Checking if there is whitespace between two elements in a String

I am working with Strings where I need to separate two chars/elements if there is a whitespace between them. I have seen a former post on SO about the same however it still has not worked for me as intended yet. As you would assume, I could just check if the String contains(" ") and then substring around the space. However my strings could possibly contains countless whitespaces at the end despite not having whitespace in between characters. Hence my question is "How do I detect a whitespace between two chars (numbers too) " ?

//Example with numbers in a String

    String test = "2 2";

    final Pattern P = Pattern.compile("^(\\d [\\d\\d] )*\\d$");

    final Matcher m = P.matcher(test);

    if (m.matches()) {
        System.out.println("There is between space!");
    }

You would use String.strip() to remove any leading or trailing whitespace, followed by String.split(). If there is a whitespace, the array will be of length 2 or greater. If there is not, it will be of length 1.

Example:

String test = "    2 2   ";
test = test.strip(); // Removes whitespace, test is now "2 2"
String[] testSplit = test.split(" "); // Splits the string, testSplit is ["2", "2"]
if (testSplit.length >= 2) {
    System.out.println("There is whitespace!");
} else {
    System.out.println("There is no whitespace");
}

If you need an array of a specified length, you can also specify a limit to split. For example:

"a b c".split(" ", 2); // Returns ["a", "b c"]

If you want a solution that only uses regex, the following regex matches any two groups of characters separated by a single space, with any amount of leading or trailing whitespace:

\s*(\S+\s\S+)\s*

Positive lookahead and lookbehind may also work if you use the regex (?<=\\w)\\s(?=\\w)

• \w : a word character [a-zA-Z_0-9]
• \\s : whitespace
• (?<=\\w)\\s : positive lookbehind, matches if a whitespace preceeded by a \w
• \\s(?=\\w) : positive lookahead, matches if a whitespace followed by a \w

List<String> testList = Arrays.asList("2 2", " 245  ");

Pattern p = Pattern.compile("(?<=\\w)\\s(?=\\w)");
for (String str : testList) {

	Matcher m = p.matcher(str);

	if (m.find()) {
		System.out.println(str + "\t: There is a space!");
	} else {
		System.out.println(str + "\t: There is not a space!");
	}
}

Output:

2 2	: There is a space!
 245  	: There is not a space!

The reason you pattern does not work as expected is because ^(\\d [\\d\\d] )*\\d$ which can be simplified to (\\d \\d )*\\d$ starts by repeating 0 or more times what is between the parenthesis.

Then it matches a digit at the end of the string. As the repetition is 0 or more times, it is optional and it would also match just a single digit.

<hr>

If you want to check if there is a single space between 2 non whitespace chars:

\\S \\S

Regex demo | Java demo

final Pattern P = Pattern.compile("\\S \\S");
final Matcher m = P.matcher(test);

if (m.find()) {
    System.out.println("There is between space!");
}

Here is the simplest way you can do it:

String testString = "   Find if there is a space.   ";
testString.trim(); //This removes all the leading and trailing spaces
testString.contains(" "); //Checks if the string contains a whitespace still

You can also use a shorthand method in one line by chaining the two methods:

String testString = "   Find if there is a space.   ";
testString.trim().contains(" ");

</details>



# 答案5
**得分**: 0

```java
String text = "2 2";
Matcher m = Pattern.compile("\\S\\s+\\S").matcher(text.trim());
if (m.find()) {
    System.out.println("Space detected.");
}

在这个Java代码示例中，text.trim() 将会移除字符串开头和结尾的空白字符，\S\s+\S 模式用于匹配非空白字符，然后是一个或多个空白字符，然后再是一个非空白字符。

String text = "2 2";
Matcher m = Pattern.compile("\\S\\s+\\S").matcher(text.trim());
if (m.find()) {
    System.out.println("Space detected.");
}