Develop Reference

Regex with error checking

I've done a bunch of searching but I'm terrible with regex statements and my google-fu in this instance as not been strong.
Scenario:
In push notifications, we're passed a URL that contains a 9-digit content ID.
Example URL: http://www.something.com/foo/bar/Some-title-Goes-here-123456789.html (123456789 is the content ID in this scenario)
Current regex to parse the content ID:
public String getContentIdFromPathAndQueryString(String path, String queryString) {
String contentId = null;
if (StringUtils.isNonEmpty(path)) {
Pattern p = Pattern.compile("([\\d]{9})(?=.html)");
Matcher m = p.matcher(path);
if (m.find()) {
contentId = m.group();
} else if (StringUtils.isNonEmpty(queryString)) {
p = Pattern.compile("(?:contentId=)([\\d]{9})(?=.html)");
m = p.matcher(queryString);
if (m.find()) {
contentId = m.group();
}
}
}
Log.d(LOG_TAG, "Content id " + (contentId == null ? "not found" : (" found - " + contentId)));
if (StringUtils.isEmpty(contentId)) {
Answers.getInstance().logCustom(new CustomEvent("eid_url")
.putCustomAttribute("contentId", "empty")
.putCustomAttribute("path", path)
.putCustomAttribute("query", queryString));
}
return contentId;
}
The problem:
This does the job but there's a specific error scenario that I need to account for.
Whoever creates the push may put in the wrong length content ID and we need to grab it regardless of that, so assume it can be any number of digits... the title can also contain digits, which is annoying. The content ID will ALWAYS be followed by ".html"

While the basic answer here would be just "replace {9} limiting quantifier matching exactly 9 occurrences with a + quantifier matching 1+ occurrences", there are two patterns that can be improved.
The unescaped dot should be escaped in the pattern to match a literal dot.
If you have no overlapping matches, no need to use a positive lookahead with a capturing group before it, just keep the capturing group and grab .group(1) value.
A non-capturing group (?:...) is still a consuming pattern, and the (?:contentId=) equals contentId= (you may remove (?: and )).
There is no need wrapping a single atom within a character class, use \\d instead of [\\d]. That [\\d] is actually a source of misunderstandings, some may think it is a grouping construct, and might try adding alternative sequences into the square brackets, while [...] matches a single char.
So, your code can look like
Pattern p = Pattern.compile("(\\d+)\\.html"); // No lookahead, + instead of {9}
Matcher m = p.matcher(path);
if (m.find()) {
contentId = m.group(1); // (1) refers to Group 1
} else if (StringUtils.isNonEmpty(queryString)) {
p = Pattern.compile("contentId=(\\d+)\\.html");
m = p.matcher(queryString);
if (m.find()) {
contentId = m.group(1);
}
}

extract String between last slash and question mark

I want to extract string between last slash and question mark using regex
String path="http://keting.amazonaws.com/media123/mediaattachments/000/000/004/original/Shoes_TVC_2013.mp4?1470248308"
I need to get "Shoes_TVC_2013.mp4" from the String.
Pattern pattern = Pattern.compile("NEED HELP HERE");
Matcher matcher = pattern.matcher(URL);
if (matcher.find()) {
System.out.println(matcher.group(1)); //prints region/country
} else {
System.out.println("Match not found");
}

.*\/(.*)\? should do the trick. See here for more.

Try this solution,
String path="http://keting.amazonaws.com/media123/mediaattachments/000/000/004/original/Shoes_TVC_2013.mp4?1470248308";
path = path.substring(path.lastIndexOf("/")+1,path.lastIndexOf("?"));

regex skips the first match

The regex
.*([0-9]{3}\\.[0-9]{2}).*
finds one match in "some short sentence 111.01 ", but it failed to match the first occurrence "111.01" in "some short sentence 111.01 & 222.02 "
I tried the lazy quantifier .*([0-9]{3}\\.[0-9]{2})?.* or .*([0-9]{3}\\.[0-9]{2}).*? for no avail.
Please help, I need to get both occurrences, here is my code.
Thank you
Pattern myPattern = Pattern.compile(".*([0-9]{3}\\.[0-9]{2}).*");
Matcher m = myPattern.matcher(mystring);
while (m.find()) {
String found = m.group(1);
}

