排序字母数字字符串java

Question

I have this array storing the suffix of some URLs the user is adding:

[U2, U3, U1, U5, U8, U4, U7, U6]

When I do this:

for (Map<String, String> map : getUrlAttachments()) {
            String tmpId = map.get("id"); //it receives the U2, in the 1st iteration, then U3, then U1,...
            if (tmpId.charAt(0) == 'U') {
                tmpId.charAt(1);//2, then 3, then 1,...
                String url = map.get("url");
                String description = map.get("description");
                URLAttachment attachment;
                String cleanup = map.get("cleanup");
                if (cleanup == null && url != null && description != null) {
                    attachment = new URLAttachmentImpl();
                    attachment.setOwnerClass(FileUploadOwnerClass.Event.toString());
                    attachment.setUrl(url);
                    attachment.setDescription(description);
                    attachment.setOwnerId(auctionHeaderID);
                    attachment.setUrlAttachmentType(URLAttachmentTypeEnum.EVENT_ATTACHMENT);
                    attachment.setDateAdded(new Date());
                    urlBPO.save(attachment);

            }

My problem:

I want to change this For condition by passing another list mapping the data sorted like [U1, U2, U3, U4, U5, U6, U7, U8] .

I'd like your help to know what's the best way I could do this.

I thought about creating an array listing the ids and then sort then, but I don't know how exactly to sort alphanumeric strings in java.

Answer 1

Just use Collections.sort() method after creating an ArrayList of your values like this:

ArrayList<String> a = new ArrayList<String>();
a.add("U2");
a.add("U1");
a.add("U5");
a.add("U4");
a.add("U3");
System.out.println("Before : "+a);
Collections.sort(a);
System.out.println("After : "+a);

Output :

Before : [U2, U1, U5, U4, U3]
After : [U1, U2, U3, U4, U5]

Answer 2

I decide to use the idea @Abu gave, but I adapted it:

1. I check the ids of the urls the user is trying to add,
2. I remove the alphabetic suffix in this id and then I create an ArrayList to store the numerical part of each id.

3. I sort this ArrayList like @Abu taught me in his answer and then I verify for each id in this sorted ArrayList in the sequence it should be added..

    ArrayList <Integer> urlSorted = new ArrayList<Integer>(); //sort the url ids for (Map<String, String> map : getUrlAttachments()) { String tmpId = map.get("id"); if (tmpId.charAt(0) == 'U') { //gets the id, removing the prefix 'U' urlSorted.add( Integer.valueOf(tmpId.substring(1))); } } //sort the urlIds to check the sequence they must be added Collections.sort(urlSorted); //checks for each url id, compares if it's on the natural order of sorting to be added. for(Integer urlId: urlSorted) { for (Map<String, String> map : getUrlAttachments()) { String sortedId = "U"+urlId; String tmpId = map.get("id"); //compare the ids to add the 1, then 2, then 3... if (map.get("id").equals(sortedId)) { //code to save according to the sorted ids. } } } 

Answer 3

Create a custom Comparator<Map<String,String>> :

public class IdComparator implements Comparator<Map<String,String>> {
  public int compare(Map<String,String> left, Map<String,String> right) {
    return orderKey(left).compareTo(orderKey(right));
  }
  static Integer orderKey(Map<String,String> m) { 
    return Integer.parseInt(m.get("id").substring(1)); 
  }
}

and then use Arrays.sort(urlAttachments, new IdComparator()); prior to iterating over it. Depending on details, you may push this sorting logic into getUrlAttachments() and keep the code you have posted exactly as it is now.

Answer 4

I think what you are asking is similar to this one :

http://www.davekoelle.com/alphanum.html

You can break string into pure string and numeric string. for eg: abc123 would be split into "abc" and "123" You can compare alphabetic string with normal comparison and then to sort "123" such kind of strings, you have two options: 1: Convert it into Integer and then compare 2: If number does not fit in Integer range, you can compare letter by letter.

for eg "123" vs "133" compare "1" and "1" = equal Compare "2" and "3" = greater so "123" < "133".

Option 2 is more accurate and less error proof.