I am new to Cloudant and I need to do a simple query on a specific document field. My documents have the following structure and I need to get only the documents with status=SIGNED
{
"_id": "3ddb4058f3b24a7a9c585f997e30ff78",
"_rev": "3-757c82c48f4e7c333911be6859aff74e",
"fileName": "Generali Architects",
"status": "SIGNED",
"user": "italy",
"_attachments": {
"Generali Architects": {
"content_type": "application/pdf",
"revpos": 3,
"digest": "md5-9hqSif7CzQ2yvKxSSbj+dw==",
"length": 323653,
"stub": true
}
}
}
Reading Cloudant documentation I created the following Design Document with a related view which returns exactly what I expected
Then from my Java application I use the following code
String cloudantView = "_design/signedDocs/status-signed-iew";
List<SignDocDocument> docs =
db.view(cloudantView).includeDocs(true).query(SignDocDocument.class);
which always returns me "org.lightcouch.NoDocumentException: Object Not Found"
Any idea which kind of mistake I am making here?
Thank you very much
Is it the typo in "_design/signedDocs/status-signed-iew"; e.g. that should be "_design/signedDocs/status-signed-view"; (depending on how your java library works...).
Always worth checking the view by direct access in your browser, too, just to make sure it's returning the expected data.
Related
I have used mongo import to import data into mongodb from csv files. I am trying to retrieve data from an Mongodb realm service. The returned data for the entry is as follows:
{
"_id": "6124edd04543fb222e",
"Field1": "some string",
"Field2": {
"$numberDouble": "145.81"
},
"Field3": {
"$numberInt": "0"
},
"Field4": {
"$numberInt": "15"
},
"Field5": {
"$numberInt": "0"
}
How do I convert this into normal JSON by removing $numberInt and $numberDouble like :
{
"_id": "6124edd04543fb222e",
"Field1": "some string",
"Field2": 145.8,
"Field3": 0,
"Field4": 15,
"Field5": 0
}
The fields are also different for different documents so cannot use Mongoose directly. Are there any solutions to this?
Also would help to know why the numbers are being stored as $numberInt:"".
Edit:
For anyone with the same problem this is how I solved it.
The array of documents is in EJSON format instead of JSON like said in the upvoted answer. To covert it back into normal JSON, I used JSON.stringify to first convert each document I got from map function into string and then parsed it using EJSON.parse with
{strict:false} (this option is important)
option to convert it into normal JSON.
{restaurants.map((restaurant) => {
restaurant=EJSON.parse(JSON.stringify(restaurant),{strict:false});
}
EJSON.parse documentation here. The module to be installed and imported is mongodb-extjson.
The format with $numberInt etc. is called (MongoDB) Extended JSON.
You are getting it on the output side either because this is how you inserted your data (meaning your inserted data was incorrect, you need to fix the ingestion side) or because you requested extended JSON serialization.
If the data in the database is correct, and you want non-extended JSON output, you generally need to write your own serializers to JSON since there are multiple possibilities of how to format the data. MongoDB's JSON output format is the Extended JSON you're seeing in your first quote.
I'm just getting started with mongoDB, and coming from a MySQL environment I'm having trouble figuring out how have my documents automatically have fields such as UPDATED, CREATED, DELETED
So when I create an entry like this:
{
"email": "some#test.com",
"name": "bob"
}
I would like it to automatically become this:
{
"email": "some#test.com",
"name": "bob",
"CREATED": 1567120458,
"UPDATED": 1567120458,
"DELETED": null
}
I have found https://docs.mongodb.com/manual/reference/operator/update/currentDate/ this resource which talks about a way of inserting dates, but I am not sure how to use this, or where to place this.
{ $currentDate: { <field1>: <typeSpecification1>, ... } }
I don't know what this piece of code means or where to use it.
I've installed Stuido 3T to help me manage the database, but I don't see any option to use this piece of code.
Construct your query use Date:
db.example.insert({"date":new Date(Date.now())})
Can anyone tell me why when I use DataExplorer for CosmoDb DB I get the following:
{
"id": "d502b51a-e70a-40f1-9285-3861880b8d90",
"Version": 1,
...
}
But when I use Robomongo I get:
{
"Version" : 1,
...
}
minus the id?
