I got lots of example to append/overwrite table in sql from AZ Databricks Notebook. But no single way to directly update, insert data using query or otherway.
ex. I want to update all row where (identity column)ID = 1143, so steps which I need to taken care are
val srMaster = "(SELECT ID, userid,statusid,bloburl,changedby FROM SRMaster WHERE ID = 1143) srMaster"
val srMasterTable = spark.read.jdbc(url=jdbcUrl, table=srMaster,
properties=connectionProperties)
srMasterTable.createOrReplaceTempView("srMasterTable")
val srMasterTableUpdated = spark.sql("SELECT userid,statusid,bloburl,140 AS changedby FROM srMasterTable")
import org.apache.spark.sql.SaveMode
srMasterTableUpdated.write.mode(SaveMode.Overwrite)
.jdbc(jdbcUrl, "[dbo].[SRMaster]", connectionProperties)
Is there any other sufficient way to achieve the same.
Note : Above code is also not working as SQLServerException: Could not drop object 'dbo.SRMaster' because it is referenced by a FOREIGN KEY constraint. , so it look like it drop table and recreate...not at all the solution.
You can use insert using a FROM statement.
Example: update values from another table in this table where a column matches.
INSERT INTO srMaster
FROM srMasterTable SELECT userid,statusid,bloburl,140 WHERE ID = 1143;
or
insert new values to rows where one of the existing column value matches
UPDATE srMaster SET userid = 1, statusid = 2, bloburl = 'https://url', changedby ='user' WHERE ID = '1143'
or just insert multiple values
INSERT INTO srMaster VALUES
(1, 10, 'https://url1','user1'),
(2, 11, 'https://url2','user2');
In SQL Server, you cannot drop a table if it is referenced by a FOREIGN KEY constraint. You have to either drop the child tables before removing the parent table, or remove foreign key constraints.
For a parent table, you can use the below query to get foreign key constraint names and the referencing table names:
SELECT name AS 'Foreign Key Constraint Name',
OBJECT_SCHEMA_NAME(parent_object_id) + '.' + OBJECT_NAME(parent_object_id) AS 'Child Table'
FROM sys.foreign_keys
WHERE OBJECT_SCHEMA_NAME(referenced_object_id) = 'dbo' AND
OBJECT_NAME(referenced_object_id) = 'PARENT_TABLE'
Then you can alter the child table and drop the constraint by its name using the below statement:
ALTER TABLE dbo.childtable DROP CONSTRAINT FK_NAME;
Related
We are facing a weird issue with Redshift and I am looking for help to debug it please. Details of the issue are following:
I have 2 tables and I am trying to perform left join as follows:
select count(*)
from abc.orders ot
left outer join abc.events e on **ot.context_id = e.context_id**
where ot.order_id = '222:102'
Above query returns ~7000 records. Looks like it is performing default join as we have only 1 record in [Orders] table with Order ID = ‘222:102’
select count(*)
from abc.orders ot
left outer join abc.events e on **ot.event_id = e.event_id**
where ot.order_id = '222:102'
Above query returns 1 record correctly. If you notice, I have just changed column for joining 2 tables. Event_ID in [Events] table is identity column but I thought I should get similar records even if I use any other column like Context_ID.
Further, I tried following query under the impression it should return all the ~7000 records as I am using default join but surprisingly it returned only 1 record.
select count(*)
from abc.orders ot
**join** abc.events e on ot.event_id = e.event_id
where ot.order_id = '222:102'
Following are the Redshift database details:
Cutdown version of table metadata:
CREATE TABLE abc.orders (
order_id character varying(30) NOT NULL ENCODE raw,
context_id integer ENCODE raw,
event_id character varying(21) NOT NULL ENCODE zstd,
FOREIGN KEY (event_id) REFERENCES events_20191014(event_id)
)
DISTSTYLE EVEN
SORTKEY ( context_id, order_id );
CREATE TABLE abc.events (
event_id character varying(21) NOT NULL ENCODE raw,
context_id integer ENCODE raw,
PRIMARY KEY (event_id)
)
DISTSTYLE ALL
SORTKEY ( context_id, event_id );
Database: Amazon Redshift cluster
I think, I am missing something essential while joining the tables. Could you please guide me in right direction?
Thank you
I am very new to sql and intermediate at python. Using sqlite3, how can I get a print() list of of primary and foreign keys (per table) in my database?
