `SP` (`SP`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=10000000000 ;
Now i want that if anybody tries to insert duplicate record then that record should be inserted into a secondary table name tblDuplicate.
I have gone through this question MySQL - ignore insert error: duplicate entry but i am not sure that instead of
INSERT INTO tbl VALUES (1,200) ON DUPLICATE KEY UPDATE value=200;
can i insert duplicate row into another table ?
what changes needed to be done in main table scheme or index column ?
**Note : Data will be inserted by importing excel or csv files and excel files generally contains 500k to 800 k records but there will be only one single column **

I believe you want to use a trigger for this. Here is the MySQL reference chapter on triggers.
Use a before insert trigger. In the trigger, check if the row is a duplicate (maybe count(*) where key column value = value to be inserted). If the row is a duplicate, perform an insert into your secondary table.

Related

Field 'category_id' doesn't have a default value MySQL

I'm learning SQL.
I'm trying to insert data. My MySQL database looks like this.
CREATE TABLE category (
category_id CHAR(100),
category_name VARCHAR(120) NOT NULL,
PRIMARY KEY (category_id)
)
I ran this command
INSERT INTO category (category_name) VALUES ("test");
But I got this error
ERROR 1364 (HY000): Field 'category_id' doesn't have a default value
Thank you in advance.
If you want to have an incrementing ID it would need to be an int. You Generally want to make ID's integers not chars to speed up lookup regardless.
CREATE TABLE IF NOT EXISTS category (
`category_id` INT NOT NULL AUTO_INCREMENT,
`category_name` VARCHAR(120) NOT NULL,
PRIMARY KEY (`category_id`)
) ENGINE=InnoDB DEFAULT CHARSET = utf8mb4 COLLATE utf8mb4_unicode_ci;
That will let you insert without adding your own ID, and automatically generate unique ID's for the records.
Issue was you set your category_id field to not have a default value and also not allow null, which means you -have- to set a value for it in the insert. If you wanted to use your existing table you would need to do this:
INSERT INTO category (category_id, category_name) VALUES ("someid", "test");

How to use more than 1 auto-increment column in MySQL

I want to create a table name Users where I should have have columns User, cookieID, sessionID, Geo and then I want to first three columns to have some random unique value assigned automatically. I tried to make all three columns AUTO_INCREMENT with User column PRIMARY and 'cookieIDandsessionIDcolumnUNIQUE`. The SQL code is:
CREATE TABLE `users` ( `User` VARCHAR(20) NOT NULL AUTO_INCREMENT ,
`cookieID` INT(20) NULL DEFAULT NULL AUTO_INCREMENT ,
`sessionID` INT(20) NULL DEFAULT NULL AUTO_INCREMENT ,
`Geo` VARCHAR(30) NULL DEFAULT NULL ,
PRIMARY KEY (`User`), UNIQUE (`cookieID`), UNIQUE (`sessionID`), UNIQUE (`Geo`));
But, it did not work because only one column can be declared as AUTO_INCREMENT which must be PRIMARY.
What is the another approach to do this?
Since the auto-increment cannot be applied to multiple to rows and there no option for sequence in MySQL. You can use triggers for the unique update of the row with datetime.
Change to table creation to be of single auto-increment row.
CREATE TABLE `users` ( `User` VARCHAR(20) NOT NULL,
`cookieID` INT(20) NULL DEFAULT NULL,
`sessionID` INT(20) NULL DEFAULT NULL AUTO_INCREMENT ,
`Geo` VARCHAR(30) NULL DEFAULT NULL,
PRIMARY KEY (`User`), UNIQUE (`cookieID`), UNIQUE (`sessionID`), UNIQUE (`Geo`));
Create a trigger on the same table as below. You can set the unique values under the SET for as many column as you want.
CREATE DEFINER=`root`#`localhost` TRIGGER `users_BEFORE_INSERT` BEFORE INSERT ON `users` FOR EACH ROW BEGIN
SET
NEW.cookieID = (SELECT curdate()+curtime());
END
Now when you insert into the table as below.
insert into `users`(`User`) values("test");
You table looks like this.
User cookieID sessionID Geo
test 20315169 0 NULL
If the value which are auto incrementing, you wanna keep both values the same. Then copy the value of one column to another during insertion time of new value.

