How to sort without considering special characters in java?

问题

I wanted to sort a list of names and I used compareTo method and I have names like this on my list

$urya, andy, prince, @rikesh,

expected output: andy, prince, $urya, @rikesh,

instead, I get: $urya @rikesh andy prince

What happens now is the name that starts with special characters come's first and then other names are sorted

Is there a way to push the names that starts with special characters to last.

if (this.firstName == name.firstName) {
  return 0;
} else if (this.firstName == null) {
  return 1;
} else if (name.firstName == null) {
  return -1;
} else {
  return this.firstName.compareToIgnoreCase(name.firstName);
}

回答1:

Java 8 steam API one liner

list.stream().filter(s -> s.replaceAll("[^a-zA-Z0-9]", "")).sorted()

回答2:

I think you want $ to be considered as S. In this case, in the compareTo method, you have to replace $ with S (similarly for others) and compare.

If you don't want to consider the $ itself(as the question says), you have to replace special characters with empty character.

回答3:

I have made a few modifications and as per your concern, i have ignored the special characters:

In the compareTo method, it has to look something like this:

if(this.name.replaceAll("[^a-zA-Z0-9]", "").compareTo(o.name.replaceAll("[^a-zA-Z0-9]", "")) >0)
                return 1;
            else if(this.name.replaceAll("[^a-zA-Z0-9]", "").compareTo(o.name.replaceAll("[^a-zA-Z0-9]", "")) < 0)
                return -1;

and the output in console is :

before sort :[10-$urya, 10-andy, 10-prince, 10-@rikesh]
after sort :[10-andy, 10-prince, 10-@rikesh, 10-$urya]

Ignore 10, 10 is age in my Person Object.

Take a look at my classes, you will understand.

import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;


public class ComaprableCheck {

    public static void main(String[] args) {


        ArrayList<Person> person = new ArrayList<Person>();
        Person p = new Person();
        p.age =10;
        p.name ="$urya";
        person.add(p);
        p = new Person();
        p.age =10;
        p.name="andy";
        person.add(p);
        p = new Person();
        p.age =10;
        p.name ="prince";
        person.add(p);
        p = new Person();
        p.age =10;
        p.name ="@rikesh";
        person.add(p);
        System.out.println("before sort :"+person);
        Collections.sort(person);
        System.out.println("after sort :"+person);

    }

}



public class Person implements Comparable<Person> {


    public  int age;
    public  String name;
    @Override
    public int compareTo(Person o) {
        if(this.age > o.age)
            return 1;
        else if(this.age<o.age)
            return -1;
        else
        {
            if(this.name.replaceAll("[^a-zA-Z0-9]", "").compareTo(o.name.replaceAll("[^a-zA-Z0-9]", "")) >0)
                return 1;
            else if(this.name.replaceAll("[^a-zA-Z0-9]", "").compareTo(o.name.replaceAll("[^a-zA-Z0-9]", "")) < 0)
                return -1;
        }
        return 0;
    }

    public String toString(){
        return this.age+"-"+this.name;
    }

}

Below piece of code does the magic. It ignores the special characters just when they are comparing and returns the sorted list accordingly :

if(this.name.replaceAll("[^a-zA-Z0-9]", "").compareTo(o.name.replaceAll("[^a-zA-Z0-9]", "")) >0)
              return 1;           else if(this.name.replaceAll("[^a-zA-Z0-9]", "").compareTo(o.name.replaceAll("[^a-zA-Z0-9]", "")) < 0)
              return -1;

来源：https://stackoverflow.com/questions/48784691/how-to-sort-without-considering-special-characters-in-java

标签

java

sorting

compareto