Using Python2.7, SQLite3, PyCharm.
sqlite3.version = 2.6.0
sqlite3.sqlite_version = 3.8.11
Also note: when I set up the database, I enabled FKs as such:
conn = sqlite3.connect(db_file)
conn.execute('pragma foreign_keys=ON')
I tried the following:
conn=sqlite3.connect(db_path)
print(conn.execute("PRAGMA table_info"))
print(conn.execute("PRAGMA foreign_key_list"))
Which returned:
<sqlite3.Cursor object at 0x0000000002FCBDC0>
<sqlite3.Cursor object at 0x0000000002FCBDC0>
I also tried the following, which prints nothing (but I think this may be because it's a dummy database with tables and fields but no records):
conn=sqlite3.connect(db_path)
rows = conn.execute('PRAGMA table_info')
for r in rows:
print r
rows2 = conn.execute('PRAGMA foreign_key_list')
for r2 in rows2:
print r2
Unknown or malformed PRAGMA statements are ignored.
The problem with your PRAGMAs is that the table name is missing. You have to get a list of all tables, and then execute those PRAGMAs for each one:
rows = db.execute("SELECT name FROM sqlite_master WHERE type = 'table'")
tables = [row[0] for row in rows]
def sql_identifier(s):
return '"' + s.replace('"', '""') + '"'
for table in tables:
print("table: " + table)
rows = db.execute("PRAGMA table_info({})".format(sql_identifier(table)))
print(rows.fetchall())
rows = db.execute("PRAGMA foreign_key_list({})".format(sql_identifier(table)))
print(rows.fetchall())
SELECT
name
FROM
sqlite_master
WHERE
type ='table' AND
name NOT LIKE 'sqlite_%';
this sql will show all table in database, for eache table run sql PRAGMA table_info(your_table_name);, you can get the primary key of the table.
Those pictures show what sql result like in my database:
first sql result
second sql result
How do you query table names and row counts for all tables in a schema using HP NonStop SQL/MX?
Thanks!
This might help you, althought this is more standard SQL and im not sure how much variation comes into sqlmx
SELECT
TableName = t.NAME,
TableSchema = s.Name,
RowCounts = p.rows
FROM
sys.tables t
INNER JOIN
sys.schemas s ON t.schema_id = s.schema_id
INNER JOIN
sys.indexes i ON t.OBJECT_ID = i.object_id
INNER JOIN
sys.partitions p ON i.object_id = p.OBJECT_ID AND i.index_id = p.index_id
WHERE
t.is_ms_shipped = 0
GROUP BY
t.NAME, s.Name, p.Rows
ORDER BY
s.Name, t.Name
Obviously this is an example, replace example data and table info with yours
Here is how to list the tables in a sql/mx schema, note that the system catalog name given here is an example, replace NONSTOP_SQLMX_SYSNAME with NONSTOP_SQLMX_xxxx where xxxx is the Expand node name of your system.
Also the definition schema name includes the schema version number, this example uses 3600. This example lists all the base table names in schema JDFCAT.T.
See chapter 10 of the SQL/MX reference manual for information on the metadata tables.
The table row counts are not stored in the system metadata, so you can't get them from there. For a table do SELECT ROW COUNT FROM TABLE;
SELECT
O.OBJECT_NAME
FROM
NONSTOP_SQLMX_SYSNAME.SYSTEM_SCHEMA.CATSYS C
INNER JOIN NONSTOP_SQLMX_SYSNAME.SYSTEM_SCHEMA.SCHEMATA S
ON (S.CAT_UID = C.CAT_UID)
INNER JOIN JDFCAT.DEFINITION_SCHEMA_VERSION_3600.OBJECTS O
on S.SCHEMA_UID = o.SCHEMA_UID
WHERE C.CAT_NAME = 'JDFCAT' AND
S.SCHEMA_NAME = 'T' AND
O.OBJECT_TYPE = 'BT'
READ UNCOMMITTED ACCESS;
If a Cloud Spanner table is created with nullable columns, is it possible to add a NOT NULL constraint on a column without recreating the table?
You can add a NOT NULL constraint to a non-key column. You must first ensure that all rows actually do have values for the column. Spanner will scan the data to verify before fully applying the NOT NULL constraint. More information about how to alter tables is here and here.