MySQL Auto increment not incrementing

I've created a table with 3 columns: postID, userID, and comment.
I have the postID as the primary key, and I am trying to make this auto-increment every time I add a new row to the table.
INSERT INTO CommentTable (postID, userID, comment) VALUES (DEFAULT, "test", "test")
When I run this query, it will run OK once but then when I run it again I get "1062 - Duplicate entry '0' for key 'PRIMARY'".
How do I properly set up an auto-increment primary key?
Here is the table structure:
DROP TABLE IF EXISTS `CommentTable`;
CREATE TABLE `CommentTable` (
`postID` int(10) NOT NULL,
`userID` varchar(10) NOT NULL,
`comment` varchar(1000) DEFAULT NULL,
PRIMARY KEY (`commentID`)
)
No need to put the field postID
INSERT INTO CommentTable (userID, comment) VALUES ("test", "test")
Edit your table as:
CREATE TABLE `CommentTable` (
`postID` int(10) NOT NULL AUTO_INCREMENT,
`userID` varchar(10) NOT NULL,
`comment` varchar(1000) DEFAULT NULL,
PRIMARY KEY (`postID`)
)
Recently I had the same issue where Auto Increment was not saved. It failed every time I tried to save it. My problem was that I had a record where the value was 0 instead of 1, so I updated that record value to a non-zero value, then tried saving the Auto Increment, and it worked.
The zero(0) value in the Primary Index field was causing the ALTER tablename to fail. Once it worked, I put the value back to 0.

What is the meaning of AUTO_INCREMENT=20018215 here in table schema

CREATE TABLE `tblspmaster` (
`CSN` bigint(20) NOT NULL AUTO_INCREMENT,
`SP` varchar(50) NOT NULL,
`FileImportedDate` date NOT NULL,
`AMZFileName` varchar(580) NOT NULL,
`CasperBatch` varchar(50) NOT NULL,
`BatchProcessedDate` date NOT NULL,
`ExpiryDate` date NOT NULL,
`Region` varchar(50) NOT NULL,
`FCCity` varchar(50) NOT NULL,
`VendorID` int(11) NOT NULL,
`LocationID` int(11) NOT NULL,
PRIMARY KEY (`CSN`)
) ENGINE=InnoDB AUTO_INCREMENT=20018215 DEFAULT CHARSET=latin1;
What is the meaning of AUTO_INCREMENT=20018215 here in table schema . as i am inserting 500k records my identity is OK from 1 to 500k but when i tried to insert next 500k records, next records identity column value is 524281 instead of 500001.
It means that the first auto-assigned value (to CSN) will be 20018215
The large initial value, 20018215, was probably the previous value of the auto increment when you did a "Send to SQL Editor" -> "Create Statement" menu selection in MySQL Workbench. This is just a safe value to skip over existing data just in case you have to reimport the previous records.
I had the same question, but after generating several "Create" edit templates from known tables, I noticed the AUTO_INCREMENT value corresponded to the quantity of existing records in those tables. I removed the large values from my templates since I want my new tables to begin with a primary key = 1.

use ON DUPLICATE KEY for table that have two primary key

I have a table with 2 primary keys(user_id,post_id)
I want to insert row only if table havn't a row with this user id and post id
And if previous data exist for this keys , only that row update with the new data
I wrote this query:
INSERT INTO trust_list(`user_id`,`post_id`,`post_per`,`comment_per`,`cat_per`)
VALUES (7,1,'000','000','000')
ON DUPLICATE KEY UPDATE `post_per`='000',`comment_per`='000',`cat_per`='000'
For example if this row exist in the table:
user_id:5
post_id:1
post_per:001
comment_per:111
cat_per:101
Then ,when i execute the above query , mysql update this row,only because post_id of this row is "1"
Whereas mysql should not update this row .
I don't understand what's the problem.
DESC trust_list
result of above query is:
Field Type Null Key Default Extra
user_id int(11) NO PRI NULL
post_id int(4) NO PRI NULL
post_per tinytext YES NULL
comment_per tinytext YES NULL
cat_per tinytext YES NULL
=================
Thanks to all of my friends
When i decide to drop table and ceate it again ,I get an export from this table
and review the .sql file,i see this:
CREATE TABLE IF NOT EXISTS `trust_list` (
`user_id` int(11) NOT NULL,
`post_id` int(11) NOT NULL,
`post_per` tinytext COLLATE utf8_estonian_ci ,
`comment_per` tinytext COLLATE utf8_estonian_ci ,
`cat_per` tinytext COLLATE utf8_estonian_ci ,
PRIMARY KEY (`idea_id`,`user_id`),
UNIQUE KEY `idea_id` (`idea_id`)
)
apparently , problem is from the UNIQUE KEY ,I remove it from file ,and then drop trust_list table ,and then import .sql file
So with this ,my problem solved
Thanks again
There is something wrong with your table schema.
Assuming that your schema defined something like this
CREATE TABLE trust_list(
`user_id` int,
`post_id` int,
`post_per` varchar(12),
`comment_per` varchar(12),
`cat_per`varchar(12),
PRIMARY KEY(`user_id`, `post_id`)
);
here is SQLFiddle that demonstrates that your INSERT statement works on it just fine.
Consider to show your CREATE TABLE statement to help you find the problem, or just change PK as showed.
I think your problem is your understanding of primary key. You can't have two primary keys in the table, only one. What you have probably is a primary key that consists of two columns. In that case you only get a key violation when both columns match.
Solution:
Introduce unique indices for both columns. Or - better - change the primary key to be only one of the two columns and set the other column to have a unique index.
Thanks to peterem here is the sqlfiddle with my solution.