Java 在正则表达式中使用backreference动态替换文本

我想像使用整数一样使用$1值。
其思想是用等效数组值替换originaltext中的所有数字，并创建一个新文本。
下面的预期结果应该是“这是DBValue4，这是DBValue2，这是DBValue7”
还有，有没有办法保存这些反向引用以供进一步使用

String[] values = {"DBValue0","DBValue1","DBValue2","DBValue3","DBValue4","DBValue5","DBValue6","DBValue7","DBValue8","DBValue9","DBValue10"};
String originaltext = "This is 4, This is 2, This is 7";
text = originaltext.replaceAll("(\\d)","$1");
// want something like
text = originaltext.replaceAll("(\\d)",values[$1]);
//or
text = originaltext.replaceAll("(\\d)",values[Integer.parseInt("$1")]);

---

您可以像这样使用

模式

和

匹配器

：

public static void main(String[] args) throws Exception {
    final String[] values = {"DBValue0", "DBValue1", "DBValue2", "DBValue3", "DBValue4", "DBValue5", "DBValue6", "DBValue7", "DBValue8", "DBValue9", "DBValue10"};
    final String originaltext = "This is 4, This is 2, This is 7";
    final Pattern pattern = Pattern.compile("(?<=This is )\\d++");
    final Matcher matcher = pattern.matcher(originaltext);
    final StringBuffer sb = new StringBuffer();
    while (matcher.find()) {
        System.out.println(matcher.group());
        final int index = Integer.parseInt(matcher.group());
        matcher.appendReplacement(sb, values[index]);
    }
    matcher.appendTail(sb);
    System.out.println(sb);
}

编辑

除了OP的注释之外，OP似乎需要替换

{name，index}

格式的

字符串

s，其中“name”是数组的名称，“index”是该数组中元素的索引

这可以通过

Map

使用

Map

将数组ping到它们的名称，然后使用

模式来实现，该模式首先捕获
名称
，然后捕获
索引

public static void main(String[] args) throws Exception {
    final String[] companies = {"Company1", "Company2", "Company3"};
    final String[] names = {"Alice", "Bob", "Eve"};
    final String originaltext = "This is {company, 0}, This is {name, 1}, This is {name, 2}";
    final Map<String, String[]> values = new HashMap<>();
    values.put("company", companies);
    values.put("name", names);
    final Pattern pattern = Pattern.compile("\\{([^,]++),\\s*+(\\d++)}");
    final Matcher matcher = pattern.matcher(originaltext);
    final StringBuffer sb = new StringBuffer();
    while (matcher.find()) {
        System.out.println(matcher.group(1));
        System.out.println(matcher.group(2));
        final int index = Integer.parseInt(matcher.group(2));
        matcher.appendReplacement(sb, values.get(matcher.group(1))[index]);
    }
    matcher.appendTail(sb);
    System.out.println(sb);
}

使用
模式
和
匹配器
你可以看看这篇文章：很抱歉，为了简单起见，这个例子不太好。您所做的工作是有效的，但我想使用更复杂的内容，如“String originaltext=“{company，3}为{name，2}从{new，11}”；“number{number，10}的所有者发送此消息”和“text=originaltext.replaceAll（\\{.+，（\\d\\d？\\}）”，值[Integer.parseInt（$1”）；”。有什么建议吗？我不明白-所以你有几个数组，你想在
字符串上选择它们吗？没有，我只是想用values[an_int]数组属性替换任何文本中的{something，an_int}。如果你做了一些更改，我的编辑应该可以帮你完成。
public static void main(String[] args) throws Exception {
    final String[] companies = {"Company1", "Company2", "Company3"};
    final String[] names = {"Alice", "Bob", "Eve"};
    final String originaltext = "This is {company, 0}, This is {name, 1}, This is {name, 2}";
    final Map<String, String[]> values = new HashMap<>();
    values.put("company", companies);
    values.put("name", names);
    final Pattern pattern = Pattern.compile("\\{([^,]++),\\s*+(\\d++)}");
    final Matcher matcher = pattern.matcher(originaltext);
    final StringBuffer sb = new StringBuffer();
    while (matcher.find()) {
        System.out.println(matcher.group(1));
        System.out.println(matcher.group(2));
        final int index = Integer.parseInt(matcher.group(2));
        matcher.appendReplacement(sb, values.get(matcher.group(1))[index]);
    }
    matcher.appendTail(sb);
    System.out.println(sb);
}

company
0
name
1
name
2
This is Company1, This is Bob, This is Eve