java - 使用Regex在字符串中的特殊字符串中查找特殊字符串

我有一个HashMap，里面有键和值。我想用字符串中的映射值替换键。

在字符串中，键写为此@keyName或@“ keyName”，应将其替换为map.get（“ keyName”）

可以说我们的地图是这个

"key1" : "2"
"key2" : "3"
"key3" : "4"
"key 4" : "5"
"key-5" : "6"

因此，如果我们处理字符串“ hello world，我今年@ key1岁。”，它将变成“ hello world，我今年2岁”。

代替@ key1，我们可以使用@“ key1”。如果我们在不带引号的情况下使用它，则键名后面应包含空格（空格字符）或EOF，并且键名中不应包含空格。但是，如果键名中有空格，则应在引号中。

并且，如果我们处理字符串“ hello world，我是@“ key @” key1“”岁。”，则第一步应替换特殊字符串中的特殊字符串，并成为“ hello world，我是@“ key2”。岁。”然后第二步应该是“你好，我今年3岁”。

我已经为一个特殊字符串完成了此操作，它无法识别特殊字符串中的特殊字符串。这是代码：

private static Pattern specialStringPattern = Pattern.compile("@\"([^\"]*?)\"|@\\S+");

/** this replaces the keys inside a string with their values.
 * for example @keyName or @"keyName" is replaced with the value of the keyName. */
public static String specialStrings(String s) {
    Matcher matcher = specialStringPattern.matcher(s);
    while (matcher.find()) {
        String text = matcher.group();
        text = text.replace("@","").replaceAll("\"","");
        s = s.replace(matcher.group(),map.get(text));
    }
    return s;
}

很抱歉，我的英语水平以及我对Regex的了解不足。我认为通过一点点修改代码来获得答案应该很容易。

这是我需要的一些例子：

There is @key1 apples on the table.
There is 2 apples on the table.

There is @"key1" apples on the table.
There is 2 apples on the table.

There is @key 4 apples on the table.
There is null 4 apples on the table.

There is @"key 4" apples on the table.
There is 5 apples on the table.

There is @key@key2 apples on the table.
There is @key3 apples on the table. (second step)
There is 4 apples on the table. (final output)

There is @"key @"key3"" apples on the table.
There is @"key 4" apples on the table. (second step)
There is 5 apples on the table. (final output)

There is @"key @key3" apples on the table.
There is @"key 4" apples on the table. (second step)
There is 5 apples on the table. (final output)

There is @"key @key3 " apples on the table.
There is @"key 4 " apples on the table. (second step)
There is null apples on the table. (final output)

There is @key-5 apples on the table.
There is 6 apples on the table.

最佳答案

我做了正则表达式，以匹配您的示例在这里：
https://regex101.com/r/nudYEl/2

@(\"[\w\s]+\")|(?!@(\w+)@(\w+))@(\w+)

并且您只需要将函数修改为递归即可：

public static String specialStrings(String s) {
    Matcher matcher = specialStringPattern.matcher(s);
    boolean findAgain = false;
    while (matcher.find()) {
        String text = matcher.group();
        text = text.replace("@","").replaceAll("\"","");
        s = s.replace(matcher.group(),map.get(text));
        findAgain = true;
    }
    if (findAgain) return specialStrings(s);
    return s;
}

[更新]
正则表达式：https://regex101.com/r/nudYEl/4

@(\"[\w\s-]+\")|(?!@([\w-]+)@([\w-]+))@([\w-]+)