I created a table in MySql
CREATE TABLE newuser(id VARCHAR(10) PRIMARY KEY,sname VARCHAR(20));
When I INSERT record it works fine
INSERT INTO newuser VALUE('abc123','monika');
But sometimes I don't want to supply id in the INSERT query and sometimes I want to supply. In case I don't supply id MySql automatically generate one.
What can I do to get both below query works?
INSERT INTO newuser VALUE('abc123','monika');
INSERT INTO newuser VALUE('nikita');
'I don't understood ANYTHING' - very new then.
Firstly second insert statement is invalid please review https://dev.mysql.com/doc/refman/8.0/en/insert.html -
'If you do not specify a list of column names for INSERT ... VALUES or INSERT ... SELECT, values for every column in the table must be provided by the VALUES list, SELECT statement, or TABLE statement.'
Secondly uuid is a function in which 'A UUID is designed as a number that is globally unique in space and time.' https://dev.mysql.com/doc/refman/8.0/en/insert.html
You can easily select uuid() to see what it produces.
You will need to increase the id size
If you wish to use it in an insert
insert into users values (uuid(),<sname>);
Related
So I read the other posts but this question is unique. So this SQL dump file has this as the last entry.
INSERT INTO `wp_posts` VALUES(2781, 3, '2013-01-04 17:24:19', '2013-01-05 00:24:19'.
I'm trying to insert this value to the table...
INSERT INTO `wp_posts` VALUES(5, 5, '2005-04-11 09:54:35', '2005-04-11 17:54:35'
it gives me the error, "Column count doesn't match value count at row 1." So I'm lost on the concept of how the column and row apply here.
Doesn't 2781,3 mean row 2781 and column 3? And doesn't 5,5 mean row 5 and column 5?
The error means that you are providing not as much data as the table wp_posts does contain columns. And now the DB engine does not know in which columns to put your data.
To overcome this you must provide the names of the columns you want to fill. Example:
insert into wp_posts (column_name1, column_name2)
values (1, 3)
Look up the table definition and see which columns you want to fill.
And insert means you are inserting a new record. You are not modifying an existing one. Use update for that.
you missed the comma between two values or column name
you put extra values or an extra column name
You should also look at new triggers.
MySQL doesn't show the table name in the error, so you're really left in a lurch. Here's a working example:
use test;
create table blah (id int primary key AUTO_INCREMENT, data varchar(100));
create table audit_blah (audit_id int primary key AUTO_INCREMENT, action enum('INSERT','UPDATE','DELETE'), id int, data varchar(100) null);
insert into audit_blah(action, id, data) values ('INSERT', 1, 'a');
select * from blah;
select * from audit_blah;
truncate table audit_blah;
delimiter //
/* I've commented out "id" below, so the insert fails with an ambiguous error: */
create trigger ai_blah after insert on blah for each row
begin
insert into audit_blah (action, /*id,*/ data) values ('INSERT', /*NEW.id,*/ NEW.data);
end;//
/* This insert is valid, but you'll get an exception from the trigger: */
insert into blah (data) values ('data1');
MySQL will also report "Column count doesn't match value count at row 1" if you try to insert multiple rows without delimiting the row sets in the VALUES section with parentheses, like so:
INSERT INTO `receiving_table`
(id,
first_name,
last_name)
VALUES
(1002,'Charles','Babbage'),
(1003,'George', 'Boole'),
(1001,'Donald','Chamberlin'),
(1004,'Alan','Turing'),
(1005,'My','Widenius');
You can resolve the error by providing the column names you are affecting.
> INSERT INTO table_name (column1,column2,column3)
`VALUES(50,'Jon Snow','Eye');`
please note that the semi colon should be added only after the statement providing values
In my case i just passed the wrong name table, so mysql couldn't find the right columns names.
So I read the other posts but this question is unique. So this SQL dump file has this as the last entry.
INSERT INTO `wp_posts` VALUES(2781, 3, '2013-01-04 17:24:19', '2013-01-05 00:24:19'.
I'm trying to insert this value to the table...
INSERT INTO `wp_posts` VALUES(5, 5, '2005-04-11 09:54:35', '2005-04-11 17:54:35'
it gives me the error, "Column count doesn't match value count at row 1." So I'm lost on the concept of how the column and row apply here.
Doesn't 2781,3 mean row 2781 and column 3? And doesn't 5,5 mean row 5 and column 5?
The error means that you are providing not as much data as the table wp_posts does contain columns. And now the DB engine does not know in which columns to put your data.
To overcome this you must provide the names of the columns you want to fill. Example:
insert into wp_posts (column_name1, column_name2)
values (1, 3)
Look up the table definition and see which columns you want to fill.
