I have a string like this:
Name: John Doe
Age: 23
Primary Language: English
Description: This is a multiline
description field that I want
to capture
Country: Canada
That's not the actual data, but you can see what I'm trying to do. I want to use regex to get an array of the "key" fields (Name, Age, Primary Language, Description, Country) and their values.
I'm using PHP.
My current attempt is this, but it doesn't work:
preg_match( '/^(.*?\:) (.*?)(\n.*?\:)/ism', $text, $matches );
Here's one solution: http://rubular.com/r/uDgXcIvhac.
\s*([^:]+?)\s*:\s*(.*(?:\s*(?!.*:).*)*)\s*
Note that I've used a negative lookahead assertion, (?!.*:). This is the only way you can check that the next line doesn't look like a new field, and at the same time continue where you left off. (This is why lookaheads and lookbehinds are known as zero-width assertions.)
EDIT: Removed bit about arbitrary-width lookaheads; I was mistaken. The above solution is fine.
Would PHP's strtok help you? You could use it with ":" as the delimeter/token and trim leading and trailing spaces to remove the unwanted new lines.
http://php.net/manual/en/function.strtok.php
Related
I want to split a string as per the parameters laid out in the title. I've tried a few different things including using preg_match with not much success so far and I feel like there may be a simpler solution that I haven't clocked on to.
I have a regex that matches the "price" mentioned in the title (see below).
/(?=.)\£(([1-9][0-9]{0,2}(,[0-9]{3})*)|[0-9]+)?(\.[0-9]{1,2})?/
And here are a few example scenarios and what my desired outcome would be:
Example 1:
input: "This string should not split as the only periods that appear are here £19.99 and also at the end."
output: n/a
Example 2:
input: "This string should split right here. As the period is not part of a price or at the end of the string."
output: "This string should split right here"
Example 3:
input: "There is a price in this string £19.99, but it should only split at this point. As I want it to ignore periods in a price"
output: "There is a price in this string £19.99, but it should only split at this point"
I suggest using
preg_split('~\£(?:[1-9]\d{0,2}(?:,\d{3})*|[0-9]+)?(?:\.\d{1,2})?(*SKIP)(*F)|\.(?!\s*$)~u', $string)
See the regex demo.
The pattern matches your pattern, \£(?:[1-9]\d{0,2}(?:,\d{3})*|[0-9]+)?(?:\.\d{1,2})? and skips it with (*SKIP)(*F), else, it matches a non-final . with \.(?!\s*$) (even if there is trailing whitespace chars).
If you really only need to split on the first occurrence of the qualifying dot you can use a matching approach:
preg_match('~^((?:\£(?:[1-9]\d{0,2}(?:,\d{3})*|[0-9]+)?(?:\.\d{1,2})?|[^.])+)\.(.*)~su', $string, $match)
See the regex demo. Here,
^ - matches a string start position
((?:\£(?:[1-9]\d{0,2}(?:,\d{3})*|[0-9]+)?(?:\.\d{1,2})?|[^.])+) - one or more occurrences of your currency pattern or any one char other than a . char
\. - a . char
(.*) - Group 2: the rest of the string.
To split a text into sentences avoiding the different pitfalls like dots or thousand separators in numbers and some abbreviations (like etc.), the best tool is intlBreakIterator designed to deal with natural language:
$str = 'There is a price in this string £19.99, but it should only split at this point. As I want it to ignore periods in a price';
$si = IntlBreakIterator::createSentenceInstance('en-US');
$si->setText($str);
$si->next();
echo substr($str, 0, $si->current());
IntlBreakIterator::createSentenceInstance returns an iterator that gives the indexes of the different sentences in the string.
It takes in account ?, ! and ... too. In addition to numbers or prices pitfalls, it works also well with this kind of string:
$str = 'John Smith, Jr. was running naked through the garden crying "catch me! catch me!", but no one was chasing him. His psychatre looked at him from the window with a circumspect eye.';
More about rules used by IntlBreakIterator here.