you need to remove ".*"s. Try this:
String mystring = "some short sentence 111.01 & 222.02 ";
Pattern myPattern = Pattern.compile("([0-9]{3}\\.[0-9]{2})");
Matcher m = myPattern.matcher(mystring);
while(m.find()) {
System.out.println("Found value: " + m.group(1) );
}
output:
Found value: 111.01
Found value: 222.02

The leading and trailing ".*" cause you to match the entire string in one match. All the lazy quantifier does in your case is controls you getting the first, not last, occurrence in the subject.

how to get text with using substring

I try to get only this part "9916-4203" in "Region Code:9916-4203 " in android. How can I do this?
I tried below code, I used substring method but it doesn't work:
firstNumber = Integer.parseInt(message.substring(11, 19));

If you know that string contains "Region Code:" couldn't you do a replace?
message = message.replace("Region Code:", "");

Assumed that you have only one phone number in your String, the following will remove any non-digit characters and parse the resulting number:
public static int getNumber(String num){
String tmp = "";
for(int i=0;i<num.length();i++){
if(Character.isDigit(num.charAt(i)))
tmp += num.charAt(i);
}
return Integer.parseInt(tmp);
}
Output in your case: 99164203
And as already mentioned, you won't be able to parse any String to Integer in case there are any non-digit characters

Im going to guess that what you want to extract is the full region code text minus the title. So maybe using regex would be a good simple fit for you?
String myString = "Region Code:9916-4203";
String match = "";
String pattern = "\:(.*)";
Pattern regEx = Pattern.compile(pattern);
Matcher m = regEx.matcher(myString);
// Find instance of pattern matches
Matcher m = regEx.matcher(myString);
if (m.find()) {
match = m.group(0);
}
Variable match will contain "9916-4203"
This should work for you.
Java code sourced from http://android-elements.blogspot.in/2011/04/regular-expressions-in-android.html

In Java the substring() method works with the first parameter being inclusive and the second parameter being exclusive. Meaning "Hello".substring(0, 2); will result in the string He.
In addition to excluding the parsing of something that isn't a number like #Opiatefuchs mentioned, your substring method should instead be message.substring(12, 21).

Android and regex to read substring

I'm trying to use regex in an Android application to read a substring. I'm using Pattern and Matcher, but I can't figure out how to do it.
My input string is: javascript:submitForm(document.voip_call_log,30,0,'xxxxx','',0,'');
where xxxxx is a variable number of digits.
How can I read 'xxxxx' using Pattern and Matcher?

I'm not an expert in RegEx, but this one should work for you:
String input = "javascript:submitForm(document.voip_call_log,30,0,'5555','',0,'');";
Pattern pattern = Pattern.compile(",'(\\d*)',");
Matcher matcher = pattern.matcher(input);
matcher.find();
String out = matcher.group(1);
System.out.println(out);
Proof http://ideone.com/WI6VB1

If you will have always that format, it might be easier to use the split(",") method and access the value you like.
Alternatively, you could try something like so:
String str = ...;
Pattern p = Pattern.compile("(\\d{3,})");
Matcher m = p.matcher(str);
while(m.find())
{
System.out.println(m.group(1));
}

Try this one:
String s = "javascript:submitForm(document.voip_call_log,30,0,'1999','',0,'');";
Pattern pattern = Pattern.compile("javascript:submitForm\\([^,]+,\\d*,\\d*,'(\\d+)','\\d*'");
Matcher m = pattern.matcher(s);
if(m.find( )) {
System.out.println(m.group(1));
}
This is following the pattern of your input string from the question description. You can find the explanation of your regex from this link. Just input your regex with single slash \ instead of double \\