Thanks
I tried to repro your scenario but it all worked correctly.
The Mongo document in Portal Data Explorer:
The Mongo document in Robo 3T:
They both have the id property.
Are you applying Projections on Robomongo / Robo 3T?
At this moment cosmodb works separately SQL API and Mongo API, each one has different implementation, SQL API use JSON and Mongo use BSON, you need to be clear this while you are creating the document.
If you create the document with a BSON-based tool like Robo3t for example, you are going to get something like this:
{
"_id": {
"$oid": "5be0d98b9cdcce3c6ce0f6b8"
},
"name": "Name",
"id": "5be0d98b9cdcce3c6ce0f6b8",
...
}
Instead, if you create your document with JSON-based like Data Explorer, you are going to get this:
{
"name": "Name",
"id": "6c5c05b4-dfce-32a5-0779-e30821e6c510",
...
}
As you can see, BSON-based needs that _id and inside $oid be implemented to works right, while JSON-based only id is required. So, you need to add the properties while you save the document (see below) or open it with the right tool, as Matias Quaranta recommend, use Azure Storage Explorer or even Data Explorer to get both protocols properly.
Also, if you use a system to create the document and you want to use BSON format, You need to add the $oid, for example in core net is something like this:
public bool TryGetMemberSerializationInfo(string memberName, out BsonSerializationInfo serializationInfo)
{
switch (memberName)
{
case "Id":
serializationInfo = new BsonSerializationInfo("_id", new ObjectIdSerializer(), typeof(ObjectId));
return true;
case "Name":
serializationInfo = new BsonSerializationInfo("name", new StringSerializer(), typeof(string));
return true;
default:
serializationInfo = null;
return false;
}
}
In mongodb i saved document like this.
"Angela_Merkel": {
"birthPlace": "Hamburg,West_Germany",
"thumbnail": "http://commons.wikimedia.org/wiki/Special:FilePath/Angela-Merkel-2014.jpg?width=300",
"almaMater": "Leipzig_University",
"birthDate": "1954-07-17",
"spouse": "Joachim_Sauer"
}
There are many person's information like this way. Now if I want to get all the information of "Angela_Merkel" or only a particular like "birthdDate" of "Angela_Merkel" then what will be the query?
Like chridam says would be more practical that you refactor your documents like this:
{"name": "Angela_Merkel",
"birthPlace": "Hamburg,West_Germany",
"thumbnail": "http://commons.wikimedia.org/wiki/Special:FilePath/Angela-Merkel-2014.jpg?width=300",
"almaMater": "Leipzig_University",
"birthDate": "1954-07-17",
"spouse": "Joachim_Sauer"
}
Inside the "people" collection (Its a convention name the collections as plurals, beeing a set of documents)
Then you could do:
db.people.find({"name":"Angela Merkel"},{"_id":0," "birthdDate":1 })
To get:
{"birthDate": "1954-07-17"}
Anyway you can get with your document what you want this way:
"birthDate":
db.person.find({"Angela_Merkel.birthDate": "1954-07-17"})
or all the info:
db.person.find({"Angela_Merkel": { $exists: true}})
But doesn't have much sense do it this way.
PD: Edited to include last part.
I understand Manual referencing being, one document containing the reference to another.
As in the MongoDB's documentation:
original_id = ObjectId()
db.places.insert({
"_id": original_id,
"name": "Broadway Center",
"url": "bc.example.net"
})
db.people.insert({
"name": "Erin",
"places_id": original_id,
"url": "bc.example.net/Erin"
})
I am able to use a find on the places to get a certain name.
db.places.find({name : "Broadway Center"});
This will give the id. And then my application can use this ID to query the people to see who live here?
But, If i dont have a application. How would I go about doing this solely in a mongo shell? I was thinking about using find, and then iterating over the cursor using a forEach? But this seems a bit hacky.
Any suggestions?
You don't need to loop through using the cursor from find(), just use findOne() because a single document is returned with this method and you have access to the document directly (no need to apply cursor methods to the result). In mongo shell you can do this:
var result = db.places.findOne({name : "Broadway Center"});
if (result) {
var place_id = result._id;
var peopleArray = db.people.find({ "places_id": place_id }).toArray();
if (peopleArray.length > 0) { printjson (peopleArray[0]); }
}