MySQL query
select s.site_id,s.site_name,u.username
FROM cloud_search.dbo.site_request s join user_db.dbo.user_info u
on u.user_id=s.site_id;
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 '.site_request s join user_db.dbo.user_info u on u.user_id=s.site_id' at line 1
What am i doing wrong here..?
Are you sure you're using dbo it's usually for MSSQL
It's probably
select s.site_id,s.site_name,u.username FROM cloud_search.site_request s join user_db.user_info u on u.user_id=s.site_id;
Related
I am using mysql 8.0.21. I am trying to execute this statement:
WITH t
AS (SELECT h.store_hsh,
Avg(s.sales) AS SALES_AVG,
Avg(s.customers) AS CSTMR_AVG
FROM sat_day_facts s
INNER JOIN link_day l
ON s.link_day_hsh = l.link_day_hsh
INNER JOIN hub_store h
ON h.store_hsh = l.store_hsh
GROUP BY h.store_hsh)
SELECT t.store_hsh,
Now() AS LOAD_DATETIME,
Now() AS LOAD_END_DATETIME,
t.sales_avg,
t.cstmr_avg,
(SELECT 1 + Count(*)
FROM t t1
WHERE t1.sales_avg > t.sales_avg) AS SALES_RANK;
but I keep having this error
ProgrammingError: 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 'WITH t as (select h.STORE_HSH, AVG(s.SALES) as SALES_AVG , AVG(s.CUSTOMERS) as' at line 1
I checked the manual of SQL but still don't understand what is going wrong.
Does anyone could help?
select
pengarang_id,buku_judul,penerbit_id
from buku
left join bukupengarang on buku.penerbit_id = bukupengarang.pengarang_id
INTERSECT
select
pengarang_id,buku_judul,penerbit_id
from buku right join bukupengarang on buku.penerbit_id = bukupengarang.pengarang_id;`
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 'INTERSECT
select pengarang_id,buku_judul,penerbit_id
from buku
right join bukup' at line 5
could anyone tell me whts problem here?
i dont understand abt this one, i just read it on inet n still didnt get it
MySQL does not support the INTERSECT operator. You can emulate the INTERSECT operator in MySQL using join clauses.
https://www.mysqltutorial.org/mysql-intersect/
It works with MariaDB since version 10.3.0.
I got an error like this "Additional information: 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 " when i running my code.Iam using webapi
string query = "SELECT `theatredetails`.*,`location`.`LocationName,`moviemaster`.`MovieMasterId`,`moviemaster`.`MovieName`,`moviemaster`.`Image` FROM `theatredetails` INNER JOIN `location` ON `theatredetails`.`LocationId`=`location`.`LocationId` INNER JOIN `moviemaster` ON `moviemaster`.`TheatreDetailsId`=`theatredetails`.`TheatreDetailsId`";
Iam using 3 inner joins,is there any error on my query?
You forgot a backtick
`location`.`LocationName
^---here
SELECT core_student.STUDENT_ID, core_student.FIRST_NAME, core_student.LAST_NAME
FROM `core_student`
INNER_JOIN `ssp_student`
WHERE ssp_student.STUDENT_ID = core_student.STUDENT_ID;
throws an error on the WHERE clause:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to user near 'ssp_student' WHERE ssp_student.STUDENT_ID = core_student.STUDENT_ID
I just want to select the fields listed in the SELECT, then join all columns from ssp_student where the ssp_student.STUDENT_ID field is the same as the core_student.STUDENT_ID field.
Correct way is
SELECT core_student.STUDENT_ID, core_student.FIRST_NAME, core_student.LAST_NAME
FROM `core_student`
INNER JOIN `ssp_student`
on ssp_student.STUDENT_ID = core_student.STUDENT_ID;
Why do I always get a syntax error on the following inner join:
SELECT 'myapp_instagram_shop'.id
FROM 'myapp_instagram_shop'
INNER JOIN 'myapp_instagram_shop_picture'
ON 'myapp_instagram_shop'.id = 'myapp_instagram_shop_picture'.shop_id;
Error is:
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 ''myapp_instagram_shop'
INNER JOIN 'myapp_instagram_shop_picture' ON 'sh' at line 1
try
SELECT myapp_instagram_shop.id FROM myapp_instagram_shop INNER JOIN myapp_instagram_shop_picture ON myapp_instagram_shop.id = myapp_instagram_shop_picture.shop_id;