I have the following string:
String p = "{requestId=146bb94xxxxxR, value=false, tier=S3,ReceivedTime=0}";
I want to extract the request Id value, so want my output to be
146bb94xxxxxR
Here is what I tried
Pattern MY_PATTERN = Pattern.compile("\\requestId=(.*?)\\,");
Matcher m = MY_PATTERN.matcher(p);
while (m.find()) {
String s = m.group(1); // s now contains "BAR"
}
However i get no output, there is a problem with the regex, but I am not sure how to correct it.
Pattern.compile("requestId=(.*?),");
You should delete redundant characters escape (\\, and \\r (it means the carriage-return character)) and it will work fine.
Your regex escapes the r in request and the , which makes no sence. Try
\\brequestId=([^,]*),
which will match a word boundary - \b, then requestId= and then capturing the ID (everything up to the ,)
You can also try this:
(?<=requestId=)(\\w*)
[] around \w? (?<=requestId=)(\\w*) will do it (including the java escape).public static void main(String[] args) {
String text = "{requestId=146bb94xxxxxR, value=false, tier=S3,ReceivedTime=0}";
Pattern p = Pattern.compile("requestId=(.*?),");
Matcher m = p.matcher(text);
m.find();
System.out.println(m.group(1));
}
Output:
146bb94xxxxxR
You don't need to surround your regex with \ (written as string "\\") like in example mentioned by you earlier in comment: Using Java to find substring of a bigger string using Regular Expression. There \ was used to escape few of regex special characters which ware [ and ].
So your regex Pattern.compile("\\requestId=(.*?)\\,") \\r part will represent \r which is carriage return (one of line separators). So remove that \\ before r. Also you don't need to escape , with \ since it is not regex special character (at least not in this context). So try with:
Pattern MY_PATTERN = Pattern.compile("requestId=(.*?),");
Matcher m = MY_PATTERN.matcher(" {requestId=146bb94xxxxxR, value=false, tier=S3,ReceivedTime=0}");
while (m.find()) {
String s = m.group(1);
System.out.println(s);
}
\ is also special not only in String literal, but also in regex syntax. So writing regex as string "\\r" will represent \r which in Java's regex engine is interpreted as one of line separators (carriage return), not \ character followed by r.
\\brequestId=([^,]*),.indexOf()you could find the indexes of the=and,. With those indexes you can easily usesubstring()to get the desired output.