{0}的Java模式匹配

我需要用hashmap中的值替换{0}。hashmap键分别为0,1,2,3。我觉得我们可以实现模式匹配器。这个值的模式是什么

Dear {0}, 
You are being contacted because the '{1}' named '{2}' has been changed by '{3}' in the Application. These changes may impact any existing reports you are using which depend upon this information. If you have not authorized these changes, please contact '{4}' or send a request to  IT Support to have the changes reversed 

输出：

Dear abc , 
You are being contacted because the ' Attribute' named 'prod1_group' has been changed by ' Guest ' in the Application. These changes may impact any existing reports you are using which depend upon this information. If you have not authorized these changes, please contact 'Guest' or send a request to IT Support to have the changes reversed 

---

您可以使用此正则表达式：

Pattern p = Pattern.compile("\\{(\\d+)}");

并使用

matcher.group（1）

作为

while（matcher.find（））{..}

循环中

HashMap

的键

其中，

matcher

是：

Matcher matcher = p.matcher( input );

---

您需要ecsape

{

字符，否则您将获得PatternSyntaxException，为了捕获数字

\\d+

。最后

匹配器。组（1）

将返回字符串，因此您需要将其强制转换为

整数

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

public class Registrar {
    public static void main(String[] args) {

        String input = "Dear {0}, \n" +
                "You are being contacted because the '{1}' named '{2}' has been changed by '{3}' in the Application. These changes may impact any existing reports you are using which depend upon this information. If you have not authorized these changes, please contact '{4}' or send a request to IT Support to have the changes reversed";


        Pattern pattern = Pattern.compile("\\{(\\d+)}");

        HashMap<Integer, String> map = new HashMap<>();
        map.put(0, "zero");
        map.put(1, "one");
        map.put(2, "two");
        map.put(3, "three");
        map.put(4, "four");

        Matcher matcher = pattern.matcher(input);

        while (matcher.find()) {
            String val = matcher.group(1);
            String group = matcher.group();
            input = input.replace(group, map.get(Integer.parseInt(val)));
        }

        System.out.println(input);
    }
}

你可以用
\{[0-9]{1}

    String testString = "{0}";
    String myPattern = "\\{[0-9]{1}}";
    Pattern pattern = Pattern.compile(myPattern);
    Matcher m = pattern.matcher(testString);
    if(m.matches()) {
        System.out.println("Correct Value");
    } else {
        System.out.println("Wrong Value");          
    }

要从字符串中匹配它，请执行以下操作

    String testString = "Dear {0},";
    String myPattern = ".*\\{[0-9]{1}}.*";

PatternSyntaxException将失败，您必须转义{字符。感谢@iMmo:I之前有
\{
和
\}
，但后来它在regex101站点上正常工作，所以我将其删除：）我看到了，但regex101不支持java，顺便说一句，不需要转义：）
    String testString = "Dear {0},";
    String myPattern = ".*\\{[0-9]{1}}.*";