2

我有一个服务方法，它接受一个字符串，然后用标签库中的项目替换字符串中的标签。如下：

for( MetaDataDTO tag : tagValues )
{
    message = message.replace( tag.getKey(), tag.getText1() );
}

明显地; 这会产生大量新字符串并且很糟糕。但是 StringBuilder 替换方法对于一个字符串中的多个字符串使用起来很麻烦。我怎样才能使我的方法更有效？

它用于文本块，例如：

亲爱的#firstName#，您的#applicationType# 申请已被#approvedRejected# 抱歉。

其中#firstName#等是元数据数据库中的键。标签也可能没有被散列字符包围。

4

2 回答 2

10

基本上你想像这样复制Matcher.replaceAll()的执行：

public static String replaceTags(String message, Map<String, String> tags) {
  Pattern p = Pattern.compile("#(\\w+)#");
  Matcher m = p.matcher(message);
  boolean result = m.find();
  if (result) {
    StringBuffer sb = new StringBuffer();
    do {
      m.appendReplacement(sb, tags.containsKey(m.group(1) ? tags.get(m.group(1)) : "");
      result = m.find();
    } while (result);
    m.appendTail(sb);
    message = sb.toString();
  }
  return message;
}

Note: I've made an assumption about the valid tag (namely \w in the regex). You will need to cater this for what's really valid (eg "#([\w_]+)#").

I've also assumed the tags above looks something like:

Map<String, String> tags = new HashMap<String, String>();
tags.add("firstName", "Skippy");

and not:

tags.add("#firstName#", "Skippy");

If the second is correct you'll need to adjust accordingly.

This method makes exactly one pass across the message string so it doesn't get much more efficient than this.

于 2009-05-03T10:25:44.040 回答
0

Thanks for your help guys. Certainly learned more about java. Here is my solution. It is this way to support different looking tags and tags within tags:

private static String replaceAllTags(String message, Map< String, String > tags)
{
    StringBuilder sb = new StringBuilder( message );
    boolean tagFound = false;
    /**
     * prevent endless circular text replacement loops
     */
    long recurrancyChecker = 5000;

    do
    {
        tagFound = false;
        Iterator it = tags.entrySet().iterator();
        while( it.hasNext() )
        {
            Map.Entry pairs = (Map.Entry) it.next();

            int start = sb.indexOf( pairs.getKey().toString() );

            while( start > -1 && --recurrancyChecker > 0 )
            {
                int length = pairs.getKey().toString().length();
                sb.replace( start, start + length, pairs.getValue().toString() );
                start = sb.indexOf( pairs.getKey().toString() );
                tagFound = true;
            }
        }
    }
    while( tagFound && --recurrancyChecker > 0 );
    return sb.toString();
}
于 2009-05-06T02:38:42.030 回答