Using Server side javascript, I need to sort a NotesDcumentCollection based on a field in the collection containing a date when the documents was created or any built in field when the documents was created.
It would be nice if the function could take a sort option parameter so I could put in if I want the result back in ascending or descending order.
the reason I need this is because I use database.getModifiedDocuments() which returns an unsorted notesdocumentcollection. I need to return the documents in descending order.
The following code is a modified snippet from openNTF which returns the collection in ascending order.
function sortColByDateItem(dc:NotesDocumentCollection, iName:String) {
try{
var rl:java.util.Vector = new java.util.Vector();
var tm:java.util.TreeMap = new java.util.TreeMap();
var doc:NotesNotesDocument = dc.getFirstDocument();
while (doc != null) {
tm.put(doc.getItemValueDateTimeArray(iName)[0].toJavaDate(), doc);
doc = dc.getNextDocument(doc);
}
var tCol:java.util.Collection = tm.values();
var tIt:java.util.Iterator = tCol.iterator();
while (tIt.hasNext()) {
rl.add(tIt.next());
}
return rl;
}catch(e){
}
}
When you construct the TreeMap, pass a Comparator to the constructor. This allows you to define custom sorting instead of "natural" sorting, which by default sorts ascending. Alternatively, you can call descendingMap against the TreeMap to return a clone in reverse order.
This is a very expensive methodology if you are dealing with large number of documents. I mostly use NotesViewEntrycollection (always sorted according to the source view) or view navigator.
For large databases, you may use a view, sorted according to the modified date and navigate through entries of that view until the most recent date your code has been executed (which you have to save it somewhere).
For smaller operations, Tim's method is great!
Related
I have an array of ids that i want to sort a given column by them in the collection.
For example,
$suppliersOrder = [8,7,5,1,3,4];
$items = Items::get()->sortBy(function($model) use ($suppliersOrder) {
return array_search($model->supplier_id, $suppliersOrder);
})->values();
This acts like ordering items as [1,3,4,5,7,8] instead of the given order. And if I try sortByDesc, likewise [8,7,5,4,3,1] but I couldn't figure out the way to actually sort them as my given array's order.
My ultimate goal is then running $items->groupBy('supplier.name') so I can have my desired order.
What Alexander Villalobos suggested in the comments, I changed my code like this:
$items = Items::get()->sortBy(function($model) use ($suppliersOrder) {
return rsort($model->supplier_id, $suppliersOrder);
});
Indirect modification of overloaded property App\Item::$supplier_id has no effect
$suppliersOrder = [8,7,5,1,3,4];
$items = Items::get()->sortBy(function($row,$key) use ($suppliersOrder) {
return array_search($row->supplier_id, $suppliersOrder);
});
This should give you sorted collection of items by the order you described in $suppliersOrder. As per Laravel docs, the parameters to the callback function include one being the row for the collection and another being the key of that row in the collection.
I am looking for the most efficient way to get the last elements of a fairly large (> 1 million docs) MongoDB collection.
Specifically, it is the oplog collection and I am looking for all entries after a given timestamp. It makes no sense to search the first million or so entries for a timestamp larger than the current one, since they are all definitely older because the collection is stored in its natural order.
Is there a way to tell MongoDB to search from the end of a collection?
I tried a linq query with Skip(N) but it's very slow. It seems it parses through all documents from the beginning and just doesn't return the first N.
The most efficient way is probably using aggregation. If your collection is sorted, you can get the last Timestamp using this aggregation:
var group = new BsonDocument
{
{
"$group", new BsonDocument
{
{"_id", 0},
{"newestTimeStamp", new BsonDocument { {"$last","$timeStamp"} } }
}
}
};
var pipeline = new[] {group};
var result = _dtCollection.Aggregate(pipeline);
}
Then you can deserialize the result into a Timestamp class. If you want to get several elements, you could create a similar expression using $match.
Also make sure to add an index to the collection on the TimeStamp field. This will probably make your LINQ-query faster if you decide to use that instead.
I have a TableView associated to a TreeView. Each time a node in the TreeView is selected, the TableView is refreshed with different data.
I am able to sort any column in the TableView, just pressing the corresponding column header. That works fine.
But: when I select a different node in the tree-view, eventhough the column headers keep showing as sorted. The data is not.
