grouping data in java - java

is there someway we can group similar data in java?
i want to group all the data with same id and print it out.
i am querying for the data using jdbc and was searching for a library i could use for this.
any idea?
thanks

Use a Map<GroupID, List<Data>>.
Map<Long, List<Data>> groups = new HashMap<Long, List<Data>>();
while (resultSet.next()) {
Long groupId = resultSet.getLong("groupId");
String col1 = resultSet.getString("col1");
String col2 = resultSet.getString("col2");
// ...
List<Data> group = groups.get(groupId);
if (group == null) {
group = new ArrayList<Data>();
groups.put(groupId, group);
}
group.add(new Data(groupId, col1, col2 /* ... */));
}
You could also just make it a property of another (parent) bean.
See also:
Collections and Maps tutorial

Ideally you should use a where clause in your SQL query to limit the returned data to the id in question:
select *
from table
where id = 'xxxxxx'
Of course if you will be printing out the data for all id's this may be a bad choice, as then your app will perform multiple sql queries, which usually will result in a performance hit.
As for grouping data in Java, take a look at java.util.HashMap (or any of the container classes that implement the Map interface). HashMap is a container of key-value pairs. In your case, the 'key' can be a String (or whichever data type applies) representing your id, and the 'value' can be an object to contain the data associated to the id key (i.e.: ArrayList of Strings, or a new class you define to help you manage the data)

Are you looking for the SQL ORDER BY clause?
SELECT columns
WHERE criteria
ORDER BY id ASC;
That will give you all the data in your criteria and will order it by the id column which naturally means that all the rows with the same id will appear consecutively.

Related

How to generate SQL from template with order by parameter using jOOQ?

I generate the SQL template like this with jOOQ 3.11.11.
DSLContext context = new DefaultDSLContext(conf);
Query query = context.select()
.from("table1")
.where(DSL.field("report_date").eq(DSL.param("bizdate")))
.orderBy(DSL.param("sort"));
String sqlTemp = context.renderNamedParams(query);
SQL template:
select *
from table1
where report_date = :bizdate
order by :sort
The SQL template is stored and the params are decided at realtime query condition.
ResultQuery resultQuery = context.resultQuery(sqlTemp, DSL.param("bizdate", "20190801"), DSL.param("sort", "id desc"));
The realtime SQL:
select *
from table1
where report_date = '20190801'
order by 'id desc'
There is something wrong with the order by clause.
So. How to replace the order by param sort with "id desc" or "name asc" and eliminate the quotes?
DSL.param() creates a bind variable, which is generated as ? in SQL, or :bizdate if you choose to use named parameters, or '20190801' if you choose to inline the bind variables. More about bind variables can be seen here.
You cannot use DSL.param() to generate column references or keywords. A column expression (e.g. a reference) is described in the jOOQ expression tree by the Field type. Keywords are described by the Keyword type, but you probably do not want to go this low level. Instead you want to handle some of the logic in your query expression. For example:
String sortField = "id";
SortOrder sortOrder = SortOrder.ASC;
Query query = context.select()
.from("table1")
.where(DSL.field("report_date").eq(DSL.param("bizdate")))
.orderBy(DSL.field(sortField).sort(sortOrder));
The mistake you're making is to think that you can use a single SQL template for all sorts of different dynamic SQL queries, but what if you're dynamically adding another predicate? Or another join? Or another column? You'd have to build a different jOOQ expression tree anyway. Just like here. You could store two SQL strings (one for each sort order), and repeat that for each sort column.
But, instead of pre-generating a single SQL string, I recommend you extract a function that takes the input parameters and generates the query every time afresh, e.g.:
ResultQuery<?> query(String bizDate, Field<?> sortField, SortOrder sortOrder) {
return context.selectFrom("table1")
.where(field("report_date").eq(bizDate))
.orderBy(sortField.sort(sortOrder));
}
Here is some further reading about using jOOQ for dynamic SQL:
https://www.jooq.org/doc/latest/manual/sql-building/dynamic-sql
https://blog.jooq.org/2017/01/16/a-functional-programming-approach-to-dynamic-sql-with-jooq

