如何将一个字符串中的特殊字符插入另一个字符串？

Question

我有一个字符串，

    string1 = "Sri Lanka National Chess Championship this year and represented Sri Lanka at represented Sri Lanka Universities at the World University Chess Championships."

我还有另一个名为“string2”的字符串，它只有被“<NOUN> and </NOUN>”标签包围的字符串，并用空格分隔。

string2 = "<NOUN>Sri Lanka National Chess Championship</NOUN> <NOUN>Sri Lanka</NOUN> <NOUN>Sri Lanka</NOUN> <NOUN>World University Chess</NOUN>"

注意第二个字符串可以有任何名词标记词（基于'string1'，例如：如果string1有3个名词，string2将有相同的3个名词，被名词标签包围）
我想要将标签添加到'string1'并使string1如下，

string1 = "<NOUN>Sri Lanka National Chess Championship</NOUN> this year and represented <NOUN>Sri Lanka</NOUN> at represented <NOUN>Sri Lanka</NOUN> Universities at the <NOUN>World University Chess</NOUN> Championships."

我使用以下代码来做到这一点，

Pattern p = Pattern.compile("<NOUN>(.*?)</NOUN>");
    Matcher m = p.matcher(string2);
    while(m.find()) {
        string1= string1.replaceAll(m.group(1),m.group(0));
    } 

但它给了我以下输出，

<NOUN><NOUN><NOUN>Sri Lanka</NOUN></NOUN> National Chess Championship</NOUN> this year and represented <NOUN><NOUN>Sri Lanka</NOUN></NOUN> at represented <NOUN><NOUN>Sri Lanka</NOUN></NOUN> Universities at the <NOUN>World University Chess</NOUN> Championships.

谁能告诉我如何正确执行此操作？
或者请告诉我如何从给定的输出中获得所需的输出？

Answer 1

而不是：

string1= string1.replaceAll(m.group(1),m.group(0));

使用：

string1= string1.replaceAll("(?<!<NOUN>)("+m.group(1)+")(?!</NOUN>)",m.group(0));

查看更多关于“向前看和向后看构造”here

Answer 2

你的例子的问题是Sri Lanka National Chess Championship是一个名词而Sri Lanka，这个字符串的一部分也是一个名词。因此，您的匹配器正在多次替换字符串。

您可以通过不替换已替换的字符串片段来解决此问题。我将每个匹配的字符串分成三个部分：之前、匹配字符串、之后。 保持断弦的顺序。 Vector 是一种非常方便的数据结构。

import java.util.Vector;
import java.util.regex.Matcher;
import java.util.regex.Pattern;


public class Check {

static String print(Vector<String> parts) {
    String str = parts.elementAt(0);

    for(int i=1; i<parts.size(); i++) {
        str += parts.elementAt(i); 
        //System.out.print(i + " : " + parts.elementAt(i) + "\n");
    }

    return str;
}

public static void main(String args[]) {
    String string1;
    String string2;
    String expected;

    string1 = "Sri Lanka National Chess Championship this year and represented Sri Lanka at represented Sri Lanka Universities at the World University Chess Championships.";
    string2 = "<NOUN>Sri Lanka National Chess Championship</NOUN> <NOUN>Sri Lanka</NOUN> <NOUN>Sri Lanka</NOUN> <NOUN>World University Chess</NOUN>";
    expected = "<NOUN>Sri Lanka National Chess Championship</NOUN> this year and represented <NOUN>Sri Lanka</NOUN> at represented <NOUN>Sri Lanka</NOUN> Universities at the <NOUN>World University Chess</NOUN> Championships.";


    Pattern p = Pattern.compile("<NOUN>(.*?)</NOUN>");
    Matcher m = p.matcher(string2);
    Vector<String> parts = new Vector<String>();
    parts.add(string1);

    while(m.find()) {
        for(int i=0; i<parts.size(); i++) {

            //search for used part
            if(parts.elementAt(i).indexOf("<NOUN>")!=-1) {
                continue;
            }

            // search for pattern
            String cur = parts.elementAt(i);
            int disp = cur.indexOf(m.group(1));
            if(disp==-1) {
                continue;
            } else {
                parts.remove(i);
                Vector<String> newParts = new Vector<String>();

                if(disp!=0) {
                    newParts.add(cur.substring(0, disp));
                }

                newParts.add(m.group(0));

                if((disp+m.group(1).length())!=cur.length()) {
                    newParts.add(cur.substring(disp+m.group(1).length()));
                }

                if(i!=0) {
                    parts.addAll(i, newParts);
                } else {
                    parts.addAll(newParts);
                }

                //System.out.print(print(parts) + "\n");
            }           
        }
    }

    string1 = print(parts);
    if(!string1.equals(expected)) {
        System.out.println("Unexpected output !!");
    } else {
        System.out.println("Correct !!");
    }
}

};

为方便起见，您可以将打印方法重命名为字符串化。