However, you can not add such a constraint to a key column. That kind of change would require rewriting all the data in the table, because the nullness of the key affects how the data is encoded. The only option for making that change is to create a new table that's set up the way you want, make code changes to support using both tables temporarily, gradually move the data from the old table to the new table, and eventually change the code to use only the new table and drop the old table. If you further then wanted the original table name, you'd have to do the whole thing again.
Unfortenately there is not way to add not null column
The way to do it:
1 add nullable column
ALTER TABLE table1 ADD COLUMN column1 STRING(255)
UPDATE table1.column1, SET NOT NULL VALUE to the column (if the table is not empty).
UPDATE TABLE table1 SET column1 = "<GENERATED DATA>"
Add constraint
ALTER TABLE table1 ADD COLUMN column1 STRING(255) NOT NULL
Thanks.
Creating a non-nullable column in Spanner on an existing table is typically a three step process:
# add new column to table
ALTER TABLE <table_name> ADD COLUMN <column_name> <value_type>;
# create default values
UPDATE <table_name> SET <column_name>=<default_value> WHERE TRUE;
# add constraint
ALTER TABLE <table_name> ALTER COLUMN <column_name> <value_type> NOT NULL;
I am using this link.
I have connected my cpp file with Eclipse to my Database with 3 tables (two simple tables
Person and Item
and a third one PersonItem that connects them). In the third table I use one simple primary and then two foreign keys like that:
CREATE TABLE PersonsItems(PersonsItemsId int not null auto_increment primary key,
Person_Id int not null,
Item_id int not null,
constraint fk_Person_id foreign key (Person_Id) references Person(PersonId),
constraint fk_Item_id foreign key (Item_id) references Items(ItemId));
So, then with embedded sql in c I want a Person to have multiple items.
My code:
mysql_query(connection, \
"INSERT INTO PersonsItems(PersonsItemsId, Person_Id, Item_id) VALUES (1,1,5), (1,1,8);");
printf("%ld PersonsItems Row(s) Updated!\n", (long) mysql_affected_rows(connection));
//SELECT newly inserted record.
mysql_query(connection, \
"SELECT Order_id FROM PersonsItems");
//Resource struct with rows of returned data.
resource = mysql_use_result(connection);
// Fetch multiple results
while((result = mysql_fetch_row(resource))) {
printf("%s %s\n",result[0], result[1]);
}
My result is
-1 PersonsItems Row(s) Updated!
5
but with VALUES (1,1,5), (1,1,8);
I would like that to be
-1 PersonsItems Row(s) Updated!
5 8
Can somone tell me why is this not happening?
Kind regards.
I suspect this is because your first insert is failing with the following error:
Duplicate entry '1' for key 'PRIMARY'
Because you are trying to insert 1 twice into the PersonsItemsId which is the primary key so has to be unique (it is also auto_increment so there is no need to specify a value at all);
This is why rows affected is -1, and why in this line:
printf("%s %s\n",result[0], result[1]);
you are only seeing 5 because the first statement failed after the values (1,1,5) had already been inserted, so there is still one row of data in the table.
I think to get the behaviour you are expecting you need to use the ON DUPLICATE KEY UPDATE syntax:
INSERT INTO PersonsItems(PersonsItemsId, Person_Id, order_id)
VALUES (1,1,5), (1,1,8)
ON DUPLICATE KEY UPDATE Person_id = VALUES(person_Id), Order_ID = VALUES(Order_ID);
Example on SQL Fiddle
Or do not specify the value for personsItemsID and let auto_increment do its thing:
INSERT INTO PersonsItems( Person_Id, order_id)
VALUES (1,5), (1,8);
Example on SQL Fiddle
I think you have a typo or mistake in your two queries.
You are inserting "PersonsItemsId, Person_Id, Item_id"
INSERT INTO PersonsItems(PersonsItemsId, Person_Id, Item_id) VALUES (1,1,5), (1,1,8)
and then your select statement selects "Order_id".
SELECT Order_id FROM PersonsItems
In order to achieve 5, 8 as you request, your second query needs to be:
SELECT Item_id FROM PersonsItems
Edit to add:
Your primary key is autoincrement so you don't need to pass it to your insert statement (in fact it will error as you pass 1 twice).
You only need to insert your other columns:
INSERT INTO PersonsItems(Person_Id, Item_id) VALUES (1,5), (1,8)