诡谲的appendReplacement和replaceAll

诡异的appendReplacement和replaceAll

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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

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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

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但如果是

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

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则会抛出异常

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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。

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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();}

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