I am using the SQL feature phpMyAdmin to add 1 single record into my table. For simplicity, the record will be blank except for the 'symbol' field.
Table structure:
token_id = auto-increment, primary key
symbol = varchar(255)
every thing else is set to allow null entires, so should be irrelevant
I have tried the following queries, but all result in the same error:
unknown column 'symbol' in 'field list'
What I have tried:
INSERT INTO tokens (symbol) VALUES ('XYZ');
INSERT INTO tokens (symbol) VALUES ("XYZ");
INSERT INTO tokens (symbol) VALUES (XYZ);
INSERT INTO tokens.symbol VALUES ('XYZ');
INSERT INTO `tokens`.`symbol` VALUES ('XYZ');
Any suggestions?
Just for reference, trying the INSERT and using all columns and setting them to null results in the same exact error.
The correct format with backticks is
INSERT INTO tokens (symbol) VALUES ('XYZ');
FYI, I tried your first query on my server and worked fine so might be a problem with table structure.
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 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>);
I've got a script that's creating a table, and then inserting a row afterwards. Here is my SQL code executing to create the table:
CREATE TABLE polls (
id INT NOT NULL UNIQUE AUTO_INCREMENT,
name VARCHAR(255) NOT NULL UNIQUE,
author VARCHAR(255) NOT NULL,
created DATETIME NOT NULL,
expires DATETIME,
PRIMARY KEY(id)
)
And here is where I add a new row:
INSERT INTO polls
VALUES ('TestPoll'),('Billy Bob'),('2013-05-01 04:17:31'),('2013-05-01 04:17:31')
or
INSERT INTO polls
VALUES ('TestPoll','Billy Bob','2013-05-01 04:17:31','2013-05-01 04:17:31')
(I get the same error regardless)
I always get this error:
<class '_mysql_exceptions.OperationalError'>, OperationalError(1136, "Column count doesn't match value count at row 1"), <traceback object at 0x7f7bed982560>
Your syntax is wrong, try:
INSERT INTO polls
VALUES ('TestPoll','Billy Bob','2013-05-01 04:17:31','2013-05-01 04:17:31')
but if your table structure changes, your code will break, a safer version is:
INSERT INTO polls (name, author, created, expires)
VALUES ('TestPoll','Billy Bob','2013-05-01 04:17:31','2013-05-01 04:17:31')
Your INSERT query is not correctly formatted.
INSERT INTO polls (name, author, created, expires)
VALUES ('TestPoll','Billy Bob','2013-05-01 04:17:31','2013-05-01 04:17:31');
For more information, visit MySQL Reference Manual for the INSERT statement.
EDIT: It's always a good idea to explicitly type each column name, in case the table structure will change in some foreseeable future.
In mysql you have to pass column name in your insert query.After assigning column name your query will look like
INSERT INTO polls (name,author,created,expires) values ('TestPoll','Billy Bob','2013-05-01 04:17:31','2013-05-01 04:17:31');
Hope it helps.
This works:
INSERT INTO people (`name`, `job`) values ('Bob', 'sales')
but this fails:
INSERT INTO people (`name`, `Sold`) values ('Bob', '56')
Giving the error:
Unknown column 'Sold' in 'field list'
All fields are of type varchar. Can't figure this out!
This error is referring to the table column name Sold in the table people.
Your two examples are meaningless because you didn't successfully insert a value into the column Sold in the first example. Neither did you successfully insert a number into a VARCHAR column in the first example.
This has nothing to do with a type mismatch, the error says this column is unknown (doesn't exist).
Try this:
INSERT INTO people (name, sold) values ('Bob', '56')
I'll need more informations about table
INSERT INTO test (`name`, `Sold`) values ('Bob', '56')
worked for me given that the table was created like this..
create table test (id int, name varchar(6),Sold varchar(4),job varchar(7));
please check your data type in the created 'Sold' column.
it could be a boolean in your case..