And insert means you are inserting a new record. You are not modifying an existing one. Use update for that.
you missed the comma between two values or column name
you put extra values or an extra column name
You should also look at new triggers.
MySQL doesn't show the table name in the error, so you're really left in a lurch. Here's a working example:
use test;
create table blah (id int primary key AUTO_INCREMENT, data varchar(100));
create table audit_blah (audit_id int primary key AUTO_INCREMENT, action enum('INSERT','UPDATE','DELETE'), id int, data varchar(100) null);
insert into audit_blah(action, id, data) values ('INSERT', 1, 'a');
select * from blah;
select * from audit_blah;
truncate table audit_blah;
delimiter //
/* I've commented out "id" below, so the insert fails with an ambiguous error: */
create trigger ai_blah after insert on blah for each row
begin
insert into audit_blah (action, /*id,*/ data) values ('INSERT', /*NEW.id,*/ NEW.data);
end;//
/* This insert is valid, but you'll get an exception from the trigger: */
insert into blah (data) values ('data1');
MySQL will also report "Column count doesn't match value count at row 1" if you try to insert multiple rows without delimiting the row sets in the VALUES section with parentheses, like so:
INSERT INTO `receiving_table`
(id,
first_name,
last_name)
VALUES
(1002,'Charles','Babbage'),
(1003,'George', 'Boole'),
(1001,'Donald','Chamberlin'),
(1004,'Alan','Turing'),
(1005,'My','Widenius');
You can resolve the error by providing the column names you are affecting.
> INSERT INTO table_name (column1,column2,column3)
`VALUES(50,'Jon Snow','Eye');`
please note that the semi colon should be added only after the statement providing values
In my case i just passed the wrong name table, so mysql couldn't find the right columns names.
I followed this tutorial with 'INSERT IGNORE' first and 'INSERT ... ON DUPLICATE KEY UPDATE' then. But it doesn't work.
I'm using NodeJS to get some data from an API, and store these data into a mySQL database. Before storing data, I want to know if this row already exists.
The ID is AUTO_INCREMENT and I d'ont know this one. Instead of using async/await or promises in NodeJS, I wanted to treat this point with mySQL without knowing the ID.
I tried this one but it adds a new row with a new ID instead of another row already exists:
INSERT INTO test (nom, date, heure) VALUES ('123456', '2018-08-23', '10:45:00') ON DUPLICATE KEY UPDATE nom='123456', date='2018-08-23', heure='10:45:00';
After that, I tested this one, same result:
INSERT IGNORE INTO test (nom, date, heure) VALUES ('123456', '2018-08-23', '10:45:00')
Thanks for your help!
Set a unique index for the 3 columns:
ALTER TABLE test ADD UNIQUE un_index(nom, date, heure);
Then execute:
INSERT IGNORE INTO test (nom, date, heure) VALUES ('123456', '2018-08-23', '10:45:00');
If the 3 values already exist, the row will not be inserted.
Table test
id - int, primary key, auto_incriment
name - varchar(100)
INSERT INTO test
(ID, Name)
VALUES
('$ID', '$Name');
Tell me please really return ID in one query when we make insert ?
as say #MichaelBerkowski you cannt return ID in one query.
For return ID make next:
INSERT INTO test
(ID, Name)
VALUES
('$ID', '$Name');
SELECT LAST_INSERT_ID();
inforamtion about LAST_INSERT_ID(); you are can read here
Since ID is auto incrementing, don't specify it in your query, you can simply do
INSERT INTO test
(Name)
VALUES
('$Name');
(Note if this is coming from PHP, be sure to watch out for SQL injection problems!)
If you specify an ID it will override MySQL's normal operations (which can be useful on occasion but not likely in this case).
You can refer back to the last inserted ID using LAST_INSERT_ID(). (note the column has to be inserted first, you can't call this and then insert with that value)
I'm trying to add a value to a table but not without checking if the value already exists. This is what I have so far:
IF NOT EXISTS (
SELECT series.seriesName
FROM series
WHERE series.seriesName='Avengers'
)
BEGIN
INSERT INTO series (seriesName) VALUES 'Avengers'
END;
Database is a MySQL db on Ubuntu
You can use IGNORE keyword here.
It could look like:
INSERT IGNORE INTO series (seriesName) VALUES 'Avengers'
The important thing is to create a unique key on seriesName field as it seems that you want it to be unique.
INSERT IGNORE doesn't make the insert when key value already exists.
If you would like to be able to get id (primary key value) for row that wasn't inserted (already existed), you can do the following trick:
INSERT IGNORE INTO series (seriesName) VALUES 'Avengers'
ON DUPLICATE KEY UPDATE seriesID= LAST_INSERT_ID(seriesID)
Then you will be able to get the ID with LAST_INSERT_ID() function no matter if the row was inserted or not.