How to convert comma-separated String to List?

Question

Is there any built-in method in Java which allows us to convert comma separated String to some container (eg array, List or Vector)? Or do I need to write custom code for that?

String commaSeparated = "item1 , item2 , item3";
List<String> items = //method that converts above string into list??

Answer 1

Convert comma separated String to List

List<String> items = Arrays.asList(str.split("\\s*,\\s*"));

The above code splits the string on a delimiter defined as: zero or more whitespace, a literal comma, zero or more whitespace which will place the words into the list and collapse any whitespace between the words and commas.

---

Please note that this returns simply a wrapper on an array: you CANNOT for example .remove() from the resulting List . For an actual ArrayList you must further use new ArrayList<String> .

Answer 2

Arrays.asList returns a fixed-size List backed by the array. If you want a normal mutable java.util.ArrayList you need to do this:

List<String> list = new ArrayList<String>(Arrays.asList(string.split(" , ")));

Or, using Guava :

List<String> list = Lists.newArrayList(Splitter.on(" , ").split(string));

Using a Splitter gives you more flexibility in how you split the string and gives you the ability to, for example, skip empty strings in the results and trim results. It also has less weird behavior than String.split as well as not requiring you to split by regex (that's just one option).

Answer 3

两步：

1. String [] items = commaSeparated.split("\\s*,\\s*");
2. List<String> container = Arrays.asList(items);

Answer 4

List<String> items= Stream.of(commaSeparated.split(","))
     .map(String::trim)
     .collect(Collectors.toList());

Answer 5

If a List is the end-goal as the OP stated, then already accepted answer is still the shortest and the best. However I want to provide alternatives using Java 8 Streams , that will give you more benefit if it is part of a pipeline for further processing.

By wrapping the result of the .split function (a native array) into a stream and then converting to a list.

List<String> list =
  Stream.of("a,b,c".split(","))
  .collect(Collectors.toList());

If it is important that the result is stored as an ArrayList as per the title from the OP, you can use a different Collector method:

ArrayList<String> list = 
  Stream.of("a,b,c".split(","))
  .collect(Collectors.toCollection(ArrayList<String>::new));

Or by using the RegEx parsing api:

ArrayList<String> list = 
  Pattern.compile(",")
  .splitAsStream("a,b,c")
  .collect(Collectors.toCollection(ArrayList<String>::new));

Note that you could still consider to leave the list variable typed as List<String> instead of ArrayList<String> . The generic interface for List still looks plenty of similar enough to the ArrayList implementation.

By themselves, these code examples do not seem to add a lot (except more typing), but if you are planning to do more, like this answer on converting a String to a List of Longs exemplifies, the streaming API is really powerful by allowing to pipeline your operations one after the other.

For the sake of, you know, completeness.

Answer 6

Here is another one for converting CSV to ArrayList:

String str="string,with,comma";
ArrayList aList= new ArrayList(Arrays.asList(str.split(",")));
for(int i=0;i<aList.size();i++)
{
    System.out.println(" -->"+aList.get(i));
}

Prints you

-->string
-->with
-->comma

Answer 7

List<String> items = Arrays.asList(commaSeparated.split(","));

那应该对你有用。

Answer 8

There is no built-in method for this but you can simply use split() method in this.

String commaSeparated = "item1 , item2 , item3";
ArrayList<String> items = 
new  ArrayList<String>(Arrays.asList(commaSeparated.split(",")));

Answer 9

This code will help,

String myStr = "item1,item2,item3";
List myList = Arrays.asList(myStr.split(","));

Answer 10

你可以结合 asList 和 split

Arrays.asList(CommaSeparated.split("\\s*,\\s*"))

Answer 11

You can use Guava to split the string, and convert it into an ArrayList. This works with an empty string as well, and returns an empty list.

import com.google.common.base.Splitter;
import com.google.common.collect.Lists;

String commaSeparated = "item1 , item2 , item3";

// Split string into list, trimming each item and removing empty items
ArrayList<String> list = Lists.newArrayList(Splitter.on(',').trimResults().omitEmptyStrings().splitToList(commaSeparated));
System.out.println(list);

list.add("another item");
System.out.println(list);

outputs the following:

[item1, item2, item3]
[item1, item2, item3, another item]

You could also use the following approach without Guava:

  public static List<String> getStringAsList(String input) {
    if (input == null || input.trim().length() == 0) {
      return Collections.emptyList();
    }

    final String separators = "[;:,-]";
    var result =
        Splitter.on(CharMatcher.anyOf(separators))
            .trimResults()
            .omitEmptyStrings()
            .splitToList(input);

    return result;
  }

Answer 12

There are many ways to solve this using streams in Java 8 but IMO the following one liners are straight forward:

String  commaSeparated = "item1 , item2 , item3";
List<String> result1 = Arrays.stream(commaSeparated.split(" , "))
                                             .collect(Collectors.toList());
List<String> result2 = Stream.of(commaSeparated.split(" , "))
                                             .collect(Collectors.toList());

Answer 13

An example using Collections .

import java.util.Collections;
 ...
String commaSeparated = "item1 , item2 , item3";
ArrayList<String> items = new ArrayList<>();
Collections.addAll(items, commaSeparated.split("\\s*,\\s*"));
 ...

