找到没有被困在双引号java正则表达式之间的特定字符串

Question

I want to find a string (say x ) that satisfies two conditions:

1. matches the pattern \\b(x)\\b
2. does not match the pattern ".*?(x).*?(?<!\\\\)"

In other words, I am looking for a value of x that is a complete word (condition1) and it is not in double quotes (condition2).

• " x /" m" not acceptable
• " x \\" " + x + " except" :only the second x is acceptable.

What Java code will find x ?

Answer 1

The first condition is straight forward. To check second condition you will have to check number of valid double quotes. If they are even then the string captured in first condition is valid.

String text = "basdf + \" asdf \\\" b\" + b + \"wer \\\"\"";
String toCapture = "b";
Pattern pattern1 = Pattern.compile("\\b" + toCapture + "\\b");
Pattern pattern2 = Pattern.compile("(?<!\\\\)\"");
Matcher m1 = pattern1.matcher(text);
Matcher m2; 
while(m1.find()){                               // if any <toCapture> found (first condition fulfilled)
    int start = m1.start();
    m2 = pattern2.matcher(text);
    int count = 0;
    while(m2.find() && m2.start() < start){     // count number of valid double quotes "
        count++;
    }
    if(count % 2 == 0) {                        // if number of valid quotes is even 
        char[] tcar = new char[text.length()];
        Arrays.fill(tcar, '-');
        tcar[start] = '^';
        System.out.println(start);
        System.out.println(text);
        System.out.println(new String(tcar));
    }
}

Output :

23
basdf + " asdf \" b" + b + "wer \""
-----------------------^-----------