Is there a way to partition a mysql table, by range columns, where the 2 columns in question are of datetime...
I need to partition a table into 4 partitions, p0 to p3: with p0 having all rows with "id"=0, and p1 having all rows with "id"=1 and "doj" before february, and p2 having "id"=1 and "doj" before march, and partition "future" with others. The value for id, is guaranteed to be either 0 or 1 always.
I've tried to come up with a solution and the following is what i arrived at:
create table temp
(id int not null primary key,
doj datetime not null primary key)
partition by range columns(id,month(doj))
(partition p0 values less than (1, 13),
partition p1 values less than (2, 2),
partition p2 values less than (2, 3),
partition p3 values less than(maxvalue, maxvalue));
I get the following error when i try and execute the above sql command:
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 '(doj))
(partition p0 values less than (1, maxvalue),
partition p1 values less t' at line 4
Any insights as to how to partition the table in a manner described above would be much appreciated.
Version of MySql used is:
5.6.19-67.0-log
mysql mysql-5.6 partitioning
asked 5 mins ago
Sai VaibhavSai Vaibhav
1
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I need to partition a table into 4 partitions, p0 to p3: with p0 having all rows with "id"=0, and p1 having all rows with "id"=1 and "doj" before february, and p2 having "id"=1 and "doj" before march, and partition "future" with others. The value for id, is guaranteed to be either 0 or 1 always.
I've tried to come up with a solution and the following is what i arrived at:
create table temp
(id int not null primary key,
doj datetime not null primary key)
partition by range columns(id,month(doj))
(partition p0 values less than (1, 13),
partition p1 values less than (2, 2),
partition p2 values less than (2, 3),
partition p3 values less than(maxvalue, maxvalue));
I get the following error when i try and execute the above sql command:
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 '(doj))
(partition p0 values less than (1, maxvalue),
partition p1 values less t' at line 4
Any insights as to how to partition the table in a manner described above would be much appreciated.
Version of MySql used is:
5.6.19-67.0-log
mysql mysql-5.6 partitioning
asked 5 mins ago
Sai VaibhavSai Vaibhav
1
New contributor
Sai Vaibhav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I need to partition a table into 4 partitions, p0 to p3: with p0 having all rows with "id"=0, and p1 having all rows with "id"=1 and "doj" before february, and p2 having "id"=1 and "doj" before march, and partition "future" with others. The value for id, is guaranteed to be either 0 or 1 always.
I've tried to come up with a solution and the following is what i arrived at:
create table temp
(id int not null primary key,
doj datetime not null primary key)
partition by range columns(id,month(doj))
(partition p0 values less than (1, 13),
partition p1 values less than (2, 2),
partition p2 values less than (2, 3),
partition p3 values less than(maxvalue, maxvalue));
I get the following error when i try and execute the above sql command:
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 '(doj))
(partition p0 values less than (1, maxvalue),
partition p1 values less t' at line 4
Any insights as to how to partition the table in a manner described above would be much appreciated.
Version of MySql used is:
5.6.19-67.0-log
mysql mysql-5.6 partitioning
asked 5 mins ago
Sai VaibhavSai Vaibhav
1
New contributor
Sai Vaibhav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I need to partition a table into 4 partitions, p0 to p3: with p0 having all rows with "id"=0, and p1 having all rows with "id"=1 and "doj" before february, and p2 having "id"=1 and "doj" before march, and partition "future" with others. The value for id, is guaranteed to be either 0 or 1 always.
I've tried to come up with a solution and the following is what i arrived at:
create table temp
(id int not null primary key,
doj datetime not null primary key)
partition by range columns(id,month(doj))
(partition p0 values less than (1, 13),
partition p1 values less than (2, 2),
partition p2 values less than (2, 3),
partition p3 values less than(maxvalue, maxvalue));
I get the following error when i try and execute the above sql command:
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 '(doj))
(partition p0 values less than (1, maxvalue),
partition p1 values less t' at line 4
Any insights as to how to partition the table in a manner described above would be much appreciated.
Version of MySql used is:
5.6.19-67.0-log
mysql mysql-5.6 partitioning
mysql mysql-5.6 partitioning
asked 5 mins ago
Sai VaibhavSai Vaibhav
1
New contributor
Sai Vaibhav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
Sai VaibhavSai Vaibhav
1
New contributor
Sai Vaibhav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
Sai VaibhavSai Vaibhav
1
New contributor
Sai Vaibhav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
Sai VaibhavSai Vaibhav
1
asked 5 mins ago
Sai VaibhavSai Vaibhav
1
1
New contributor
Sai Vaibhav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Sai Vaibhav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Sai Vaibhav is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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