Question

I am having problem with android regex pattern.

My target is to find {#} inside one sentence and replaces them with an array of String.

For example, my raw string will be

{#} students go to {#}.

My replacement array will be

[“5″,”park”]

After replacing, final string should be “5 students go to park.” My current snippet work just fine until the replacement array comes up with $ sign such as

[“$5″,”$park”]

Here is my current snippet.

 Pattern pattern = Pattern.compile("(\\{#\\})");
 String raw_q = "Pete has {#}. He bought {#}.";
 String[] var = ["$70","6 pens at $5 each"];
 String act_q = "";
 for (int m = 0; m < var.length(); m++) {
    try{
       act_q = raw_q.replaceFirst(pattern.pattern(),var[m]);
    }catch(Exception e){
       e.printStackTrace();
    }

Result is a blank string which I believe try block never get executed. Stack trace shows arrayindexoutofbound exception. I don’t really get what it’s happening. Any help would be appreciated.

Note – the length of replacement array is always equal to the number of {#} occurrences.

Answer 1

For your problem, see the javadoc. String.replaceFirst() use Matcher.replaceFirst(), which refers to Matcher.appendReplacement():

Note that backslashes () and dollar signs ($) in the replacement
string may cause the results to be different than if it were being
treated as a literal replacement string. Dollar signs may be treated
as references to captured subsequences as described above, and
backslashes are used to escape literal characters in the replacement
string.

That is, you must escape all $ and \.

---

But you have other problem…. You have compile a pattern and NOT use it! replaceFirst() always compile the pattern again. Try this for better performance.

Matcher m = pattern.matcher(raw_q);
StringBuffer sb = new StringBuffer();
int i = 0;
while (m.find()) {
     m.appendReplacement(sb, var[i].replaceAll("\\$", "\\\\$"));
     i++;
}
m.appendTail(sb);
act_q = sb.toString();