Regex match till the end of text in Java

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I want to fetch all the email addresses of From field using regex like get all lines of text that starts with "From:" and end with "/n" new line.

Here is the complete text on which I want to apply this regex,

Sent: Tue Mar 05 15:42:11 IST 2019
From: xtest@xyz.co.in
To: akm@xyz.com
Subject: Re: Foausrnisfseur invadlide (030000000000:3143)
Message: 


----------------------------

Sent: Tue Mar 05 15:40:51 IST 2019
From: ytest@xyz.com
To: bpcla@xpanxion.com
Subject: Foausrnisfseur invadlide (O4562000888885456:3143)
Message:
This is not right please correct
Termes de paiement Foausrnisfseur non spécifiés
impact potentiel: 3 000,00
You should write From field with abc@xyz.com
and not From: field with abc@xyz.com in the column
Date détecté: 2019-02-26 12:55:03


---- Please do not delete or modify this line. (2423000000000149:3143) ----

-------------------------
Sent: Tue Mar 05 15:40:51 IST 2019
From: ytest@xyz.co.in
To: bpcla@xpanxion.com
Subject: Foausrnisfseur invadlide (O4562000888885456:3143)

I have tried following patterns but it did not work,

[^.?!]*(?<=[.?\s!])string(?:(?=[\s.?!])[^.?!]*(?:[.?!].*)?)?$    
/^([\w\s\.]*)string([\w\s\.]*)$/    
"^\\w*\\s*((?m)Name.*$)"

The desired result expected from above text is :

xtest@xyz.co.in, ytest@xyz.com, ytest@xyz.co.in,

PS. I want regex for Java logic

Try this pattern: ^From:\s*(\S+)$

It first matches beginning of a line with ^, then matches From: literally, then matches 0 or more whitespaces with \s*, then matches one or more non-whitespeaces and stores it in capturing group, $ matches end of a line.

To get e-mail address, just use value of first capturing group.

Demo

Java regex word boundary, 3. Java regex word boundary – Match word at the end of content. The anchors "\Z" and "\z" always match at the very end of the content, after the last character. Place "\Z" or "\z" at the end of your regular expression to test whether the content ends with the text you want to match. In regex, anchors are not used to match characters. Rather they match a position i.e. before, after, or between characters. To match start and end of line, we use following anchors: Caret (^) matches the position before the first character in the string.

  String test = "   Sent: Tue Mar 05 15:42:11 IST 2019  "
            + "   From: xtest@xyz.co.in  "
            + "   To: akm@xyz.com  "
            + "   Subject: Re: Foausrnisfseur invadlide (030000000000:3143)  "
            + "   Message:   "
            + "     "
            + "     "
            + "   ----------------------------  "
            + "     "
            + "   Sent: Tue Mar 05 15:40:51 IST 2019  "
            + "   From: ytest@xyz.com  "
            + "   To: bpcla@xpanxion.com  "
            + "   Subject: Foausrnisfseur invadlide (O4562000888885456:3143)  "
            + "   Message:  "
            + "   This is not right please correct  "
            + "   Termes de paiement Foausrnisfseur non spécifiés  "
            + "   impact potentiel: 3 000,00  "
            + "   You should write From field with abc@xyz.com  "
            + "   and not From: field with abc@xyz.com in the column  "
            + "   Date détecté: 2019-02-26 12:55:03  "
            + "     "
            + "     "
            + "   ---- Please do not delete or modify this line. (2423000000000149:3143) ----  "
            + "     " + "   -------------------------  "
            + "   Sent: Tue Mar 05 15:40:51 IST 2019  " + "   From: ytest@xyz.co.in  "
            + "   To: bpcla@xpanxion.com  "
            + "  Subject: Foausrnisfseur invadlide (O4562000888885456:3143)  ";

