What is the effect of FeedOptions.EnableLowPrecisionOrderBy Property

What is the effect of FeedOptions.EnableLowPrecisionOrderBy Property

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What is the effect of FeedOptions.EnableLowPrecisionOrderBy Property
Tag : development , By : robinator
Date : November 27 2020, 03:01 PM

This might help you Azure Cosmos DB supports varying the precision of your index to reduce the storage footprint of indexing (default is full precision). For example, with a numeric precision of 5, the service would index the first 5 bytes of your number.
By default, sorting a property with a lower precision is disallowed. But you can opt-in to perform an ORDER BY by setting EnableLowPrecisionOrderBy. Let's say you choose a numeric precision of 5 (for an 8-byte number). The query results will be in order based on the most significant 5 bytes of the number, but guarantees no order for the remaining 3 bytes. This option lets you perform fast order by without requiring full precision indexing, for example, when you're performing ad-hoc exploration of data.

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Is specifying the partition key in the SQL equivalent to using FeedOptions.PartitionKey?

Tag : chash , By : Ryan Adriano
Date : March 29 2020, 07:55 AM
it helps some times Using the SQL api you can specify the partition key in the SQL statement e.g. SELECT * FROM c WHERE c.MyPartitionKey = 'KeyValue' or using FeedOptions.PartitionKey. , I did some tests.
Test One: 100 documents
for (int i = 1; i <=100 ; i++) {
     Document doc = new Document();
     doc.setId(i + "");
     if(i%2 == 0)
        doc.set("name", "white");
        doc.set("name", "black");
     documentClient.createDocument("dbs/db/colls/part", doc, null, true);
     System.out.println("insert document No. " + i);
String sql ="SELECT * FROM c where c.name='black'";
FeedOptions options = new FeedOptions();
FeedResponse<Document> queryResults = documentClient.queryDocuments("dbs/db/colls/part",sql,options);
FeedOptions options = new FeedOptions();
PartitionKey partitionKey = new PartitionKey("black");
String sql ="SELECT * FROM c";
FeedResponse<Document> queryResults = documentClient.queryDocuments("dbs/db/colls/part",sql,options);
for (int i = 1; i <=1000 ; i++) {
         Document doc = new Document();
         doc.setId(i + "");
         if(i%2 == 0)
            doc.set("name", "white");
            doc.set("name", "black");
         documentClient.createDocument("dbs/db/colls/part", doc, null, true);
         System.out.println("insert document No. " + i);

CosmosDB - what does FeedOptions.DisableRUPerMinuteUsage actually do?

Tag : azure , By : LUK
Date : March 29 2020, 07:55 AM
it should still fix some issue I got an answer from microsoft employee ,

Difference between partition key indicated in query expression and feedoptions in cosmos db

Tag : java , By : Steve M
Date : March 29 2020, 07:55 AM
Hope that helps I suggest you indicate partition key in FeedOptions.I did a micro test to observe the performance of both of the solutions.
First One:
String name = "A";
FeedResponse<Document> feedResponse = client
                        "SELECT * FROM c WHERE c.name ='" + name + "'", null);
FeedOptions queryOptions = new FeedOptions();
PartitionKey partitionKey = new PartitionKey("/A");
FeedResponse<Document> feedResponse1 = client
                    "SELECT * FROM c ", queryOptions);

Does setting FeedOptions.EnableCrossPartitionQuery to false need to be combined with FeedOptions.PartitionKey?

Tag : chash , By : Jesenko Mehmedbasic
Date : March 29 2020, 07:55 AM
wish help you to fix your issue Yes, you also need to specify the partition key explicitly, using FeedOptions and its PartitionKey property as you've noted. Here's an example:
feedOptions.PartitionKey = new PartitionKey("yourPartitionKey");

var query = documentClient.CreateDocumentQuery<ArticleEntity>(collectionLink, feedOptions)
    .Where(entity => entity.Type == someType)

Two objects which have the same property. What is the effect of deleting the property of one object?

Tag : javascript , By : snk
Date : March 29 2020, 07:55 AM
I hope this helps . In your first example, you are assigning a string b['name'] = a['name']. String's are passed by value in javascript, so effectively a new copy of that string is created in memory. a['name'] is not equal to b['name'] in terms of the actual in-memory location they are referencing.
You can think of this as having two copies of a string. When you delete the string, you delete one copy, but not the other.
let tag_obj = {'name': 'gavin'};
let a = { 'tag': tag_obj };
let b = {};
b['tag'] = a['tag'];

// a and b both now equal { tag: { name: 'gavin'} }

console.log(a.tag === tag_object) // True, it references tag object
console.log(b.tag === tag_object) // Also true, as it references the same object.

delete a.tag.name; 
let j = {'tag': {'name': 'james'}};
let k = j
console.log(k === j) // Will log true, they are the same object

let a = {'tag': {'name': 'gavin'}};
let b = Object.assign({}, a);
console.log(a === b) // Will log false, a new object has been created.
let a = {'tag': {'name': 'gavin'}};
let b = Object.assign({}, a);
console.log(a.tag === b.tag) // Will log true, they share the same tag object
let j = {'tag': {'name': 'james'}};
let k = JSON.parse(JSON.stringify(j));
console.log(j === k); // Will log false, a new object has been created.
console.log(j.tag === k.tag); // Will log false, a new object has been created.
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