You could simply use this regex:
\.
Since you only have a space after the first sentence (and not a price), this should work just as well, right?
I have a regex in PHP that replaces everything I don't want with spaces
/[^a-z0-9\p{L}]/siu
But there is this one exception, I want to keep punctuations for abbreviations.
Example:
F.B.I.Federal.Bureau.of.Investigation => 'F B I Federal Bureau of
Investigation'
S.W.A.T.Team => 'S W A T Team'
Should be:
F.B.I.Federal.Bureau.of.Investigation => 'F.B.I. Federal Bureau of
Investigation'
S.W.A.T.Team => 'S.W.A.T. Team'
PHP code:
$s = "F.B.I.Federal.Bureau.of.Investigation";
return preg_replace('/[^a-z0-9\p{L}]/siu', " ", $s);
so the logic is, that it should check the second char of first match, and if it's an '.' char, then don't replace.
Not sure if this is possible with regex, then I would appreciate an alternative with PHP.
Actually, there are many types of abbreviations, and as Jon Stirling says, there is no really 100% working solution here since you need a whole list of possible abbreviations to filter out. You may have a peek at some fancy regex solution by #ndn and grab the pattern part related to abbreviations there.
If you need to only handle patterns like in the question, you may consider using
'~(\b(?:\p{Lu}\.){2,})|[^0-9\p{L}]~u'
or - if D.Word should also be treated as an abbreviation:
'~(\b(?:\p{Lu}\.)+)|[^0-9\p{L}]~u'
and replace with '$1 '. See the regex demo.
Pattern details:
(\b(?:\p{Lu}\.)+) - Group 1 (later referenced with $1 backreference): 1 or more consequent occurrences of any Unicode uppercase letter and a dot after it
| - or
[^0-9\p{L}] - any char that is not an ASCII digit and a Unicode letter.
And here is a variant of a regex with #ndn's abbreviations:
'~\b((?:[Ee]tc|St|Gen|Hon|Prof|Dr|Mr|Ms|Mrs|[JS]r|Col|Maj|Brig|Sgt|Capt|Cmnd|Sen|Rev|Rep|Revd|pp|[Vv]iz|i\.?\s*e|[Vvol]|[Rr]col|maj|Lt|[Ff]ig|[Ff]igs|[Vv]iz|[Vv]ols|[Aa]pprox|[Ii]ncl|Pres|[Dd]ept|min|max|[Gg]ovt|lb|ft|c\.?\s*f|vs|\p{Lu}(?:\.\p{Lu})+)\.)|[^0-9\p{L}]~'
See the regex demo.
If you do not want to remove -, ( and ), just make sure to add them to the negated character class, replace [^0-9\p{L}] with [^0-9\p{L}()-].
Feel free to update by adding more abbreviations or enhance by shrinking the alternatives.
I am trying to retrieve matches from a comma separated list that is located inside parenthesis using regular expression. (I also retrieve the version number in the first capture group, though that's not important to this question)
What's worth noting is that the expression should ideally handle all possible cases, where the list could be empty or could have more than 3 entries = 0 or more matches in the second capture group.
The expression I have right now looks like this:
SomeText\/(.*)\s\(((,\s)?([\w\s\.]+))*\)
The string I am testing this on looks like this:
SomeText/1.0.4 (debug, OS X 10.11.2, Macbook Pro Retina)
Result of this is:
1. [6-11] `1.0.4`
2. [32-52] `, Macbook Pro Retina`
3. [32-34] `, `
4. [34-52] `Macbook Pro Retina`
The desired result would look like this:
1. [6-11] `1.0.4`
2. [32-52] `debug`
3. [32-34] `OS X 10.11.2`
4. [34-52] `Macbook Pro Retina`
According to the image above (as far as I can see), the expression should work on the test string. What is the cause of the weird results and how could I improve the expression?