      String emailRegex = "[a-zA-Z0-9._%+-]+@[A-Za-z0-9.-]+\\.[a-zA-Z]{2,6}";
    Pattern pattern = Pattern.compile("From\\:\\s(" + emailRegex + ")");// From\\:\\s same as Form : and () here i added Email Id  regex or you also change to (.*\n) but not recommended
    Matcher match = pattern.matcher(test);
    while (match.find()) {
        System.out.println(match.group(1));
    }

output :

 xtest@xyz.co.in
ytest@xyz.com
ytest@xyz.co.in

Regex - Match Start or End of String (Line Anchors), Regex – Match Start or End of String (Line Anchors) Drop me your questions related to programs for regex starts with and ends with java. Match: First Start: 0 End: 5 Match: Capital Start: 8 End: 15 Match: Words Start: 16 End: 21 Match: I Start: 37 End: 38 more. Here we can see that each match contains only the words we're expecting. The start property shows the zero-based index of the match within the string. The end shows the

Use this regular expression for your case:

From:\s+([\w-]+@([\w-]+\.)+[\w-]+)

I have tried this regular expression with https://www.freeformatter.com/java-regex-tester.html#ad-output and it is matching what you require.

Your required match is in capture Group 1.

Working Demo: https://regex101.com/r/dGaPbD/4

Regex Tutorial, In a regular expression, the caret matches the concept “start of string”, while the Because “start of string” must be matched before the match of \d+, and “end of string” must Delphi, Java, and the JGsoft flavor treat CRLF as an indivisible pair . The regex will stop at the first position where it encounters an invalid sequence, so in my example code below, I make sure that the position of the last match is at the end of the string. Full example program (Java 7, but you can remove the named capturing group to make it run in previous versions of Java):

    String emailRegex = "[^\\s]+"; // Replace with a better one
    Matcher m = Pattern.compile("(?m)^From:\\s*(" + emailRegex + ")\\s*$").matcher(yourString);

    List<String> allMatches = new ArrayList<String>();
    while(m.find())
      System.out.println(m.group(1));

Regex Tutorial - \b Word Boundaries, In regular expressions, \b anchors the regex at a word boundary or the position between a word Java supports Unicode for \b but not for \w. Again, the \b fails to match and continues to do so until the second space is reached. It also matches at the end of the string if the last character in the string is a word character. How to create the javascript regular expression for number with some special symbols javascript,regex what can be the java-script regular expression which gives the numbers with some symbols For example following condition must be pass. Number can start with $ Can have the . or , : symbols between and % sign at the send.

Boundary Matchers (The Java™ Tutorials > Essential Classes , Until now, we've only been interested in whether or not a match is found at some the match is taking place on a word boundary, or at the end of the previous match. Enter your regex: ^dog$ Enter input string to search: dog No match found. The appropriate regex would be the ' char followed by any number of any chars [including zero chars] ending with an end of string/line token: '.*$ And if you wanted to capture everything after the ' char but not include it in the output, you would use: (?<=').*$ This basically says give me all characters that follow the ' char until the end of the line.

Special pattern matching character operators, Match any character (except newline) $ Match the end of the line (or before newline the ^ character is guaranteed to match at only the beginning of the string, the case modification (think vi) \Q quote regular expression metacharacters till \E. i need to split the above paragraph as blocks of text starting with Name and ending with !!! . Here I dont want to use !!! as the only delimiter to split the paragraph.I need to include the starting sequence (Name) also in my regex. ie., my result api should looks like SplitAsBlocks("Paragraph","startswith Name","endswith !!!")

Using Regex for Text Manipulation in Python, The above regex expression will match the text string, since we are trying to match a string of any length and any character. To check whether a string ends with a specific word or not, we can use the Till now we have been using regex to find if a pattern exists in a string. Guide to Java Streams: forEach() with Examples� (?<!a) is a negated lookbehind assertion that ensures, that before the end of the string (or row with m modifier), there is not the character "a". See it here on Regexr You can also easily extend this with other characters, since this checking for the string and isn't a character class.

Comments
  • Your regex for email address is poor, Please, have a look at these sites: TLD list; valid/invalid addresses; regex for RFC822 email address
  • @Toto changing to valid one Ip Address Regex
  • I dont want to check email address validity. Please dont misjudge the question, I havent mentioned anywhere that emailId is to be validated. Its just "From: to /n" thats it !!
  • Ahmedabad and then you replace to (.*\n)
  • The regex is too poor.
  • Can you please elaborate?
  • Your solution seems correct, Ill mark it once Ill run all my test cases on it
  • @shreymathuria If my answer solved your problem, click the big checkbox to accept it as the answer.