用Regex在字符串中查找特殊字符串中的特殊字符串

Question

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，键被写成这个@keyName或@"keyName" --它们应该被替换为

我们的地图是这样的

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

所以，如果我们处理字符串“你好世界，我是@key1 1岁”，，它将变成“你好世界，我2岁”。

我们可以使用@key1代替@"key1"。如果我们使用它没有引号，空格(空格字符)或EOF应该在键名后面，键名不应该有空格。但是如果键名中有一个空格，那么它应该是引号。

如果我们处理字符串"hello world，我是@"key@"key1"“几岁”。，在第一步，它应该取代特殊字符串中的特殊字符串，成为"hello world，I‘am @"key2“。”，然后第二步应该是"hello world，我3岁“。

我已经做了一个特殊的字符串，它不识别特殊字符串中的特殊字符串。以下是代码：

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;
}

对不起，我的英语，我缺乏Regex知识。我认为通过稍微修改代码就可以很容易地得到答案。

以下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.

Answer 1

我在这里做了一个与你的例子相匹配的正则表达式：https://regex101.com/r/nudYEl/2

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

您只需将函数修改为递归：

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

regex：https://regex101.com/r/nudYEl/4

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

Answer 2

不要使用regex：

for (boolean hit = true; hit;) {
    hit = false;
    for (String key : map.keySet()) {
        if (str.contains("@\"" + key + "\"")) {
            str = str.replace("@\"" + key + "\"", map.get(key));
            hit = true;
        } else if (str.contains("@" + key  )) {
            str = str.replace("@" + key + "", map.get(key));
            hit = true;
        }
    }
}