如何检查是否有新行？答案

【问题标题】：How can I check if there is a new line?如何检查是否有新行？
【发布时间】：2014-05-20 20:43:14
【问题描述】：

这个程序应该转换如下：

\hyp76{a,1+a-b}b{xy}

到：

\HyperpFq{7}{6}@@{a,1+a-b}{b}{xy}.

如您所见，如果组的内容是一个字符，则没有括号。但是，当带括号的组之一位于行尾时，它会跳过该组并将其视为单个字符（不带括号）的情况。我怎样才能避免这种情况？谢谢。这是我的代码：

static int checkNestedBrackFront(String line, int pos){
    int count=0;
    for(int i=pos;i<line.length();i++ ){
        if(line.charAt(i)=='{')
            count++;
        if(line.charAt(i)=='}'&&count==0)
            return i;
        if(line.charAt(i)=='{')
            count--;
    }
    return 0;
}

line = new Scanner(new File("KLSadd.tex")).useDelimiter("\\Z").next();
    PrintWriter writer = new PrintWriter("Converted.tex");
    while(line.contains("\\hyp76")){
            int posHyp = line.indexOf("\\hyp76");
            String beforeHyp = line.substring(0,posHyp);
            int start = posHyp+7;   
            String firstArgs = line.substring(start, line.indexOf("}", start));
            if(line.charAt((line.indexOf("}", start)+1))!='{'&&!(line.substring(start, start+4).contains("\n"))){ //this is to check for single characters
                secArgs = line.substring(line.indexOf("}", start) + 1,line.indexOf("}", start) + 2 );
                posSec=line.indexOf("}", start)+1;
            }
            else {
                int posBrack = line.indexOf("{", line.indexOf("}", start));
                posSec = checkNestedBrackFront(line, posBrack+1);
                secArgs = line.substring(posBrack+1, posSec);
            }
            if((line.charAt(posSec+1)!='{')&&!(line.substring(posSec,posSec+4).contains("\n"))){ //this is to check for single characters
                System.out.println(line.charAt(posSec+1)+"hello");
                 thirdArgs= line.substring(posSec+1,posSec+2);
                 afterHyp=line.substring(posSec+2);
            }
            else{
            int posThirdBrack = line.indexOf("{", posSec);
            int posThird = checkNestedBrackFront(line, posThirdBrack+1);
            thirdArgs = line.substring(posThirdBrack+1,posThird);
            afterHyp = line.substring(posThird+1);
            }
            line=beforeHyp+"\\HyperpFq{7}{6}@@{"+firstArgs+"}{"+secArgs+"}{"+thirdArgs+"}"+afterHyp;
        }
      writer.print(line);
    writer.close();

【问题讨论】：

非常混乱。试着澄清你的问题。什么是单字符大小写？什么是括号案例？这个返回 0 的无用函数 'checkNestedBrackFront' 是什么？

标签： java string file

【解决方案1】：

有点令人困惑，但据我了解，您想检测新行：

 PrintWriter writer = new PrintWriter("Converted.tex");
        Scanner scanner = new Scanner(new File("KLSadd.txt"))
                .useDelimiter("\n");
        while (scanner.hasNext()) {
        String  lne = scanner.next();
        String[] lines =  lne.split("YOUR DELIMETER");
        for(int j=0;j<lines.length;j++){
            line = lines[j];
            //the rest of your logic  goes here
            }

【讨论】：