诡异的appendReplacement和replaceAll

 

一、起源

    这段代码的作用是将字符串中${param}替换为map中的数据

private static String replaceVariantOldVersion(String str, Map<String,String> variantMap){
		Matcher m = Pattern.compile("\\$\\{.*?\\}").matcher(str);
		StringBuffer rtn = new StringBuffer();
		while(m.find()){
			String foundStr = m.group();
			String key = foundStr.substring("${".length(),foundStr.length()-1).trim();
			String value = variantMap.get(key);
			if(null == value){
				value = "{UNKNOWN_VALUE}";
			}
			m.appendReplacement(rtn, value);
		}
		m.appendTail(rtn);
		return rtn.toString();
	}

   例如

 

String str = "my name is ${name}";
Map<String,String> map = new HashMap<String,String>();
map.put("name", "lazy");
System.out.println(replaceVariantOldVersion(str,map));//输出my name is lazy

 

 

但如果是

 

		
String str = "my name is ${name}";
Map<String,String> map = new HashMap<String,String>();
map.put("name", "$.lazy");
System.out.println(replaceVariantOldVersion(str,map));

 

则会抛出异常

 

Exception in thread "main" java.lang.IllegalArgumentException: Illegal group reference

at java.util.regex.Matcher.appendReplacement(Unknown Source)

at lgc.tools.struts2.ConventionUtil.replaceVariantOldVersion(ConventionUtil.java:241)

at lgc.tools.struts2.ConventionUtil.main(ConventionUtil.java:269)

二、 分析

分析下appendReplacement源代码

 public Matcher appendReplacement(StringBuffer sb, String replacement) {

        // If no match, return error
        if (first < 0)
            throw new IllegalStateException("No match available");

        // Process substitution string to replace group references with groups
        int cursor = 0;
        String s = replacement;
        StringBuffer result = new StringBuffer();

        while (cursor < replacement.length()) {
            char nextChar = replacement.charAt(cursor);
            if (nextChar == '\\') {
                cursor++;
                nextChar = replacement.charAt(cursor);
                result.append(nextChar);
                cursor++;
            } else if (nextChar == '$') {
                // Skip past $
                cursor++;

                // The first number is always a group
                int refNum = (int)replacement.charAt(cursor) - '0';
                if ((refNum < 0)||(refNum > 9))
                    throw new IllegalArgumentException(
                        "Illegal group reference");
                cursor++;

                // Capture the largest legal group string
                boolean done = false;
                while (!done) {
                    if (cursor >= replacement.length()) {
                        break;
                    }
                    int nextDigit = replacement.charAt(cursor) - '0';
                    if ((nextDigit < 0)||(nextDigit > 9)) { // not a number
                        break;
                    }
                    int newRefNum = (refNum * 10) + nextDigit;
                    if (groupCount() < newRefNum) {
                        done = true;
                    } else {
                        refNum = newRefNum;
                        cursor++;
                    }
                }

                // Append group
                if (group(refNum) != null)
                    result.append(group(refNum));
            } else {
                result.append(nextChar);
                cursor++;
            }
        }

        // Append the intervening text
        sb.append(getSubSequence(lastAppendPosition, first));
        // Append the match substitution
        sb.append(result.toString());

        lastAppendPosition = last;
	return this;
    }

 原来第二个参数replacement可以使用$n来引用分组!所以‘$’和‘\’都被当做特殊字符处理！replacement中的字符串$1代表匹配的分组1！

什么是分组？

看程序

Matcher m = Pattern.compile("((\\d\\d)(\\w))").matcher("11a22b");
StringBuffer sb = new StringBuffer();
while(m.find()){
	m.appendReplacement(sb, "$0,$1,$2,$3;");
}
m.appendTail(sb);
System.out.println(sb);

 输出的结果是11a,11a,11,a;22b,22b,22,b;

正则表达式((\d\d)(\w))有三个分组，每个括号包含的内容称作一个分组，并按照左括号出现的顺序给每个分组给予编号1,2,3,...,编号为0的分组代表整个被匹配的字符串。

