Develop Reference

Regular expression for hgsv notation in java

HGSV nomenclature has a pattern:
xxxxx.yyyy:charactersnumbercharacters
I would like to make a regex in java and fetch the all the tokens from above eg:
it should have 5 tokens :
{ 'xxxxx', 'yyyy', 'characters', 'number' , 'characters'}
I have used simple split methodology to fetch the tokens, but I don't find its an optimal solution:
my current code is :
String hgsv = "BRAF.p:V600E";
String[] tokens = hgsv.split(".");
this.symbol = tokens[0];
String type = tokens[1].split(":")[0];
I would like to use Pattern and Matcher in Java. No idea, how to make regex for the above token.
Any clue how to do that?
(even to separate characters, numbers, characters I will be using regex). So why not to use REGEX for entire token.
I found link but this is in Python, I need similar in Java.

I think what you're probably looking for is to use capture groups, like this:
String s = "BRAF.p:V600E";
Pattern p = Pattern.compile("(\\w+)\\.(\\w+):([a-zA-Z]+)(\\d+)([a-zA-Z]+)");
Matcher m = p.matcher(s);
if (m.matches()) {
String[] parts = {m.group(1),
m.group(2),
m.group(3),
m.group(4),
m.group(5)};
// Prints "[BRAF, p, V, 600, E]"
System.out.println(Arrays.toString(parts));
} else {
// The input String is invalid.
}
That's really just a lot like a split, but it's more stable because you're using the pattern to validate the String beforehand.
Note that I have no idea if that is the exact right pattern that you should be using. I don't know the exact details of the HGSV notation you're talking about and your description is actually pretty vague. (What are e.g. xxxxx and yyyy? What are "characters"?) If you link me to some sort of specification or detailed description of this notation I can try to write a regex that's more definitely correct.
Anyhow, my example shows the basic idea. You might also see http://www.regular-expressions.info/brackets.html for more information.

Java Regex with Pattern and Matcher

I am using Pattern and Matcher classes from Java ,
I am reading a Template text and I want to replace :
src="scripts/test.js" with src="scripts/test.js?Id=${Id}"
src="Servlet?Template=scripts/test.js" with src="Servlet?Id=${Id}&Template=scripts/test.js"
I'm using the below code to execute case 2. :
//strTemplateText is the Template's text
Pattern p2 = Pattern.compile("(?i)(src\\s*=\\s*[\"'])(.*?\\?)");
Matcher m2 = p2.matcher(strTemplateText);
strTemplateText = m2.replaceAll("$1$2Id=" + CurrentESSession.getAttributeString("Id", "") + "&");
The above code works correctly for case 2. but how can I create a regex to combine both cases 1. and 2. ?
Thank you

You don't need a regular expression. If you change case 2 to
replace Servlet?Template=scripts/test.js with Servlet?Template=scripts/test.js&Id=${Id}
all you need to do is to check whether the source string does contain a ? if not add ?Id=${Id} else add &Id=${Id}.
After all
if (strTemplateText.contains("?") {
strTemplateText += "&Id=${Id}";
}
else {
strTemplateText += "?Id=${Id}";
}
does the job.
Or even shorter
strTemplate += strTemplateText.contains("?") ? "&Id=${Id}" : "?Id=${Id}";

Your actual question doesn't match up so well with your example code. The example code seems to handle a more general case, and it substitutes an actual session Id value instead of a reference to one. The code below takes the example code to be more indicative of what you really want, but the same approach could be adapted to what you asked in the question text (using a simpler regex, even).
With that said, I don't see any way to do this with a single replaceAll() because the replacement text for the two cases is too different. You could nevertheless do it with one regex, in one pass, if you used a different approach:
Pattern p2 = Pattern.compile("(src\\s*=\\s*)(['\"])([^?]*?)(\\?.*?)?\\2",
Pattern.CASE_INSENSITIVE);
Matcher m2 = p2.matcher(strTemplateText);
StringBuffer revisedText = new StringBuffer();
while (m2.find()) {
// Append the whole match except the closing quote
m2.appendReplacement(revisedText, "$1$2$3$4");
// group 4 is the optional query string; null if none was matched
revisedText.append((m2.group(4) == null) ? '?' : '&');
revisedText.append("Id=");
revisedText.append(CurrentESSession.getAttributeString("Id", ""));
// append a copy of the opening quote
revisedText.append(m2.group(2));
}
m2.appendTail(revisedText);
strTemplateText = revisedText.toString();
That relies on BetaRide's observation that query parameter order is not significant, although the same general approach could accommodate a requirement to make Id the first query parameter, as in the question. It also matches the end of the src attribute in the pattern to the correct closing delimiter, which your pattern does not address (though it needs to do to avoid matching text that spans more than one src attribute).
Do note that nothing in the above prevents a duplicate query parameter 'Id' being added; this is consistent with the regex presented in the question. If you want to avoid that with the above approach then in the loop you need to parse the query string (when there is one) to determine whether an 'Id' parameter is already present.

