extract word from string that have combination of letter and number in java?

Question

this is my first question and sorry for my bad English

I want to extract only word from String that have combination of letter and number and store it in array

I try this code, but I don't get what I want

String temp = "74 4F 4C 4F 49 65  brown fox jump over the fence";
String [] word = temp.split("\\W");

this is the result that I want (only word and no empty array)

brown
fox
jump
over
the
fence

Please help, Thank you !

Answer 1

You can use:

String temp = "74 4F 4C 4F 49 65  brown fox jump over the fence";
List<String> arr = new ArrayList<String>();
Pattern p = Pattern.compile("(?i)(?:^|\\s+)([a-z]+)");
Matcher m = p.matcher(temp);
while (m.find())
    arr.add(m.group(1));

// convert to String[]
String[] word = arr.toArray(new String[0]);
System.out.println( Arrays.toString(word) );

OUTPUT:

[brown, fox, jump, over, the, fence]

Answer 2

Based on @anubhava's answer, you could do something like

String temp = "74 4F 4C 4F 49 65  brown fox jump over the fence";
Pattern pattern = Pattern.compile("\\b[A-Za-z]+\\b");
Matcher matcher = pattern.matcher(temp);

while (matcher.find()) {
  System.out.println("Matched " + matcher.group());
}

Answer 3

Works for:

android java

---

There are already some answers here but this is the one I'd preferred to do that. Try this code:

List<String> wordsOnlyList = new ArrayList<>(); // This list contains all the words without numbers or special chars
String sentence = "I bought a A7A for $36,000"; // This is a sample sentence to test the result
String[] words = sentence.split(" "); // split into each word

for(String word : words){
   if (!(Pattern.compile("[^a-z ]", Pattern.CASE_INSENSITIVE).matcher(word).find())) { //checking if it has only alphabets
      // it has only alphabets
      wordsOnlyList.add(word); // add the word to the list
   }
}

System.out.println(wordsOnlyList.toString()); // display in console

Result:

[I, bought, a, for]

You can also test the code from here.