例子程序中,

“11a”被 ((\d\d)(\w))匹配，$0是整个字符串，等于11a，此时sb是：11a

$1是(\d\d)(\w)撇配的内容，等于11a，此时sb是11a,11a

$2是(\d\d)匹配的内容，等于11，此时sb是11a,11a,11

$3是(\w)匹配的内容，等于a，此时sb是11a,11a,11,a;

相信大家到此应该大致明白分组的意义，如有问题请留言。

三、结论

回到正题，既然 appendReplacement(StringBuffer sb, String replacement)的replacement参数中'$'和'\'，那么我就对这两个字符进行转义。

(其中'\'字符是被当做特殊字符处理是因为，java将\$当做普通的$进行处理，所以也'\'被当做了特殊字符。你可以尝试运行

String str = "my name is ${name}";
Map<String,String> map = new HashMap<String,String>();
map.put("name", "\\.lazy");
System.out.println(replaceVariantOldVersion(str,map));

同样会因为特殊字符而报错。

)

最终，我修正后的代码是

private static String replaceVariant(String str, Map<String,String> variantMap){
		Matcher m = Pattern.compile("\\$\\{.*?\\}").matcher(str);
		StringBuffer rtn = new StringBuffer();
		while(m.find()){
			String foundStr = m.group();
			String key = foundStr.substring("${".length(),foundStr.length()-1).trim();
			String value = variantMap.get(key);
			if(null == value){
				value = "{UNKNOWN_VALUE}";
			}
			String valueReplacement = value.replaceAll("\\\\","\\\\\\\\").replaceAll("\\$", "\\\\\\$");
			m.appendReplacement(rtn, valueReplacement);
		}
		m.appendTail(rtn);
		return rtn.toString();
	}

我们重点关注

String valueReplacement = value.replaceAll("\\\\","\\\\\\\\").replaceAll("\\$", "\\\\\\$");

 其实就是将一个 \ 变成 \\ ，一个 $ 变成 \$

可能会有人问，replaceAll的第一个参数是正则表达式，所以需要4个\，为什么第二个参数也需要那么多个\?这也是我开始写代码是遇到的疑惑，实际上，replaceAll最终也是调用我们刚分析过的函数

 public Matcher appendReplacement(StringBuffer sb, String replacement)

replaceAll的第二个参数将会传到appendReplacement的replacement，所以

    replaceAll的第二个参数中的\和$也属于特殊字符串!

这也解释了，我们将文件路径分隔符/替换为\的时候，为什么需要那么多\\\\了。

String s = "E:/mydir/mydir2/mdir3";
//System.out.println(s.replaceAll("/","\\"));报错
System.out.println(s.replaceAll("/","\\\\"));//正确

使用replaceAll的时候，我们需要记住的是，不论是第一个参数还是第二参数，4个\才代表自然字符串中的一个\。在这里我也期待java能推出自然字符串的语法，像python的r"\n"一样。

 

评论

2 楼 lazy_ 2012-09-05  

urfriend 写道

java.util.regex已经自带了解决这个问题的函数

String java.util.regex.Matcher.quoteReplacement(String s)

Returns a literal replacement String for the specified String. This method produces a String that will work as a literal replacement s in the appendReplacement method of the Matcher class. The String produced will match the sequence of characters in s treated as a literal sequence. Slashes ('\') and dollar signs ('$') will be given no special meaning.

谢谢您宝贵的回复！

看了下源代码，跟我的想法一致，就是在$或\前加上一个\

 
public static String quoteReplacement(String s) {
        if ((s.indexOf('\\') == -1) && (s.indexOf('$') == -1))
            return s;
        StringBuffer sb = new StringBuffer();
        for (int i=0; i<s.length(); i++) {
            char c = s.charAt(i);
            if (c == '\\') {
                sb.append('\\'); sb.append('\\');
            } else if (c == '$') {
                sb.append('\\'); sb.append('$');
            } else {
                sb.append(c);
            }
        }
        return sb.toString();
}

1 楼 urfriend 2012-09-04  

java.util.regex已经自带了解决这个问题的函数

String java.util.regex.Matcher.quoteReplacement(String s)

Returns a literal replacement String for the specified String. This method produces a String that will work as a literal replacement s in the appendReplacement method of the Matcher class. The String produced will match the sequence of characters in s treated as a literal sequence. Slashes ('\') and dollar signs ('$') will be given no special meaning.