You can do the following:
//strTemplateText is the Template's text
String strTemplateText = "src=\"scripts/test.js\"";
strTemplateText = "src=\"Servlet?Template=scripts/test.js\"";
java.util.regex.Pattern p2 = java.util.regex.Pattern.compile("(src\\s*=\\s*[\"'])(.*?)((?:[\\w\\s\\d.\\-\\#]+\\/?)+)(?:[?]?)(.*?\\=.*)*(['\"])");
java.util.regex.Matcher m2 = p2.matcher(strTemplateText);
System.out.println(m2.matches());
strTemplateText = m2.replaceAll("$1$2$3?Id=" + CurrentESSession.getAttributeString("Id", "") + (m2.group(4)==null? "":"&") + "$4$5");
System.out.println(strTemplateText);
It works on both cases.
If you are using java > 1.6; then, you could use custom-named group-capturing features for making the regex exp. more human-readable and easier to debug.

Java regex throwing exception for no match found when pattern found in line

I am dying trying to figure out why a regex won't match. Any help is much appreciated. I'm going line by line of a web page (that works fine), but I need to pull out the links for each line. The application will check to see if there is a link in the line, but I need to actually pull out the URL. help?
Pattern p = Pattern.compile("^.*href=\"([^\"]*)");
Matcher m = p.matcher(result);
String urlStr = m.group();
links.add(urlStr);
The error message I keep getting is this:
Exception in thread "main" java.lang.IllegalStateException: No match found
at java.util.regex.Matcher.group(Matcher.java:485)
Even though 'result' has a link reference (hxxp://www.yahoo.com) in it.
links is an ArrayList fyi. Thanks in advance!

first call
m.find();
or
m.matches();
and then you'll be able to use m.group() if matcher succeeded.

Ok, so I figured it out. The only issue was, my regex had to be
".*href=\"([^\"]*?)\".*"
Which then made the code
private String regex = ".*href=\"([^\"]*?)\".*";
private Pattern p = Pattern.compile(regex);
Matcher m = p.matcher(result);
if (m.matches()) {
String urlStr = m.group(1);
links.add(urlStr);
}
So that was my issue with the regex, I had to use the '?' to not be greedy I guess!

Regex string modifications

I have the following String and I want to filter the MBRB1045T4G out with a regular expression in Java. How would I achieve that?
String:
<p class="ref">
<b>Mfr Part#:</b>
MBRB1045T4G<br>
<b>Technologie:</b>
Tab Mount<br>
<b>Bauform:</b>
D2PAK-3<br>
<b>Verpackungsart:</b>
REEL<br>
<b>Standard Verpackungseinheit:</b>
800<br>

As Wrikken correctly says, HTML can't be parsed correctly by regex in the general case. However it seems you're looking at an actual website and want to scrape some contents. In that case, assuming space elements and formatting in the HTML code don't change, you can use a regex like this:
Mfr Part#:</b>([^<]+)<br>
And collect the first capture group like so (where string is your HTML):
Pattern pt = Pattern.compile("Mfr Part#:</b>\s+([^<]+)<br>",Pattern.MULTILINE);
Matcher m = pt.matcher(string);
if (m.matches())
System.out.println(m.group(1));

Negating a Regular Expression for string replacement

I have the following code that can replace the email address in a String in Java:
addressStr.replaceFirst("([a-zA-Z0-9_\\-\\.]+)#((\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.)|(([a-zA-Z0-9\\-]+\\.)+))([a-zA-Z]{2,4}|[0-9]{1,3})", "")
So, a string with John Smith <john#smith.com> would become John Smith <>. How do I negate it so that it will instead replace all that doesn't match the email address and have the final result as just john#smith.com?
I tried to put in the ^ and ?<= at the front but it doesn't work.

Well, it's not the regex you need to change but the calling code. Your regex matches the e-mail address (in a weird way), and the replace() removes it from the string.
So just use
Pattern regex = Pattern.compile("([a-zA-Z0-9_\\-\\.]+)#((\\[[0-9]{1,3}\\.[0-9]{1,3}\\.[0-9]{1,3}\\.)|(([a-zA-Z0-9\\-]+\\.)+))([a-zA-Z]{2,4}|[0-9]{1,3})");
Matcher regexMatcher = regex.matcher(addressStr);
if (regexMatcher.find()) {
address = regexMatcher.group();
}

The complete Java regex for catching e-mails would be as follows:
"(?:[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*|\"(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21\\x23-\\x5b\\x5d-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])*\")#(?:(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?|\\[(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|[a-z0-9-]*[a-z0-9]:(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21-\\x5a\\x53-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])+)\\])"
Take a look at https://www.rfc-editor.org/rfc/rfc2822#section-3.4.1 for more info on this.
A bit complicated but it is valid for all known and valid emails formats (yours do not allows mails like bob+bib#gmail.com which are valid).
For your problem, as stated multiple times, just find (stealing Tim Pietzcker piece of code):
Pattern regex = Pattern.compile("(?:[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*|\"(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21\\x23-\\x5b\\x5d-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])*\")#(?:(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?|\\[(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|[a-z0-9-]*[a-z0-9]:(?:[\\x01-\\x08\\x0b\\x0c\\x0e-\\x1f\\x21-\\x5a\\x53-\\x7f]|\\\\[\\x01-\\x09\\x0b\\x0c\\x0e-\\x7f])+)\\])");
Matcher regexMatcher = regex.matcher(addressStr);
foundMatch = regexMatcher.find();

You can try:
String mailId = Pattern.compile(regexp, Pattern.LITERAL).matcher(addressStr).group();
Idea here is to get the matched string rather than trying to replace everything else with blank. You can extract the pattern into a field if this operation is repetitive.

Just don't replace.... use match(es) instead.