Τη Δευτέρα, 27 Μαρτίου 2017 - 2:27:31 π.μ. UTC+3, ο χρήστης INADA Naoki έγραψε:
> > i dont have to update table set column1 = this value, column2=that value and
> > so on
>
> Why do you think so? Did you really read the manual?
>
> mysql> create table test_update (a int primary key, b int, c int);
> Query OK, 0 rows affected (0.02 sec)
>
> mysql> insert into test_update values (1, 2, 3);
> Query OK, 1 row affected (0.00 sec)
>
> mysql> update test_update set (b, c) values (4, 5) where a = 1;
> 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 '(b, c) values (4, 5) where a = 1' at line 1
>
> mysql> update test_update set b=4, c=5 where a = 1;
> Query OK, 1 row affected (0.01 sec)
> Rows matched: 1 Changed: 1 Warnings: 0
>
> >
> > It's just when the LIKE clause jumps in that is causing all this trouble....
>
> Your MariaDB said:
>
> > check the manual that corresponds to your MariaDB server version for the
> > right syntax to use near '(pagesID, host, ...
>
> MariaDB / MySQL shows part of your SQL from where they failed to parse.
> In your case, your MariaDB can't parse from '('
> LIKE clause is not problem for this issue?
Yes indeed it is.
I was just so sure that UPDATE was working like INSERT and i was persistent
that the WHERE LIKE clause was cauing this.
I'am still surprised that:
> mysql> update test_update set (b, c) values (4, 5) where a = 1;
is failign to parse. It just seems so undoubtly straightforward and correct
syntactically.
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