Is there a way to programmatically enforce the sort order made by the user each time the data changes?
Ok, I found how to do it. I will summarize it here in case it is useful to others:
Before you update the contents of the TableView, you must save the sortcolum (if any) and the sortType:
TableView rooms;
...
TableColumn sortcolumn = null;
SortType st = null;
if (rooms.getSortOrder().size()>0) {
sortcolumn = (TableColumn) rooms.getSortOrder().get(0);
st = sortcolumn.getSortType();
}
Then, after you are done updating the data in the TableView, you must restore the lost sort-column state and perform a sort.
if (sortcolumn!=null) {
rooms.getSortOrder().add(sortcolumn);
sortcolumn.setSortType(st);
sortcolumn.setSortable(true); // This performs a sort
}
I do not take into account the possibility of having multiple columns in the sort, but this would be very simple to do with this information.
I had the same problem and found out that after an update of the data you only have to call the function sort() on the table view:
TableView rooms;
...
// Update data of rooms
...
rooms.sort()
The table view knows the columns for sorting thus the sort function will sort the new data in the wanted order.
This function is only available in Java 8.
If your TableView is not reinitialized, you can also do the following:
TableColumn<BundleRow, ?> sortOrder = rooms.getSortOrder().get(0);
rooms.getSortOrder().clear();
rooms.getSortOrder().add(sortOrder);
The example of fornacif works, but not if there is more than one sort order (try shift-click on a second column to create secondary sort order).
To do a re-sort on all columns you would need to do something like this:
List<TableColumn<Room, ?>> sortOrder = new ArrayList<>(roomTable.getSortOrder());
roomTable.getSortOrder().clear();
roomTable.getSortOrder().addAll(sortOrder);
If you use the TableView.setItems() method, it appears to reset several aspects of the TableView. Leave the ObservableList in the TableView in place, clear its contents, and then add your new items. Then, TableView.sort() will still know which columns were previously sorted and it will work. Like this:
tableView.getItems().clear();
tableView.getItems().addAll(newTableData);
tableView.sort();
Marco Jakob's answer is good for most cases, but I found that I needed to create a comparator that matches the table sort order for more flexibility. You can then use any method that takes a comparator to do sorting, searching, etc. To create the comparator, I extended that ComparatorChain class from apache's Common-Collections to easily do multiple column sorting. It looks like this.
public class TableColumnListComparator extends ComparatorChain {
public TableColumnListComparator(ObservableList<? extends TableColumn> columns) {
// Get list of comparators from column list.
for (TableColumn column : columns) {
addComparator(new ColumnComparator(column));
}
}
/**
* Compares two items in a table column as if they were being sorted in the TableView.
*/
private static class ColumnComparator implements Comparator {
private final TableColumn column;
/**
* Default Constructor. Creates comparator based off given table column sort order.
*
* #param column
*/
public ColumnComparator(TableColumn column) {
this.column = column;
}
#Override
public int compare(Object o1, Object o2) {
// Could not find a way to do this without casts unfortunately
// Get the value of the column using the column's cell value factory.
final ObservableValue<?> obj1 = (ObservableValue) column.getCellValueFactory().call(
new TableColumn.CellDataFeatures(column.getTableView(), column, o1));
final ObservableValue<?> obj2 = (ObservableValue) column.getCellValueFactory().call(
new TableColumn.CellDataFeatures(column.getTableView(), column, o2));
// Compare the column values using the column's given comparator.
final int compare = column.getComparator().compare(obj1.getValue(), obj2.getValue());
// Sort by proper ascending or descending.
return column.getSortType() == TableColumn.SortType.ASCENDING ? compare : -compare;
}
}
}
You can then sort at anytime with
Collections.sort(backingList, new TalbeColumnListComparator(table.getSortOrder());
I use this to sort multiple lists with the same sort, sort on background threads, do efficient updates without resorting the whole list, etc. I think there are going to be some improvements to table sorting in Javafx 8 so this won't be necessary in the future.
You can also use a SortedList.
SortedList<MatchTableBean> tableItems = new SortedList<>(
observableList, Comparator.comparing(MatchTableBean::isMarker).reversed().thenComparing(MatchTableBean::getQueryRT));
tableItems.comparatorProperty().bind(table.comparatorProperty());
table.setItems(tableItems);
This way the table is sorted, even when the content changes or is completely replaced.
