Duplicate entry '20' for key 'user_id'
$userid = its foreign key (20)
$query = mysql_query("insert into qualification (q_id,course_name,institute_name,pass_year,user_id) values ('','$coursename','$institutename','$passyear','$userid')") or die(mysql_error());
Does anybody have an idea on how to skip this error and add multiple record on the same userid ?
When you get this error, you have a UNIQUE or PRIMARY key constraint for user_id in your database. As Marc B already said. Remove the unique constraint from the user_id.
You asked Marc:
Why remove unique constraint?
Simply because then you can't have two records with the same user_id in your table. Then you also can't use it as a key for table. You can still use an index on user_id to speed up queries for this field.
Related
I want to add complex unique key to existing table. Key contains from 4 fields (user_id, game_id, date, time).
But table have non unique rows.
I understand that I can remove all duplicate dates and after that add complex key.
Maybe exist another solution without searching all duplicate data. (like add unique ignore etc).
UPD
I searched, how can remove duplicate mysql rows - i think it's good solution.
Remove duplicates using only a MySQL query?
You can do as yAnTar advised
ALTER TABLE TABLE_NAME ADD Id INT AUTO_INCREMENT PRIMARY KEY
OR
You can add a constraint
ALTER TABLE TABLE_NAME ADD CONSTRAINT constr_ID UNIQUE (user_id, game_id, date, time)
But I think to not lose your existing data, you can add an indentity column and then make a composite key.
The proper syntax would be - ALTER TABLE Table_Name ADD UNIQUE (column_name)
Example
ALTER TABLE 0_value_addition_setup ADD UNIQUE (`value_code`)
I had to solve a similar problem. I inherited a large source table from MS Access with nearly 15000 records that did not have a primary key, which I had to normalize and make CakePHP compatible. One convention of CakePHP is that every table has a the primary key, that it is first column and that it is called 'id'. The following simple statement did the trick for me under MySQL 5.5:
ALTER TABLE `database_name`.`table_name`
ADD COLUMN `id` INT NOT NULL AUTO_INCREMENT FIRST,
ADD PRIMARY KEY (`id`);
This added a new column 'id' of type integer in front of the existing data ("FIRST" keyword). The AUTO_INCREMENT keyword increments the ids starting with 1. Now every dataset has a unique numerical id. (Without the AUTO_INCREMENT statement all rows are populated with id = 0).
Set Multiple Unique key into table
ALTER TABLE table_name
ADD CONSTRAINT UC_table_name UNIQUE (field1,field2);
I am providing my solution with the assumption on your business logic. Basically in my design I will allow the table to store only one record for a user-game combination. So I will add a composite key to the table.
PRIMARY KEY (`user_id`,`game_id`)
Either create an auto-increment id or a UNIQUE id and add it to the natural key you are talking about with the 4 fields. this will make every row in the table unique...
For MySQL:
ALTER TABLE MyTable ADD MyId INT AUTO_INCREMENT PRIMARY KEY;
If yourColumnName has some values doesn't unique, and now you wanna add an unique index for it. Try this:
CREATE UNIQUE INDEX [IDX_Name] ON yourTableName (yourColumnName) WHERE [id]>1963 --1963 is max(id)-1
Now, try to insert some values are exists for test.
I have a SQL query as follows-
"INSERT INTO users(id, rank) SELECT v.user, v.vote FROM votes v WHERE
v.assertion = '$ID' ON DUPLICATE KEY UPDATE
rank = ( CASE WHEN v.vote = '1' THEN rank+50 WHEN v.vote = '-1'
THEN rank-200 WHEN v.vote = '3' THEN rank+100 ELSE rank END)"
applied on a database with a table users with and id and rank field, and a votes table with a user and vote field. I have to update the rank of the users in the users table based on their vote.
I really like this kind of query, but I've noticed a problem: every time I execute this from my PHP script the query adds a row to the users table completely empty (with only an ID, which is A_I, and a rank of 1, when usually there would be other field as well). I can't really wrap my head around why this happens.
