How can I successfully parse the text below in that format to parse just
To: User <test#test.com>
and
To: <test#test.com>
When I try to parse the text below with
/To:.*<[A-Z0-9._+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}>/mi
It grabs
Message-ID <CC2E81A5.6B9%test#test.com>,
which I dont want in my answer.
I have tried using $ and \z and neither work. What am I doing wrong?
Information to parse
To: User <test#test.com> Message-ID <CC2E81A5.6B9%test#test.com>
To:
<test#test.com>
This is my parsing information in Rubular http://rubular.com/r/DQMQC4TQLV
Since you haven't specified exactly what your tool/language is, assumptions must be made.
In general regex pattern matching tends to be aggressive, matching the longest possible pattern. Your pattern starts off with .*, which means that you're going to match the longest possible string that ENDS WITH the remainder of your pattern <[A-Z0-9._+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}>, which was matched with <CC2E81A5.6B9%test#test.com> from the Message-ID.
Both Apalala's and nhahtdh's comments give you something to try. Avoid the all-inclusive .* at the start and use something that's a bit more specific: match leading spaces, or match anything EXCEPT the first part of what you're really interested in.
You need to make the wildcard match non greedy by adding a question mark after it:
To:.*?<[A-Z0-9._+-]+#[A-Z0-9.-]+\.[A-Z]{2,4}>
Related
I have a basic regex for checking if string has at least one digit -
if (!("^(?=.*[0-9])".toRegex().matches(registerInput.input))) {
// Warn user
}
However, even with an input 123, it fails and goes into // Warn user section.
I used online validation tools and it seems nothing is wrong with my regex. Any help please?
I'm not familiar with kotlin, but I looked up kotlin regex to see whether you really needed to anchor your regex to the start of the string.
The description for the match method, is:
Indicates whether the regular expression matches the entire input.
Which isn't going to be true for your example. It looks like containsMatchIn is a better option:
if (!("[0-9]".toRegex().containsMatchIn(registerInput.input))) {
// Warn user
}
If you're just searching for a match, you don't need to anchor your regex to the start of a string, or use fancy look-ahead groups or anything like that.
Following SpoonMeiser's answer, the positive lookahead test will match if you add what is need to get the entire input matched:
if (!("^(?=.*[0-9]).*$".toRegex().matches("123")))
Because ^(?=.*[0-9]) matches only the beginning of the line (looked-ahead characters aren't taken, or "consumed").
So to match the whole line, just add .*$ after.
Of course, this is overkill in this context, SpoonMeiser'answer is way much better.
I am trying to regex the following string:
https://www.amazon.com/Tapps-Top-Apps-and-Games/dp/B00VU2BZRO/ref=sr_1_3?ie=UTF8&qid=1527813329&sr=8-3&keywords=poop
I want only B00VU2BZRO.
This substring is always going to be a 10 characters, alphanumeric, preceded by dp/.
So far I have the following regex:
[d][p][\/][0-9B][0-9A-Z]{9}
This matches dp/B00VU2BZRO
I want to match only B00VU2BZRO with no dp/
How do I regex this?
Here is one regex option which would produce an exact match of what you want:
(?<=dp\/)(.*)(?=\/)
Demo
Note that this solution makes no assumptions about the length of the path fragment occurring after dp/. If you want to match a certain number of characters, replace (.*) with (.{10}), for example.
Depending on your language/method of application, you have a couple of options.
Positive look behind. This will make your regex more complicated, but will make it match what you want exactly:
(<=dp/)[0-9A-Z]{10}
The construct (<=...) is called a positive look behind. It will not consume any of the string, but will only allow the match to happen if the pattern between the parens is matched.
Capture group. This will make the regex itself slightly simpler, but will add a step to the extraction process:
dp/([0-9A-Z]{10})
Anything between plain parens is a capture group. The entire pattern will be matched, including dp/, but most languages will give you a way of extracting the portion you are interested in.
Depending on your language, you may need to escape the forward slash (/).
As an aside, you never need to create a character class for single characters: [d][p][\/] can equally well be written as just dp\/.
Quite a simple one in theory but can't quite get it!
I want a regex in ant which matches anything as long as it has a slash on the end.
Below is what I expect to work
<regexp id="slash.end.pattern" pattern="*/"/>
However this throws back
java.util.regex.PatternSyntaxException: Dangling meta character '*' near index 0
*/
^
I have also tried escaping this to \*, but that matches a literal *.
Any help appreciated!
Your original regex pattern didn't work because * is a special character in regex that is only used to quantify other characters.
The pattern (.)*/$, which you mentioned in your comment, will match any string of characters not containing newlines, however it uses a possibly unnecessary capturing group. .*/$ should work just as well.
If you need to match newline characters, the dot . won't be enough. You could try something like [\s\S]*/$
On that note, it should be mentioned that you might not want to use $ in this pattern. Suppose you have the following string:
abc/def/
Should this be evaluated as two matches, abc/ and def/? Or is it a single match containing the whole thing? Your current approach creates a single match. If instead you would like to search for strings of characters and then stop the match as soon as a / is found, you could use something like this: [\s\S]*?/.
I am trying to form a regular expression that will match strings that do NOT end a with a DOT FOLLOWED BY NUMBER.
eg.
abcd1
abcdf12
abcdf124
abcd1.0
abcd1.134
abcdf12.13
abcdf124.2
abcdf124.21
I want to match first three.