I know there are other ways of solving this problem, but I would like to use a single regular expression if possible. Please don't suggest other options.
When dealing with a varying number of groups, regex ain't the best. Solve it in two steps.
First, break down the statement using a simple regex:
SomeText\/([\d.]*) \(([^)]*)\)
1. [9-14] `1.0.4`
2. [16-55] `debug, OS X 10.11.2, Macbook Pro Retina`
Then just explode the second result by ',' to get your groups.
Probably the \G anchor works best here for binding the match to an entry point. This regex is designed for input that is always similar to the sample that is provided in your question.
(?<=SomeText\/|\G(?!^))[(,]? *\K[^,)(]+
(?<=SomeText\/|\G) the lookbehind is the part where matches should be glued to
\G matches where the previous match ended (?!^) but don't match start
[(,]? *\ matches optional opening parenthesis or comma followed by any amount of space
\K resets beginning of the reported match
[^,)(]+ matches the wanted characters, that are none of ( ) ,
Demo at regex101 (grab matches of $0)
Another idea with use of capture groups.
SomeText\/([^(]*)\(|\G(?!^),? *([^,)]+)
This one without lookbehind is a bit more accurate (it also requires the opening parenthesis), of better performance (needs fewer steps) and probably easier to understand and maintain.
SomeText\/([^(]*)\( the entry anchor and version is captured here to $1
|\G(?!^),? *([^,)]+) or glued to previous match: capture to $2 one or more characters, that are not , ) preceded by optional space or comma.
Another demo at regex101
Actually, stribizhev was close:
(?:SomeText\/([^() ]*)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\))
Just had to make that one class expect at least one match
(?:SomeText\/([0-9.]+)\s*\(|(?!^)\G),?\s*([^(),]+)(?=[^()]*\)) is a little more clear as long as the version number is always numbers and periods.
I wanted to come up with something more elegant than this (though this does actually work):
SomeText\/(.*)\s\(([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?([^\,]+)?\,?\s?\)
Obviously, the
([^\,]+)?\,?\s?
is repeated 6 times.
(It can be repeated any number of times and it will work for any number of comma-separated items equal to or below that number of times).
I tried to shorten the long, repetitive list of ([^\,]+)?\,?\s? above to
(?:([^\,]+)\,?\s?)*
but it doesn't work and my knowledge of regex is not currently good enough to say why not.
This should solve your problem. Use the code you already have and add something like this. It will determine where commas are in your string and delete them.
Use trim() to delete white spaces at the start or the end.
$a = strpos($line, ",");
$line = trim(substr($line, 55-$a));
I hope, this helps you!
I'd like to capture up to four groups of text between <p> and </p>. I can do that using the following regex:
<h5>Trivia<\/h5><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p>
The text to match on:
<h5>Trivia</h5><p>Was discovered by a freelance photographer while sunbathing on Bournemouth Beach in August 2003.</p><p>Supports Southampton FC.</p><p>She has 11 GCSEs and 2 'A' Levels.</p><p>Listens to soul, R&B, Stevie Wonder, Aretha Franklin, Usher Raymond, Michael Jackson and George Michael.</p>
It outputs the four lines of text. It also works as intended if there are more trivia items or <p> occurrences.
But if there are less than 4 trivia items or <p> groups, it outputs nothing since it cannot find the fourth group. How do I make that group optional?
I've tried: <h5>Trivia<\/h5><p>(.*?)<\/p>(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)? and that works according to http://gskinner.com/RegExr/ but it doesn't work if I put it inside PHP code. It only detects one group and puts everything in it.
The magic word is either 'escaping' or 'delimiters', read on.
The first regex:
<h5>Trivia<\/h5><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p><p>(.*)<\/p>
worked because you escaped the / characters in tags like </h5> to <\/h5>.
But in your second regex (correctly enclosing each paragraph in a optional non-capturing group, fetching 1 to 5 paragraphs):
<h5>Trivia</h5><p>(.*?)</p>(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?
you forgot to escape those / characters.