Any help/idea?
Your table does not have a primary key first provide a primary key to id
run this sql query
alter table user add primary key (id)
and than try it will work
There are two possible reasons :
The id column is not the primary key, and probably you table doesn't have a primary key at all.
Create a primary key like this :
alter table user add primary key (id)
If you insert an value of 0 in an auto increment column, a new id is generated. An auto incremented column must not contain the value 0.
There is also a more general problem with your approach : in fact you only insert the user id and the rank, other compulsory fields in the table (username) are missing. The insert part does not seem to be valid for this reason. If you use an insert on duplicate key update, you must make sure that the result is correct which ever of insert and update is executed.
I have two tables users, and posts. The posts table has a column with the foreign key user_id which corresponds to the id in the users table. If another row is inserted to the posts table and it has the same value in the user_id column as another row, I get an error:
"Duplicate entry '4' for key in 'user_id'"
How can I allow for duplicate entries in this user_id column?
I went into the relational view, edit keys, and changed it from unique, to index. Thanks to Cory.
I have a mysql on duplicate key statement.
mysql_query("INSERT INTO statistics (classify, apply) VALUES ('$classify', 1)
ON DUPLICATE KEY UPDATE apply = apply + 1");
id classify apply
1 A 1
but it didn't update the existing row and it keep add another row, Where is the problem?
It's probably the column classify is not unique. You need to have a UNIQUE field in the table to make ON DUPLICATE KEY UPDATE work. If you have not set one, you can execute this statement below.
ALTER TABLE statistics ADD CONSTRAINT tb_uq UNIQUE (classify)
ON DUPLICATE KEY will update a row only when you try to insert a record that would throw a duplicate keys error (like the name states). So this happens only if you are a using a unique key or a primary key for that column. It looks like you didn't created a unique key for the classify column.
I am relatively new in php and mysql.The problem that i am facing while i inserting value in my leave table.My leave table containing following column..
1.lid(INT primary key)
2.empname(varchar)
3.username(varchar)
4.nod(INT)
5.sdate(DATE)
6.edate(DATE)
7.reason(varchar)
8.action(varchar)
9.empID (INT FOREIGN KEY)
here empID is the foreign key references from users table. The problem that im facing while inserting values into the leave table.ERROR is given below
Cannot add or update a child row: a foreign key constraint fails (db_attendance1.leave, CONSTRAINT leave_ibfk_1 FOREIGN KEY (empID) REFERENCES users (empID))
here i just send the query and here it is..
mysql_query("INSERT INTO `leave`
(`empname`, `username`,
`nod`, `sdate`, `edate`,
`reason`,`action`)
VALUES ('$empname', '$username',
'$nod', '$sdate',
'$edate', '$reason','');",
$dbCon) or die(mysql_error());
You can put
SET FOREIGN_KEY_CHECKS=0;
and run your query. Once you are done again set it back to 1 by
SET FOREIGN_KEY_CHECKS=1;
A foreign key constraint means that you one table doesn't accept inserts, updates or deletes that would 'break' the foreign key. This means, you can't update a EmpID if the new EmpID doesn't exist in the users. You can't add a new EmpID if it doesn't exist in the users table, etcetera.
So to solve this issue, you need to make sure that the EmpID you're trying to add to table 'leave', first exists in table 'users'.
Foreign keys can be a real powerful item, but can be a real pain too. Since the DB you're working on had foreign key constraints, I suggest you read on them a bit: http://en.wikipedia.org/wiki/Foreign_key
Borniet, you helped me solve my similar problem.
#OP - All I had to do to fix this was create a corresponding row in the table so that the foreign key would exist.
E.g. Table 1 has column Name
Table 2 has column friends_name, a foreign key tied to Name in table 1.
I got this error because I was trying to insert a row into table 2, where the friends_name referenced a non existing Name in table 1. So I created the name and we're off to the races :).