I am using codeigniter insert_batch($table, $data);
I have a column name as "Comment posted by the customer".
When I try to batch insert data into it. The insert query changes as:
INSERT INTO `table` (`id`, `Name`, `Status`,`Comment posted by the` `customer`, `isActive`) VALUES (),(),()...;
As seen the column splits up into 2 parts -
1.Comment posted by the
2.customer
And this throws an error. How do I overcome this? I need to insert it considering it as a single column.
Try to format the query like this
INSERT INTO table (id, Name, Status, `Comment posted by the customer`, isActive) VALUES (),(),()
UPDATE
You can omit the column names and specify the values for all columns
INSERT INTO table VALUES (), (), ()
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 have a table named student which consists of (rollno, name, sem, branch)
If I want to INSERT only one value (i.e only name has to be entered) what is the query?
To insert values into specific columns, you first have to specify which columns you want to populate. The query would look like this:
INSERT INTO your_table_name (your_column_name)
VALUES (the_value);
To insert values into more than one column, separate the column names with a comma and insert the values in the same order you added the column names:
INSERT INTO your_table_name (your_column_name_01, your_column_name_02)
VALUES (the_value_01, the_value_02);
If you are unsure, have a look at W3Schools.com. They usually have explanations with examples.
insert into student(name) values("The name you wan to insert");
Be careful not to forget to insert the primary key.
First, if the table is all empty, just wanna make the column 'name' with values, it is easy to use INSERT INTO. https://www.w3schools.com/sql/sql_insert.asp.
`INSERT INTO TABLE_NAME (COLUMN_NAME) VALUES ("the values")`
Second, if the table is already with some values inside, and only wanna insert some values for one column, use UPDATE. https://www.w3schools.com/sql/sql_update.asp
UPDATE table_name SET column1 = value1 WHERE condition;
insert into student (name)
select 'some name'
or
insert into student (name)
values ('some name')
Following works if other columns accept null or do have default value:
INSERT INTO Student (name) VALUES('Jack');
Further details can be found from the Reference Manual:: 13.2.5 INSERT Syntax.
Execute this query, if you want the rest of columns as "#", do insert # inside the single quote which is left blank in query.
Also, there is no need of defining column name in the query.
insert into student values('','your name','','');
Thanks...
i want to set query if row not exist
IF NOT EXISTS (SELECT id FROM table WHERE id=1)
INSERT INTO table (id, name) VALUES (1,'abx');
and that if id!=1 then values are inserted and Check if row already exists
if any solution ?
thanks in advance ...
finally i soved it
INSERT INTO table (id, name) VALUES (1,'abx') ON DUPLICATE KEY UPDATE id = 1;
thanks for your suport
You have problem with data types - in SELECT id is compared to number (id = 1 - without apostrophes) and in INSERT id is written as a string (in apostrophes: '1').
INSERT IGNORE INTO `table_name` (`id`, `name`)
VALUES (1, 'abx');
MySQL does not have "IF NOT EXISTS".
For more info: How to 'insert if not exists' in MySQL?
You'll see that there are workarounds, but using MsSQL syntax won't work.
Using IGNORE should help: INSERT IGNORE INTO
Alternatively, simply let it fail.
I think that:
INSERT IGNORE INTO table (id, name) VALUES (1,'abx');
does that you want. It'll fail silently if the INSERT was unsuccessful.
Note that the key doesn't need quotes, it's an integer, not a string
I'm dealing with a relational table and I've been wondering if there's a way to lower the number of queries I need to make when inserting data to the tables..
Here are the queries I currently use:
I insert the "main" values.
INSERT INTO products
(title, description, status, url)
VALUES
('some title', 'description of doom', 1, 'some-title');
We make it insert the value only if it doesn't exist already.
INSERT IGNORE INTO values
(value)
VALUES
('example value');
Since I'm not sure if the query was actually inserted, I get the id..
SELECT id
FROM
values
WHERE
value = 'example value';
Where "?" is the ID I got from the last query.
INSERT INTO link
( id_product, id_catalog, id_value )
VALUES
( 33, 1, ? );
This means that each extra value I need to add will cost 3 queries. So my question is: Is there a more efficient way to do this?
You can do this to at least drop one of the queries:
INSERT INTO link
( id_product, id_catalog, id_value )
VALUES
( 33, 1, (SELECT id
FROM values
WHERE value = 'example value') );
I basically am replacing the '?' with a sub select of the second query to get the id.
"Is there a more efficient way to do this?"
No. Not really. Creating three things takes three inserts.
You should be able to tell whether the insert succeeded with the ROW___COUNT() function from inside MySQL. If calling from another language (e.g. PHP), the mysql_query or equivalent function will return the row count.
You could use an INSERT INTO ... ON DUPLICATE KEY UPDATE statement.
This does, however, require that the primary key be one of the values for the insert, so it doesn't work on tables with an auto-increment.