Start with the rightmost point (the point with the biggest x coordinate) Get all edges from this point Follow the edge with the smallest angle to the positive x-axis, and also add it to the solution set From the point reached, follow and add the edge with the smallest angle to the edge you came … Read more
I experienced similar behavior with an older version of Spring Data and this was my take on it: https://arnoldgalovics.com/how-much-projections-can-help/ I had a talk with Oliver Gierke (Spring Data lead) and he made some improvements (that’s why you get so “good” results 🙂 ) but basically there will be always a cost on having abstractions vs … Read more
Quite on old question but literally none of the proposed answers are good. TLDR: You can’t use $elemMatch in a $project stage. but you can achieve the same result using other aggregation operators like $filter. db.itens.aggregate([ { $project: { compList: { $filter: { input: “$complist”, as: “item”, cond: {$eq: [“$$item.a”, 1]} } } } } … Read more
with spring data you can cut the middle-man and simply define public interface IdsOnly { Integer getId(); String getOtherId(); } and use a native query like; @Query(value = “Id, OtherId from TestTable where CreationDate > ?1 and Type in (?2)”, nativeQuery = true) public Collection<IdsOnly> findEntriesAfterDate(Date creationDate, List<Integer> types); check out https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#projections
Perhaps the following illustration showing the relationship between the various spaces will help: Depending if you’re using the fixed-function pipeline (you are if you call glMatrixMode(), for example), or using shaders, the operations are identical – it’s just a matter of whether you code them directly in a shader, or the OpenGL pipeline aids in … Read more
If you want use the annotation @Query with Spring Data Projections you have to use field alias and you need to make sure you alias the projects matching the projection fields. The following code should work for question 1: interface PersonRepository extends CrudRepository<Person, Long> { @Query(“select p.firstName as firstname, p.address as address from Person p … Read more
Look at http://qfox.nl/notes/116 var overlay = new google.maps.OverlayView(); overlay.draw = function() {}; overlay.setMap(map); var point = overlay.getProjection().fromLatLngToDivPixel(latLng); Ugly indeed. Much easier in v2 – another flaw of google api v3!
Select only columnA and columnB from your table, and move further processing in memory: return context.myTable .Select(o => new { o.columnA, o.columnB }) // only two fields .AsEnumerable() // to clients memory .Select(o => new KeyValuePair<int, string>(o.columnA, o.columnB)) .ToList(); Consider also to create dictionary which contains KeyValuePairs: return context.myTable.ToDictionary(o => o.columnA, o => o.columnB).ToList();
The output of a vertex shader is a four component vector, vec4 gl_Position. From Section 13.6 Coordinate Transformations of core GL 4.4 spec: Clip coordinates for a vertex result from shader execution, which yields a vertex coordinate gl_Position. Perspective division on clip coordinates yields normalized device coordinates, followed by a viewport transformation (see section 13.6.1) … Read more
Note the example given is a left-handed, row major matrix. So the operation is: Translate to the origin first (move by –eye), then rotate so that the vector from eye to At lines up with +z: Basically you get the same result if you pre-multiply the rotation matrix by a translation –eye: [ 1 0 … Read more