OrderByDecending(LINQ) equivalent in Java

Question

Hi I am a c# developer getting into Java. The question is simple:

How can i write the below c# code into Java and still get it to work:

myCoffeeList.OrderByDescending(x => x.Name?.ToLower()?.Trim() == sender.Text.ToLower()?.Trim()));

My sender.Text is basically a textbox.Text with a value like Cappuccino.

What does the above code do?

Well I have a collection of coffee and I want that the name of the coffee that exactly matches the sender.text should be on top and the rest can be of the order returned from the server.

I've checked out the below:

list.streams to use something like a LINQ
Anotation type OrderBy but it's related to databases

The question context seems similar like this one: https://stackoverflow.com/questions/1217228/what-is-the-java-equivalent-for-linq. You can found something useful things there. — Tetsuya Yamamoto, Nov 08 '17 at 09:25
You can use a custom `Comparable`, here is a good place to start learning about them https://docs.oracle.com/javase/tutorial/collections/interfaces/order.html — Titus, Nov 08 '17 at 09:28
@TetsuyaYamamoto No the question is quite different from that, reason being, I don't want to use LINQ. I just want to get an idea on how can I achieve what the above codeline does in Java — iam.Carrot, Nov 08 '17 at 09:33
@Titus thanks I was looking at the `Comparable`, can you please help me out with a sample?\ — iam.Carrot, Nov 08 '17 at 09:33
GGO, I think the C# tag should stay, as there are a few recent/relatively advanced C# mechanisms at work here and I wouldn't expect a core java guru with passing knowledge of C# to reasonably know what all these things are.. — Caius Jard, Nov 08 '17 at 09:36
@iam.Carrot unless I misunderstand the way `entity?.method` works, I don't think it's required on .ToLower() - ToLower() will never return null, and cannot be called on a null, so stopping at x.Name? would be fine => if name is null the runtime will cease trying to call further methods and evaluate the expression as null — Caius Jard, Nov 08 '17 at 09:38
@CaiusJard Yeah thanks for that man. It just got as a habit to put that `?` maybe I over used it here. — iam.Carrot, Nov 08 '17 at 09:54

J.R · Accepted Answer · 2017-11-08T09:57:20.273

3

List<Coffee> sorted = lst.stream()
                .sorted(Comparator.comparing(s -> !s.trim().equalsIgnoreCase(sender.trim())))
                .collect(Collectors.toList());

edited Nov 08 '17 at 09:57

answered Nov 08 '17 at 09:34

J.R

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I can't use your answer as it's using a string while My object is of type Coffee and the object of coffee has a peroperty/field named Name of type string – iam.Carrot Nov 08 '17 at 09:40
Ok. I just given an idea. Anyway just consider the string as a instance of Coffee object. And get the coffee name by getter method. – J.R Nov 08 '17 at 09:42
can you please share a sample where in we take a coffee object and match the name to the string? – iam.Carrot Nov 08 '17 at 09:44
I've edited my answer. Considering sender as String. – J.R Nov 08 '17 at 09:44
1

You can use `...equalsIgnoreCase(...)` to replace those `...toLowerCase()` calls. – Titus Nov 08 '17 at 09:53
You are welcome. I've edited as per the suggestion given by Titus. – J.R Nov 08 '17 at 09:58
need null checking to perfectly convert into java code – SungJin Steve Yoo Nov 08 '17 at 10:08

score -1 · Answer 2 · answered Nov 08 '17 at 09:30

Use Comparable and the sort method:

List<String> coffee = new ArrayList<>();
coffee.add("C");
coffee.add("A");
coffee.add("B");
coffee.add("D");

String top = "D";
coffee.sort((a, b) -> a.equals(top) ? -1 : 0);
coffee.forEach(System.out::println);

EDIT: for natural order use:

coffee.sort((a, b) -> a.equals(top) ? -1 : a.compareTo(b));

OrderByDecending(LINQ) equivalent in Java

What does the above code do?

I've checked out the below:

2 Answers2