Answer 14

在 groovy 中，您可以使用 tokenize(Character Token) 方法：

list = str.tokenize(',')

Answer 15

Same result you can achieve using the Splitter class.

var list = Splitter.on(",").splitToList(YourStringVariable)

(written in kotlin)

Answer 16

While this question is old and has been answered multiple times, none of the answers is able to manage the all of the following cases:

• "" -> empty string should be mapped to empty list
• " a, b , c " -> all elements should be trimmed, including the first and last element
• ",," -> empty elements should be removed

Thus, I'm using the following code (using org.apache.commons.lang3.StringUtils , eg https://mvnrepository.com/artifact/org.apache.commons/commons-lang3/3.11 ):

StringUtils.isBlank(commaSeparatedEmailList) ?
            Collections.emptyList() :
            Stream.of(StringUtils.split(commaSeparatedEmailList, ','))
                    .map(String::trim)
                    .filter(StringUtils::isNotBlank)
                    .collect(Collectors.toList());

Using a simple split expression has an advantage: no regular expression is used, so the performance is probably higher. The commons-lang3 library is lightweight and very common.

Note that the implementation assumes that you don't have list element containing comma (ie "a, 'b,c', d" will be parsed as ["a", "'b", "c'", "d"] , not to ["a", "b,c", "d"] ).

Answer 17

Java 9 introduced List.of():

String commaSeparated = "item1 , item2 , item3";
List<String> items = List.of(commaSeparated.split(" , "));

Answer 18

您可以先使用String.split(",")拆分它们，然后使用Arrays.asList(array)将返回的 String array转换为ArrayList

Answer 19

List commaseperated = new ArrayList();
String mylist = "item1 , item2 , item3";
mylist = Arrays.asList(myStr.trim().split(" , "));

// enter code here

Answer 20

I usually use precompiled pattern for the list. And also this is slightly more universal since it can consider brackets which follows some of the listToString expressions.

private static final Pattern listAsString = Pattern.compile("^\\[?([^\\[\\]]*)\\]?$");

private List<String> getList(String value) {
  Matcher matcher = listAsString.matcher((String) value);
  if (matcher.matches()) {
    String[] split = matcher.group(matcher.groupCount()).split("\\s*,\\s*");
    return new ArrayList<>(Arrays.asList(split));
  }
  return Collections.emptyList();

Answer 21

List<String> items = Arrays.asList(s.split("[,\\s]+"));

Answer 22

In Kotlin if your String list like this and you can use for convert string to ArrayList use this line of code

var str= "item1, item2, item3, item4"
var itemsList = str.split(", ")

Answer 23

In java, It can be done like this

String catalogue_id = "A, B, C";
List<String> catalogueIdList = Arrays.asList(catalogue_id.split(", [ ]*"));

Answer 24

You can do it as follows.

This removes white space and split by comma where you do not need to worry about white spaces.

    String myString= "A, B, C, D";

    //Remove whitespace and split by comma 
    List<String> finalString= Arrays.asList(myString.split("\\s*,\\s*"));

    System.out.println(finalString);

Answer 25

This method will convert your string to an array, taking two parameters: the string you want converted, and the characters that will separate the values in the string. It converts it, and then returns the converted array.

private String[] convertStringToArray(String stringIn, String separators){

    // separate string into list depending on separators
    List<String> tempList = Arrays.asList(stringIn.split(separators));
    
    // create a new pre-populated array based on the size of the list
    String[] itemsArray = new String[tempList.size()];
    
    // convert the list to an array
    itemsArray = tempList.toArray(itemsArray);
    
    return itemsArray;
}

Answer 26

String -> Collection conversion: (String -> String[] -> Collection)

//         java version 8

String str = "aa,bb,cc,dd,aa,ss,bb,ee,aa,zz,dd,ff,hh";

//          Collection,
//          Set , List,
//      HashSet , ArrayList ...
// (____________________________)
// ||                          ||
// \/                          \/
Collection<String> col = new HashSet<>(Stream.of(str.split(",")).collect(Collectors.toList()));

Collection -> String[] conversion:

String[] se = col.toArray(new String[col.size()]);

String -> String[] conversion:

String[] strArr = str.split(",");

And Collection -> Collection :

List<String> list = new LinkedList<>(col);

Answer 27

convert Collection into string as comma seperated in Java 8

listOfString object contains ["A","B","C" ,"D"] elements-

listOfString.stream().map(ele->"'"+ele+"'").collect(Collectors.joining(","))

Output is :- 'A','B','C','D'

And Convert Strings Array to List in Java 8

    String string[] ={"A","B","C","D"};
    List<String> listOfString = Stream.of(string).collect(Collectors.toList());

Answer 28

ArrayList<HashMap<String, String>> mListmain = new ArrayList<HashMap<String, String>>(); 
String marray[]= mListmain.split(",");