You can also do this for 0 or more Sort-Columns:
List<TableColumn<Room, ?>> sortColumns = new LinkedList<>(rooms.getSortOrder());
// rooms.setItems(...)
rooms.getSortOrder().addAll(sortColumns);
The reason why you create a new LinkedList is that you don't wanna just point at rooms.getSortOrder() like this:
List<TableColumn<Room, ?>> sortColumns = rooms.getSortOrder();
because this way both rooms.getSortOrder() and sortColumns will become empty after you call rooms.setItems(...) which seems to clear the rooms.getSortOrder().
I am using a CDbCriteria with its own conditions, with & order clauses. However, the order i want to give to the elements in the array is way too complex to specify in the order clause.
The solution i have in mind consists of obtaining the active records with the defined criteria like this
$theModelsINeed = MyModel::model()->findAll($criteria);
and then rearrange the order from my php code. How can i do this? I mean, i know how to iterate through its elements, but i don´t know if it is possible to actually change them.
I have been looking into this link about populating active records, but it seems quite complicated and maybe someone could have some better advice.
Thanks
There is nothing special about Yii's active records. The find family of methods will return an array of objects, and you can sort this array like any other array in PHP.
If you have complex sort criteria, this means that probably the best tool for this is usort. Since you will be dealing with objects, your user-defined comparison functions will look something like this:
function compare($x, $y)
{
// First sort criterion: $obj->Name
if ($x->Name != $y->Name) {
return $x->Name < $y->Name ? -1 : 1; // this is an ascending sort
}
// Second sort criterion: $obj->Age
if ($x->Age != $y->Age) {
return $x->Age < $y->Age ? 1 : -1; // this is a descending sort
}
// Add more criteria here
return 0; // if we get this far, the items are equal
}
If you do want to get an array as a result, you can use this method for fetching data that supports dbCriteria:
$model = MyModel::model()->myScope();
$model->dbCriteria->condition .= " AND date BETWEEN :d1 AND :d2";
$model->dbCriteria->order = 'field1 ASC, field2 DESC';
$model->dbCriteria->params = array(':d1'=>$d1, ':d2'=>$d2);
$theModelsINeed = $model->getCommandBuilder()
->createFindCommand($model->tableSchema, $model->dbCriteria)
->queryAll();
The above example shows using a defined scope and modifying the condition with named parameters.
If you don't need Active Record, you could also look into Query Builder, but the above method has worked pretty well for me when I want to use AR but need an array for my result.
As I can see in the Spine.js sources the Model.each() function returns Model's records in the order of their IDs. This is completely unreliable in scenarios where ordering is important: long person list etc.
Can you suggest a way to keep original records ordering (in the same order as they've arrived via refresh() or similar functions) ?
P.S.
Things are even worse because by default Spine.js internally uses new GUIDs as IDs. So records order is completely random which unacceptable.
EDIT:
Seems that in last commit https://github.com/maccman/spine/commit/116b722dd8ea9912b9906db6b70da7948c16948a
they made it possible, but I have not tested it myself because I switched from Spine to Knockout.
Bumped into the same problem learning spine.js. I'm using pure JS, so i was neglecting the the contact example http://spinejs.com/docs/example_contacts which helped out on this one. As a matter of fact, you can't really keep the ordering from the server this way, but you can do your own ordering with javascript.
Notice that i'm using the Element Pattern here. (http://spinejs.com/docs/controller_patterns)
First you set the function which is gonna do the sorting inside the model:
/*Extending the Student Model*/
Student.extend({
nameSort: function(a,b) {
if ((a.name || a.email) > (b.name || b.email))
return 1;
else
return -1
}
});
Then, in the students controller you set the elements using the sort:
/*Controller that manages the students*/
var Students = Spine.Controller.sub({
/*code ommited for simplicity*/
addOne: function(student){
var item = new StudentItem({item: student});
this.append(item.render());
},
addAll: function(){
var sortedByName = Student.all().sort(Student.nameSort);
var _self = this;
$.each(sortedByName, function(){_self.addOne(this)});
},
});
And that's it.