如何检测括号（在行尾或新行）？

Question

This is my code.

while(in.hasNext()){
        String line = in.nextLine();
        in.hasNextLine();
        if (line.length() >0){
            int k = -1;
            k = -1;
            while(true){
                k = line.indexOf('\n' + "{", k+1);
                if(k<0)break;
                bracketNewLine++;
            }
            k = -1;
            while(true){
                k = line.indexOf(" {", k+1);
                if(k<0)break;
                bracketWithSpace++;
            }
        }
    }

if I have text file

if (...) {
}

and 

if (...)
{
}

output is:

• bracet end of line are: 1
• bracet new line are: 1

Thank you for answering.

Answer 1

You read the file line by line. Hence, there is no chance to find the character \\n followed by another character in the same String. Hence, '\\n' + "{" will never be found.

You can do it with simple regex :

for(String line : Files.readAllLines(Paths.get("/path/to/input.txt"))) {
  if(line.matches("\\{.*")) {
    bracketNewLine++;
  }

  if(line.matches(".* \\{")) {
    bracketWithSpace++;
  }
}

Answer 2

You could use some regex like this:

String patternInLine = ".+\\{$";
String patternNewLine = "^\\{":

Pattern p1 = new Pattern(patternInLine);
Pattern p2 = new Pattern(patternNewLine);

while(in.hasNext()) {
    String line = in.nextLine();
    in.hasNextLine();

    Matcher m1 = p1.match(line);
    Matcher m2 = p2.match(line);
    if(m1.match())
    {
        //inLine++;
    }
    else if (m2.match())
    {
        //newLine++;
    }
    else
    {
        //other cases
    }
}

Answer 3

When you use nextLine() method, you already have the source line by line. The only thing you should do is checking those lines in every loop with existing String methods : startsWith() , endsWith() . If we assume that you are correctly obtaining line string in each loop, inside of that while block should be like that:

if(line.startsWith("{"))
    bracketNewLine++;
if(line.endsWith("{"))
    bracketWithSpace++;

PS1 hasNext() method does not assure us that we have one more new line.

PS2 searching a string with fixed size space is not a true approach. You may use regex instead of that : ^[\\s]*{