My json data
[{"cameraid":"000000001","timestamp":"2016-06-17 23:08","filename":"7e3800fbd0557c683874ed2f41ed7057"},
{"cameraid":"000000002","timestamp":"2016-06-17 23:08","filename":"b260cc730da88a6af4e5038d6e1e32db"}]
How can i have cameraid to link to a specific name?
Like example my cameraid "000000001" to be call as bedok.
Anybody got any idea on how to do it?
Create another collection with names(lets call it db_name) and link the cameraid to the _id of db_name collection. This way you can fetch the names using cameraid. This is more like primary key and foreign key concept in relational database(RDBMS).
More on this with code here: Primary Key and Foreign Key Concept in MongoDB
Assuming you want to add a name field for specific cameraid,
Try following query:-
db.collname.update({"cameraid":"000000001"},{$set : {name : 'bedok'}});
EDIT:-
The above query will update only one record which matches the query {"cameraid":"000000001"}.
Add multi:true to the query, to multiple records.
db.collname.update({"cameraid":"000000001"},{$set : {name : 'bedok'}},{multi : true});
Now it will update all the records that matches the query {"cameraid":"000000001"}.
Related
I have a flow-files with the below structure
{
"PN" : "U0-WH",
"INPUT_DATE" : "44252.699895833335",
"LABEL" : "Marker",
"STATUS" : "Approved",
}
and I need to execute an update statement using some fields
update table1 set column1 = 'value' where pn=${PN}
I found convertJsonToSQL but am not sure how to use it in this case
You can use a processor namely ConvertjSONToSQL. Using this you can convert your json into an update query.
ConvertjSONToSQL Description
It takes the following parameters :
1. JDBC Connection Pool : Create a JDBC pool which takes DB connection information as input.
2. Statement Type : Here you need to provide type of statement you want to create. In your case its 'UPDATE'.
3. Table Name : Name of the table for which update query needed to be created
4. Schema Name : Name of the schema of your database.
5. Translate Field Names : If true, the Processor will attempt to translate JSON field names into the appropriate column names for the table specified. If false, the JSON field names must match the column names exactly, or the column will not be updated
6. Unmatched Field Behaviour : if an incoming JSON element has a field that does not map to any of the database table's columns, this property specifies how to handle the situation
7. Unmatched Column Behaviour : If an incoming JSON element does not have a field mapping for all of the database table's columns, this property specifies how to handle the situation
8. Update Keys : A comma-separated list of column names that uniquely identifies a row in the database for UPDATE statements. If the Statement Type is UPDATE and this property is not set, the table's Primary Keys are used. In this case, if no Primary Key exists, the conversion to SQL will fail if Unmatched Column Behaviour is set to FAIL. This property is ignored if the Statement Type is INSERT
Supports Expression Language: true (will be evaluated using flow file attributes and variable registry)
Read the description above and try to use the properties given. Detailed description of the processor is given in the link.
ConvertjSONToSQL Description
Ok i found myself in a simple but annoying problem. My mongo documents are using java.util.Date as id, and as you might guess the id gets converted (spring converters) to ObjectId, I can't update these documents because every time a new ObjectId(Date) is created get a completely different id even though the date is the same...
how do i force mongo to just use java.util.Date as an id?
providing the sample code:
public void updateNode(...node..) {
final MongoTemplate mongoTemplate = ...
final String collectionName = ...
final Query query = (new Query()).addCriteria(Criteria.where("time").is(node.getTime()));
final Update update = Update.update("time", node.getTime()).set("top", node.getTop())
.set("bottom", node.getBottom()).set("mid", node.getMid())
.set("startTime", node.getStartTime()).set("potential", node.isPotential());
mongoTemplate.upsert(query, update, MyClassNode.class, collectionName);
}
if I ran this code for the first time the objects are inserted into the database but with ObjectId... if the node.getTime() is a java.sql.Date then everything is fine.
if the node.getTime() is not a java.sql.Date I cannot update the document if it exists: why? because everytime the document is prepared it creates a new ObjectId the update and query will have two different _id field values and update fails.
On checking the documentation , i found the following details :
In MongoDB, each document stored in a collection requires a unique _id
field that acts as a primary key. If an inserted document omits the
_id field, the MongoDB driver automatically generates an ObjectId for the _id field.
This also applies to documents inserted through update operations with
upsert: true.
The following are common options for storing values for _id:
Use an ObjectId.
Use a natural unique identifier, if available. This saves space and
avoids an additional index.
Generate an auto-incrementing number.
What i understood from the documentation was that to avoid inserting the same document more than once, only use upsert: true if the query field is uniquely indexed.So, if this flag is set , you will find your id converted using ObjectId() to make it unique.
I am new in writing aggregate queries in Mongo DB + Spring
Scenario: We are storing birthDate(Jjava.uti.Date) in mongo db which got stored as ISO date. Now we are trying to look for the records which are matching with the dayOfMonth and Month only. So that we can corresponding object from the list.
I had gone through few solutions and here is the way I am trying but this is giving me a null set of records.
