There are several ways to do this: $collection = Mage::getModel(‘…’) ->getCollection() ->setPageSize(20) ->setCurPage(1); Will get first 20 records. Here is the alternative and maybe more readable way: $collection = Mage::getModel(‘…’)->getCollection(); $collection->getSelect()->limit(20); This will call Zend Db limit. You can set offset as second parameter.
Iterating Backwards through the List sounds like a better approach, because if you remove an element and other elements “fall into the gap”, that does not matter because you have already looked at those. Also, you do not have to worry about your counter variable becoming larger than the .Count. List<int> test = new List<int>(); … Read more
Let’s take a closer look at the two of them: // a) col.addAll(Arrays.asList(1, 2, 3, 4, 5)); Here’s what happens: varags + autoboxing creates Integer[] Arrays.asList creates a List<Integer> backed by the array addAll iterates over a Collection<Integer> using Iterator<Integer> // b) Collections.addAll(col, 1, 2, 3, 4, 5); Here’s what happens: varargs + autoboxing creates … Read more
Collectors.partitioningBy: Map<Boolean, List<Employee>> partitioned = listOfEmployees.stream().collect( Collectors.partitioningBy(Employee::isActive)); The resulting map contains two lists, corresponding to whether or not the predicate was matched: List<Employee> activeEmployees = partitioned.get(true); List<Employee> formerEmployees = partitioned.get(false); There are a couple of reasons to use partitioningBy over groupingBy (as suggested by Juan Carlos Mendoza): Firstly, the parameter of groupingBy is a Function<Employee, … Read more
You can cast through the untyped List interface: List<A> a = new ArrayList<A>(); List<B> b = (List)a;
You can use LinkedHashMap like this You can remove by LRU or FIFO. public static <K, V> Map<K, V> createLRUMap(final int maxEntries) { return new LinkedHashMap<K, V>(maxEntries*10/7, 0.7f, true) { @Override protected boolean removeEldestEntry(Map.Entry<K, V> eldest) { return size() > maxEntries; } }; }
One choice is to add them and remove the repeated ones: var list = new List<Car>(); list.AddRange(GetGreenCars()); list.AddRange(GetBigCars()); list.AddRange(GetSmallCars()); list = list.Distinct().ToList();
I agree with Stuart Marks that list == null ? Stream.empty() : list.stream() is the right way to do this (see his answer), or at least the right way to do this pre-Java 9 (see edit below), but I’ll leave this answer up to demonstrate usage of the Optional API. <T> Stream<T> getStream(List<T> list) { … Read more
If you’ve already checked that the keys are the same, you can just use: var dict3 = dict2.Where(entry => dict1[entry.Key] != entry.Value) .ToDictionary(entry => entry.Key, entry => entry.Value); To explain, this will: Iterate over the key/value pairs in dict2 For each entry, look up the value in dict1 and filter out any entries where the … Read more
HashMap does not call hashcode when null is passed as key and null Key is handled as special case. Put Method HashMap puts null key in bucket 0 and maps null as key to passed value. HashMap does it by linked list data structure. HashMap uses linked list data structure internally. Linked list data structure … Read more