DynamoDB: how to query with multiple filter

I have a table and the structure looks like this:
my table structure
Here correlationId is my hashKey.
I can perform simple query using hashKey:
DynamoDBMapper mapper = new DynamoDBMapper(dynamoDB);
Pickup itemRetrieved = mapper.load(Pickup.class, key);
Now I want to query on basis of fields i.e correlationId, partnerId to get transactionId.
How should I do that?
Here is the sample code with multiple filter.
List<Pickup> pickupList = null;
DynamoDBMapper dynamoDBMapper = new DynamoDBMapper(dynamoDBClient);
Pickup pickup = new Pickup();
pickup.setCorrelationId(correlationId);
Map<String, AttributeValue> attributeValues = new HashMap<>();
attributeValues.put(":partnerId", new AttributeValue(partnerId));
DynamoDBQueryExpression<Pickup> queryExpression = new DynamoDBQueryExpression<Pickup>().withHashKeyValues(pickup)
.withFilterExpression("partnerId = :partnerId")
.withExpressionAttributeValues(attributeValues);
pickupList = dynamoDBMapper.query(Pickup.class, queryExpression);
pickupList.stream().forEach(i -> System.out.println(i.toString()));
Your partition key(correlation Id) is one keys on which you want to retrieve transactionid but it's missing partnerid.
Hence do these 3 steps
Step 1 - build a global secondary index on partnerid
Step 2 - filter on partition id
Step 3 - get transaction id
Query Filtering
DynamoDB’s Query function retrieves items using a primary key or an index key from a Local or Global Secondary Index. Each query can use Boolean comparison operators to control which items will be returned.
With today’s release, we are extending this model with support for query filtering on non-key attributes. You can now include a QueryFilter as part of a call to the Query function. The filter is applied after the key-based retrieval and before the results are returned to you. Filtering in this manner can reduce the amount of data returned to your application while also simplifying and streamlining your code.
The QueryFilter that you pass to the Query API must include one or more conditions. Each condition references an attribute name and includes one or more attribute values, along with a comparison operator. In addition to the usual Boolean comparison operators, you can also use CONTAINS, NOT_CONTAINS, and BEGINS_WITH for string matching, BETWEEN for range checking, and IN to check for membership in a set.
In addition to the QueryFilter, you can also supply a ConditionalOperator. This logical operator (either AND or OR) is used to connect each of the elements in the QueryFilter.

How to use 'LIKE' function to select array of Strings with JOOQ

I now want to use 'like' function with JOOQ to select data including array of string data by not case sensitive and partitial-match.
Table schema is:
CREATE TABLE favorites (
id int,
items varchar(100)[]
);
Sample data is:
INSERT INTO favorites (id, items)
VALUES (1, '{orange, lemon, banana}');
INSERT INTO favorites (id, items)
VALUES (2, '{apple, grape}');
To get first data, SQL is like:
SELECT id, items FROM favorites WHERE 'orange' = ANY (items);
My Goal is to select data by case-sensitive and partitial-match like: For example, using likeIgnoreCase("OraNge") or like("%ang%") ?
To develop below code with LIKE function:
Connection connection = ...;
DSLContext context = DSL.using(connection, ...);
List<Table> table = context.select().from(TABLE).fetchInto(Table.class);
How can I use like function?
Thank you in Advance.
The PostgreSQL value = ANY (array) operator cannot match values like the LIKE predicate. You will need to resort to an actual LIKE predicate instead. In SQL, you'd write:
SELECT id, items
FROM favorites
WHERE EXISTS (SELECT * FROM unnest(items) AS t(item) WHERE item ILIKE '%OraNge%')
Or, with jOOQ:
context.select(FAVORITES.ID, FAVORITES.ITEMS)
.from(FAVORITES)
.whereExists(
selectFrom(unnest(FAVORITES.ITEMS).as("t", "item")
.where(field(name("item", String.class)).likeIgnoreCase("%OraNge"))
)
.fetch();
The jOOQ version, as always, assumes you have this static import:
import static org.jooq.impl.DSL.*;
In addition, here are a few ways to use LIKE. You can always use the jOOQ LIKE predicates, see their documentation. In my second example, I use sql syntax in a string, just to prove you can. You can also use contains/startsWith/endsWith like you would with strings.
jooq.dsl()
.select()
.from(MY_TABLE)
.where(Employee.EMPLOYEES.LAST_NAME.like("ER")));
jooq.dsl()
.select()
.from(EMPLOYEES)
.where(Employee.EMPLOYEES.LAST_NAME.like("ER"))
.and("first_name like ?", "ST"));