Aggregation agg = Aggregation.newAggregation(
Aggregation.project().andExpression("dayOfMonth(birthDate)").as("day").andExpression("month(birthDate)")
.as("month"),
Aggregation.group("day", "month"));
AggregationResults<Employee> groupResults = mongoTemplate.aggregate(agg, Employee.class, Employee.class);
I also tried applying a a query with the help of Criteria but this is also giving me a Employee object which all null content.
Aggregation agg = Aggregation.newAggregation(Aggregation.match(Criteria.where("birthDate").lte(new Date())), Aggregation.project().andExpression("dayOfMonth(birthDate)").as("day").andExpression("month(birthDate)")
.as("month"),
Aggregation.group("day", "month"));
AggregationResults<Employee> groupResults = mongoTemplate.aggregate(agg, Employee.class, Employee.class);
I must missing some important thing which is giving me these null data.
Additional Info: Employee object has only birthDate(Date) and email(String) in it
Please try to specify the fields to be included in the $project stage.
project("birthDate", "...").andExpression("...
The _id field is, by default, included in the output documents. To include any other fields from the input documents in the output documents, you must explicitly specify the inclusion in $project.
see: MongoDBReference - $project (aggregation)
I've created DATAMONGO-2200 to add an option to project directly onto the fields of a given domain type via something like project(Employee.class).
I'm using Parse.com and I am running a query that obtains objects in a many-to-many relational table (call this table 'RelationTable'). Obviously this table has links to objects in another table (let's call this SubObject). Now, from this query, I need to filter results by searching on a field contained within the SubObject (call this SearchField).
Any ideas on how to do this? I already have the includeKey and am trying the '.' operator in SQL to access a field in the subclass, but it's not working. Below is the code I have so far:
ParseQuery<ParseObject> query = ParseQuery.getQuery("RelationTable);
query.include("subObject"); //subObject is field name where SubObject is stored. Note CAPS difference
query.whereContains("SubObject.SearchField", searchString);
You can create a subquery on the user object, and use whereMatchesQuery on your RelationTable query :
ParseQuery<ParseObject> query = ParseQuery.getQuery("RelationTable);
query.include("subObject");
ParseQuery<ParseObject> innerQuery = ParseQuery.getQuery("SubObject");
innerQuery.whereContains("SearchField", searchString);
query.whereMatchesQuery("subObject", innerQuery);
how can I build a table of "orders" containing "IdOrder", "Description" and "User"?... the "User" field is a reference to the table "Users", which has "IdUser" and "Name". I'm using repositories.
I have this repository:
Repository<Orders> ordersRepo = new OrderRepo<Orders>(unitOfWork.Session);
to return all Orders to View, I just do:
return View(ordersRepo.All());
But this will result in something like:
IdOrder:1 -- Description: SomeTest -- User: UserProxy123ih12i3123ih12i3uh123
-
When the expected result was:
IdOrder:1 -- Description: SomeTest -- User: Thiago.
PS: I don't know why it returns this "UserProxy123ih12i3123ih12i3uh123". In Db there is a valid value.
The View:
It is showed in a foreach (var item in Model).
#item.Description
#item.User //--> If it is #item.User.Name doesn't work.
What I have to do to put the Name on this list? May I have to do a query using LINQ - NHibernate?
Tks.
What type of ORM are you using? You mention "repositories" but does that mean LinqToSql, Entity Framework, NHibernate, or other?
It looks like you are getting an error because the User field is not loaded as part of the original query. This is likely done to reduce the size of the result set by excluding the related fields from the original query for Orders.
There are a couple of options to work around this:
Set up the repository (or context, depending on the ORM) to include the User property in the result set.
Explicitly load the User property before you access it. Note that this would be an additional round-trip to the database and should not be done in a loop.
In cases where you know that you need the User information it would make sense to ensure that this data in returned from the original query. If you are using LinqToSql take a look at the DataLoadOptions type. You can use this type to specify which relationships you want to retrieve with the query:
var options = new DataLoadOptions();
options.LoadWith<Orders>(o => o.User);
DataContext context = ...;
context.LoadOptions = options;
var query = from o in context.Orders
select o;
There should be similar methods to achive the same thing whatever ORM you are using.
In NHibernate you can do the following:
using (ISession session = SessionFactory.OpenSession())
{
var orders = session.Get<Order>(someId);
NHibernateUtil.Initialize(orders.User);
}
This will result in only two database trips (regardless of the number of orders returned). More information on this can be found here.
In asp.net MVC the foreign key doesn't work the way you are using it. I believe you have to set the user to a variable like this:
User user = #item.User;
Or you have to load the reference sometimes. I don't know why this is but in my experience if I put this line before doing something with a foreign key it works
#item.UserReference.load();
Maybe when you access item.User.Name the session is already closed so NHib cannot load appropriate user from the DB.
You can create some model and initialize it with proper values at the controller. Also you can disable lazy loading for Orders.User in your mapping.
But maybe it is an other problem. What do you have when accessing "#item.User.Name" from your View?