I tried modifying this post but it didn't work for me as the number may have variable length.
Can someone help?
You can use something like this:
^((?!\.[\d]+)[\w.])+$
It anchors at the start and end of a line. It basically says:
Anchor at the start of the line
DO NOT match the pattern .NUMBERS
Take every letter, digit, etc, unless we hit the pattern above
Anchor at the end of the line
So, this pattern matches this (no dot then number):
This.Is.Your.Pattern or This.Is.Your.Pattern2012
However it won't match this (dot before the number):
This.Is.Your.Pattern.2012
EDIT: In response to Wiseguy's comment, you can use this:
^((?!\.[\d]+$)[\w.])+$ - which provides an anchor after the number. Therefore, it must be a dot, then only a number at the end... not that you specified that in your question..
If you can relax your restrictions a bit, you may try using this (extended) regular expression:
^[^.]*.?[^0-9]*$
You may omit anchoring metasymbols ^ and $ if you're using function/tool that matches against whole string.
Explanation: This regex allows any symbols except dot until (optional) dot is found, after which all non-numerical symbols are allowed. It won't work for numbers in improper format, like in string: abcd1...3 or abcd1.fdfd2. It also won't work correctly for some string with multiple dots, like abcd.ab123cd.a (the problem description is a bit ambigous).
Philosophical explanation: When using regular expressions, often you don't need to do exactly what your task seems to be, etc. So even simple regex will do the job. An abstract example: you have a file with lines are either numbers, or some complicated names(without digits), and say, you want to filter out all numbers, then simple filtering by [^0-9] - grep '^[0-9]' will do the job.
But if your task is more complex and requires validation of format and doing other fancy stuff on data, why not use a simple script(say, in awk, python, perl or other language)? Or a short "hand-written" function, if you're implementing stand-alone application. Regexes are cool, but they are often not the right tool to use.
I would just use a simple negative look-behind anchored at the end:
.*(?<!\\.\\d+)$
Let's say I have a regular expression that works correctly to find all of the URLs in a text file:
(http://)([a-zA-Z0-9\/\.])*
If what I want is not the URLs but the inverse - all other text except the URLs - is there an easy modification to make to get this?
You could simply search and replace everything that matches the regular expression with an empty string, e.g. in Perl s/(http:\/\/)([a-zA-Z0-9\/\.])*//g
This would give you everything in the original text, except those substrings that match the regular expression.
If for some reason you need a regex-only solution, try this:
((?<=http://[a-zA-Z0-9\/\.#?/%]+(?=[^a-zA-Z0-9\/\.#?/%]))|\A(?!http://[a-zA-Z0-9\/\.#?/%])).+?((?=http://[a-zA-Z0-9\/\.#?/%])|\Z)
I expanded the set of of URL characters a little ([a-zA-Z0-9\/\.#?/%]) to include a few important ones, but this is by no means meant to be exact or exhaustive.
The regex is a bit of a monster, so I'll try to break it down:
(?<=http://[a-zA-Z0-9\/\.#?/%]+(?=[^a-zA-Z0-9\/\.#?/%])
The first potion matches the end of a URL. http://[a-zA-Z0-9\/\.#?/%]+ matches the URL itself, while (?=[^a-zA-Z0-9\/\.#?/%]) asserts that the URL must be followed by a non-URL character so that we are sure we are at the end. A lookahead is used so that the non-URL character is sought but not captured. The whole thing is wrapped in a lookbehind (?<=...) to look for it as the boundary of the match, again without capturing that portion.
We also want to match a non-URL at the beginning of the file. \A(?!http://[a-zA-Z0-9\/\.#?/%]) matches the beginning of the file (\A), followed by a negative lookahead to make sure there's not a URL lurking at the start of the file. (This URL check is simpler than the first one because we only need the beginning of the URL, not the whole thing.)
Both of those checks are put in parenthesis and OR'd together with the | character. After that, .+? matches the string we are trying to capture.
Then we come to ((?=http://[a-zA-Z0-9\/\.#?/%])|\Z). Here, we check for the beginning of a URL, once again with (?=http://[a-zA-Z0-9\/\.#?/%]). The end of the file is also a pretty good sign that we've reached the end of our match, so we should look for that, too, using \Z. Similarly to a first big group, we wrap it in parenthesis and OR the two possibilities together.
The | symbol requires the parenthesis because its precedence is very low, so you have to explicitly state the boundaries of the OR.
This regex relies heavily on zero-width assertions (the \A and \Z anchors, and the lookaround groups). You should always understand a regex before you use it for anything serious or permanent (otherwise you might catch a case of perl), so you might want to check out Start of String and End of String Anchors and Lookahead and Lookbehind Zero-Width Assertions.
Corrections welcome, of course!
If I understand the question correctly, you can use search/replace...just wildcard around your expression and then substitute the first and last parts.
s/^(.*)(your regex here)(.*)$/$1$3/
im not sure if this will work exactly as you intend but it might help:
Whatever you place in the brackets [] will be matched against. If you put ^ within the bracket, i.e [^a-zA-Z0-9/.] it will match everything except what is in the brackets.
http://www.regular-expressions.info/