Hibernate Criteria -- return records where column is distinct

Sample database table:
ID = 1, msgFrom = 'Hello', foobar = 'meh'
ID = 2, msgFrom = 'Goodbye', foobar = 'comments'
ID = 3, msgFrom = 'Hello', foobar = 'response'
Sample desired output (generated by hibernate query):
ID = 1, msgFrom = 'Hello', foobar = 'meh'
ID = 2, msgFrom = 'Goodbye', foobar = 'comments'
In the above example, the third record would be excluded from the results since the msgFrom column is the same. Let's say the Java/Hibernate class is called Message. I would like the results to be returned as a list of Message objects (or Objects that can be cast to Message, anyway). I want to use the Criteria API if possible. I saw this example on SO and it seems similar but I cannot implement it correctly as of yet.
select e from Message e
where e.msgFrom IN (select distinct m.msgFrom
from Message m
WHERE m.msgTo = ?
AND m.msgCheck = 0");
The reason I am doing this is to have the filtering of distinct records done on the database, so I am not interested in answers where I have to filter anything on the application server.
edit: Article showing basically what I want to do. http://oscarvalles.wordpress.com/2008/01/28/sql-distinct-on-one-column-only/
Please try this and let me know
DetachedCriteria msgFromCriteria = DetachedCriteria.forClass(Message.class);
ProjectionList properties = Projections.projectionList();
properties.add(Projections.groupProperty("messageFrom"));
properties.add(Projections.min("id"),"id");
msgFromCriteria.setProjection(properties);
Criteria criteria = s.createCriteria(Message.class);
criteria.add(Subqueries.propertiesIn(new String[]{"messageFrom","id"},
msgFromCriteria));
List<Message> list = criteria.list();
for(Message message:list){
System.out.println(message.getId()
+"-------"
+message.getMessageFrom()
+"-----"
+message.getFoobar());
}
The difficulty with this query is not so much with Hibernate, per se, but with the relational model in general. In the example, you say you expect rows 1 and 2, but why wouldn't you just as easily expect rows 2 and 3? It would be an arbitrary decision whether to return row 1 or row 3 since they both have the same value in the msgFrom field. Databases won't make arbitrary decisions like this. That's why distinct must be applied to the entire list of select columns, not a subset. There are database-specific ways of grabbing the first matching rows. For example, have a look at
SELECT DISTINCT on one column
Sometimes there will be a date column that you can use to decide which of the matching rows to return, but again the queries get somewhat complex:
How can I SELECT rows with MAX(Column value), DISTINCT by another column in SQL?
Fetch the row which has the Max value for a column
If you don't care about any of the other columns, you can just use a simple distinct, combined with Hibernate's constructor syntax (not tested):
select new Message(msgFrom) from (select distinct msgFrom from Message)
but you have to accept throwing away all the other columns.
In the end, I often end up just doing this in code as a post query filter. Another option is to create a another table, say CurrentMessage, that includes msgFrom as part of the key. There will be more work in keeping this table up to date (you need to update a row everytime you add a row to the Message table) but querying will be much easier.
DetachedCriteria msgFromCriteria = DetachedCriteria.forClass(Message.class);
msgFromCriteria.setProjection(Projections.distinct(Projections.property("msgFrom")));
....
Criteria criteria = getSession().createCriteria(Message.class);
criteria.add(Subqueries.propertyIn("msgFrom", msgFromCriteria));
criteria.list();

Problem with processing count column with Hibernate

I want to process the count column in my action how to do with help of hiberante.in my application every class is mapped to every table here I'm using billing table. In billing table details and other columns are there.
If i pass the query into execute sql query method it getting all the details, but it returns the count column as that corresponding dao class. How to process that column. Here is my query.
select u,b,b,count(b.details) from com.cod.model.Billing b,com.cod.model.User u where b.accountId=u.id and b.details not like '%Monthly Package With Usage Value Rs:0.0%' and b.details not like '%A/C Opened:%' and b.details not like '%Voucher Recharged%' and b.details not like '%default0%' group by u.username,b.details
Here it's getting user and billing table values but count column also comes as billing table object.
Don't make the results list type safe or hibernate will grab everything he can and push it in an instance of that object. When you make your results list not type safe hibernate will just return an List over which you can iterate and retrieve the fields you need.
List.get(0)[3] should result in you having your count.
String hql = "...";
Query query = session.createQuery(hql);
//List< com.cod.model.Billing> results = query.list();
List results = query.list();

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