Unclosed Character Class Error?

Question

Here is the error:

Exception in thread "main" java.util.regex.PatternSyntaxException: Unclosed character class near index 3
], [
   ^
    at java.util.regex.Pattern.error(Pattern.java:1924)
    at java.util.regex.Pattern.clazz(Pattern.java:2493)
    at java.util.regex.Pattern.sequence(Pattern.java:2030)
    at java.util.regex.Pattern.expr(Pattern.java:1964)
    at java.util.regex.Pattern.compile(Pattern.java:1665)
    at java.util.regex.Pattern.<init>(Pattern.java:1337)
    at java.util.regex.Pattern.compile(Pattern.java:1022)
    at java.lang.String.split(String.java:2313)
    at java.lang.String.split(String.java:2355)
    at testJunior2013.J2.main(J2.java:31)

This is the area of the code that is causing the issues.

String[][] split = new String[1][rows];

split[0] = (Arrays.deepToString(array2d)).split("], ["); //split at the end of an array row

What does this error mean and what needs to be done to fix the code above?

Answer 1

TL;DR

You want:

.split("\\], \\[")`

Escape each square bracket twice — once for each context in which you need to strip them from their special meaning: within a Regular Expression first, and within a Java String secondly.

Consider using Pattern#quote when you need your entire pattern to be interpreted literally .

---

Explanation

String#split works with a Regular Expression but [ and ] are not standard characters, regex-wise: they have a special meaning in that context.

In order to strip them from their special meaning and simply match actual square brackets , they need to be escaped , which is done by preceding each with a backslash — that is, using \\[ and \\] .

However, in a Java String , \\ is not a standard character either, and needs to be escaped as well .

Thus, just to split on [ , the String used is "\\\\[" and you are trying to obtain:

.split("\\], \\[")

---

A sensible alternative

However, in this case, you're not just semantically escaping a few specific characters in a Regular Expression , but actually wishing that your entire pattern be interpreted literally : there's a method to do just that 🙂

Pattern#quote is used to signify that the:

Metacharacters [...] in your pattern will be given no special meaning.

(from the Javadoc linked above)

I recommend, in this case, that you use the following, more sensible and readable:

.split(Pattern.quote("], ["))

Answer 2

Split receives a regex and [, ] characters have meaning in regex, so you should escape them with \\\\[ and \\\\] .

The way you are currently doing it, the parser finds a ] without a preceding [ so it throws that error.

Answer 3

String.split() takes a regular expression , not a normal string as an argument. In a regular expression, ] and [ are special characters, which need to be preceded by backslashes to be taken literally. Use .split("\\\\], \\\\[") . (the double backslashes tell Java to interpret the string as "\\], \\[").

Answer 4

  .split("], [")
             ^---start of char class
                  end----?

Change it to

.split("], \[")
           ^---escape the [

Answer 5

Try to use it

 String stringToSplit = "8579.0,753.34,796.94,\"[784.2389999999999,784.34]\",\"[-4.335912230999999, -4.3603307895,4.0407909059, 4.08669583455]\",[],[],[],0.1744,14.4,3.5527136788e-15,0.330667850653,0.225286999939,Near_Crash";
 String [] arraySplitted = stringToSplit.replaceAll("\"","").replaceAll("\\[","").replaceAll("\\]","").trim().split(",");