It should then have been:
$pattern = '/<h5>Trivia<\/h5><p>(.*?)<\/p>(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?(?:<p>(.*?)<\/p>)?/';
The above is assuming you were putting your regex between two / "delimiters" characters (out of conventional habit).
To dive a little deeper into the rabbit-hole, one should note that in php the first and last character of a regular expression is usually a "delimiter", so one can add modifiers at the end (like case-insensitive etc).
So instead of escaping your regex, you could also use a ~ character (or #, etc) as a delimiter.
Thus you could also use the same identical (second) regex that you posted and enclose for example like this:
$pattern = '~<h5>Trivia</h5><p>(.*?)</p>(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?~';
Here is a working (web-based) example of that, using # as delimiter (just because we can).
You can use the question mark to make each <p>...</p> optional:
$pattern = '~<h5>Trivia</h5>(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?(?:<p>(.*?)</p>)?~';
Use the Dom is a good option too.
i'm pretty new on regex, i have learned something by the way, but is still pour knowledge!
so i want ask you for clarification on how it work!
assuming i have the following strings, as you can see they can be formatted little different way one from another but they are very similar!
DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE
now i want replace everything between the first A-Z block and the colon so for example i would keep
DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART
so on my very noobs knowledge i have worked out this shitty regex! :-(
preg_replace( '/^[A-Z](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );
but why i'm sure this regex will not work!? :-)
Pls help me!
PS: the title of question is pretty explaned, i want also know how for example use a well know string block for match another...
preg_replace( '/^[DTSTART](?!;[A-Z]=[\w\W]+):$/m' , '' , $data );
..without delete DTSTART
Thanks for the time!
Regards
Luca Filosofi
You could use a relatively simple regex like the following.
$subject = 'DTSTART;TZID="America/Chicago":20030819T000000
DTEND;TZID="America/Chicago":20030819T010000
DTSTART;TZID=US/Pacific
DTSTART;VALUE=DATE';
echo preg_replace('/^[A-Z]+\K[^:\n]*/m', '', $subject) . PHP_EOL;
It looks for a series of capital letters at the start of a line, resets the match starting point (that's what \K does) to the end of those and matches anything not a colon or newline (i.e. the parts you want to remove). Those matched parts are then replaced with an empty string.
The output from the above would be
DTSTART:20030819T000000
DTEND:20030819T010000
DTSTART
DTSTART
If the lines that you are interested in will only ever start with DTSTART or DTEND then we could be more precise about what to match (e.g. ^DT(?:START|END)) but [A-Z] obviously covers both of those.
If you want to retain part of the matched pattern in a substitution, you put parentheses around it and then refer to it by $1 (or whichever grouping it is).
For example:
s/^(this is a sentence) to edit/$1/
gives "this is a sentence"
You can check out this example work similarly as your problem
\w+): (?P\d+)/', $str, $matches);
/* This also works in PHP 5.2.2 (PCRE 7.0) and later, however
* the above form is recommended for backwards compatibility */
// preg_match('/(?\w+): (?\d+)/', $str, $matches);
print_r($matches);
?>
The above example will output:
Array
(
[0] => foobar: 2008
[name] => foobar
[1] => foobar
[digit] => 2008
[2] => 2008
)
so if u need only digit u need to print $matches[digit]
You want to remove everything between a semicolon and either a colon or the end of the line, right? So use that as your expression. You're overcomplicating things.
preg_replace('/(?:;.+?:)|(?:;.+?$)/m','',$data);
It's a pretty simple expression. Either match (?:;.+?:) or (?:;.+?$), which differ only by their terminator (the first one matches up to a colon, the second one matches up to the end of the line).
Each is a non-capturing group that starts with a semicolon, reluctantly reads in all characters, then stops at the terminator. Everything matched by this is removable according to your description.