I am upgrading my database from mssql to mysql, I am struct on creating foreign keys
In MSSQL I was using
alter table ac_master add constraint 'ac_master_table_conf' foreign key (ac_code) references table_conf (ac_code)
MySql is giving error on this
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use
near ''ac_master_table_conf' foreign key (ac_code) references
table_conf(ac_code)' at line 1
Remove the quotes from around your constraint name e.g.
ALTER TABLE ac_master ADD CONSTRAINT ac_master_table_conf FOREIGN KEY (ac_code) REFERENCES table_conf (ac_code)
As far as I can tell, neither SQL Server nor MySQL use single quote syntax for identifiers. You probably want ac_master_table_conf rather than 'ac_master_table_conf'.
Other than that, the ALTER TABLE syntax is documented, you don't need to guess.
no quotes are involved in mysql if it is a name of a field or a table.
the only need of quotes is to specify a string as below
select * from abc where column = 'string';
so be careful next time.
Related
I try to fix some error in Jira as mention in this page but when I try to run an SQL query, I got the below error:
ERROR 1064 (42000): You have an error in your SQL syntax; check the
manual that corresponds to your MySQL server version for the right
syntax to use near 'CONSTRAINT
fk_ao_8542f1_ifj_obj_attr_val_object_attribute_id' at line 1
My Query is:
ALTER TABLE AO_8542F1_IFJ_OBJ_ATTR_VAL
DROP CONSTRAINT `fk_ao_8542f1_ifj_obj_attr_val_object_attribute_id`
How I can fix this Query? I use MySQL version 5.7
Drop MySQL foreign key constraints
To drop a foreign key constraint, you use the ALTER TABLE statement:
ALTER TABLE table_name
DROP FOREIGN KEY constraint_name;
Judging by the name of the constraint, I believe it's a foreign key and MySQL has a different approach to drop foreign keys:
ALTER TABLE AO_8542F1_IFJ_OBJ_ATTR_VAL
DROP FOREIGN KEY `fk_ao_8542f1_ifj_obj_attr_val_object_attribute_id`
I have a table named 'Index'. I realize that this is a keyword in MySQL, and was wondering how I can reference this table in queries?
My error:
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''Index' (ID) )' at line 9
Use backticks:
references `Index` (ID)
You actually already did this when you created the Index table (and you must have used backticks or else you could not have created the table):
CREATE TABLE `Index`
You should avoid naming databases, tables, and columns using MySQL keywords.
you can use backticks instead of quotes
CONSTRAINT grants_fk_index
FOREIGN KEY (Index_fk)
references `Index` (ID)
I'm trying to execute the following SQL command in MySQLWorkbench but it's giving me an error.
Command:
ALTER TABLE `ABC`.`GroupMembers`
ADD CONSTRAINT `FK_PROFILES`
FOREIGN KEY ()
REFERENCES `ABC`.`profiles` ()
ON DELETE NO ACTION
ON UPDATE NO ACTION;
Error:
Operation failed: There was an error while applying the SQL script to the database.
ERROR 1064: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ')
REFERENCES `ABC`.`profiles` ()
ON DELETE NO ACTION
ON UPDATE NO ACTION' at line 3
SQL Statement:
ALTER TABLE `ABC`.`GroupMembers`
ADD CONSTRAINT `FK_PROFILES`
FOREIGN KEY ()
REFERENCES `ABC`.`profiles` ()
ON DELETE NO ACTION
ON UPDATE NO ACTION
Not sure what's up. This script was generated by MySQLWorkbench
There has to be a column (or columns) in both lists. Those can't be empty.
For example:
FOREIGN KEY (profile_id)
-- ^^^^^^^^^^
REFERENCES `ABC`.`profiles` (id)
-- ^^
The datatypes of the columns much match exhactly. And values stored in the foreign key column must match the value in a row in the referenced table. (In this example, all values in profile_id must match the value of the id column in the profiles table.
MYSQL is saying its an syntax issue, and as I can see, there is a couple of things missing in you code.
Please check this link and learn more about the sintaxis in mysql regarding constraints.
Hope it helps.
Edit:
Ok, so just to enforce spencer7593's answer (which should be marked as answer if it solves your issue):
...
/profile_id would be the name you will set to the foreign key
FOREIGN KEY (profile_id)
...
/The references should be of the same value.
/`table_name`.`column_name` is the referenced column
/ id is the column on the table which will hold foreign key
REFERENCES `ABC`.`profiles` (id)
I have created a database with two tables user and userdiary.
In the "user" table I have id and email fields.
In the "userdiary" table I have a field called email. I want to link this field with the one in the "user" table.
I'm using phpmyAdmin to add sql statements.
I tried doing ALTER TABLE userdiary FOREIGN KEY (email) REFERENCES user(email)
but I get the error #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FOREIGN KEY (email) REFERENCES user(email)' at line 1
I looked at many forums but could not find a solution to this issue.
It will be great if any of you could help me resolve this issue.
Thanks.
You are missing the keyword add:
ALTER TABLE userdiary ADD FOREIGN KEY (email) REFERENCES user(email)
----------------------^
I keep getting this sql error
"#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Option (OptionId)' at line 1"
when I try and add a foreign key to the OptionId field from the Question Table to the OptionId(pk) field in the Option field. I don't get wy I keep getting the error because I don't see what is wrong with it.
Below is the foreign key constraint using ALTER TABLE:
ALTER TABLE Question ADD CONSTRAINT FK_OptionId FOREIGN KEY (OptionId) REFERENCES Option (OptionId)
Table names and syntax are correct, I made sure by double checking.
Why is it not working?
option is a reserved word in MySQL and must be surrounded by backticks.
ALTER TABLE Question
ADD CONSTRAINT FK_OptionId FOREIGN KEY (OptionId)
REFERENCES `Option` (OptionId)