24

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?

ylun.ca
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5 Answers5

63

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("], ["))
ccjmne
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  • This is just the answer i was looking for, thank you. In what cases should I use `Pattern.quote` instead of the standard `.split("\\], [\\")`? – ylun.ca Feb 16 '14 at 21:30
  • You "should" always use `Pattern.quote`, because it's equivalent to escaping the regex string yourself... but using this, you're sure that you won't forget anything and besides that, it's more readable, in my opinion :) – ccjmne Feb 16 '14 at 21:36
6

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.

xp500
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2

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 "\], \[").

McLovin
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1
  .split("], [")
             ^---start of char class
                  end----?

Change it to 

.split("], \[")
           ^---escape the [
Marc B
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0

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(",");
Skizzo
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