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Home/ Questions/Q 6595493
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T17:56:49+00:00 2026-05-25T17:56:49+00:00

And i have no idea why! I bascially have a STRING (yes, not an

  • 0

And i have no idea why!

I bascially have a STRING (yes, not an array), that has the following contents:

[something, something else, somoething, trallala, something]

And i want to turn it into a String[]. So first off i substring() off the first and the last character to get rid of the brackets []. Then i use the split() function to split by comma. I tried using both “\|” and “,” and “\,” with the same results.

This is what i get:

[Ljava.lang.String;@186d4c1

Here’s the code for it. I made it into a one-liner:

String[] urlArr = ((matcher.group(3).toString()).substring(1, (matcher.group(3).length()-1))).split(",");

As you can see the first part is (matcher.group(3).toString()), and it DOES return a valid string (like the example i posted above). So i don’t get why it’s not working.

Any ideas?

EDIT:

I clarified the code a bit:

String arrString = matcher.group(3).toString();
int length = arrString.length();
String[] urlArr = (arrString.substring(1, length-1)).split(",");
System.out.println(urlArr);
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  1. Editorial Team
    Editorial Team
    2026-05-25T17:56:50+00:00Added an answer on May 25, 2026 at 5:56 pm

    You are getting a valid array of Strings, but trying to print it directly does not do what you would expect it to do. Try e.g.

    System.out.println(Arrays.asList(urlArr));
    

    Update

    If you want to process the parsed tokens one by one, you can simply iterate through the array, e.g.

    for (String url : urlArr) {
      // Do something with the URL, e.g. print it separately
      System.out.println("Found URL " + url);
    }
    
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