I'm parsing an external feed which contains location and date inside post title which I want to get rid of, so:
This happened on Date in Location
I need to find on (space on space) and remove everything till the end of the line, same for in(space in space).
I googled a bit, but regex is really unfathomable for me so I'd appreciate any help.
Thanks!
Well, a literal "on" does match exactly. Then tell the regex engine to match everything after: ".*". (Note, that the . doesn't match newlines, so it works as needed.)
In the case of "in" you need an alternative, which is marked by parentheses () and the vertical bar |: "(on|in)". You could also make that a bit tighter with character classes []: "[oi]n".
With that we arrive at this regex:
/ [oi]n .*/
To the end of the line? Then I suppose:
preg_replace("/(?:on|in).*?(\n|$)/", "", 'This happened on Date in Location');
Would do it.
Use a negative lookbehind if you want to remove everything after the on and in but not the on and in themselves.
(?<=\son\s).*
and
(?<=\sin\s).*
http://regexr.com?30ops
Related
I am looking for a pattern that matches everything until the first occurrence of a specific character, say a ";" - a semicolon.
I wrote this:
/^(.*);/
But it actually matches everything (including the semicolon) until the last occurrence of a semicolon.
You need
/^[^;]*/
The [^;] is a character class, it matches everything but a semicolon.
^ (start of line anchor) is added to the beginning of the regex so only the first match on each line is captured. This may or may not be required, depending on whether possible subsequent matches are desired.
To cite the perlre manpage:
You can specify a character class, by enclosing a list of characters in [] , which will match any character from the list. If the first character after the "[" is "^", the class matches any character not in the list.
This should work in most regex dialects.
Would;
/^(.*?);/
work?
The ? is a lazy operator, so the regex grabs as little as possible before matching the ;.
/^[^;]*/
The [^;] says match anything except a semicolon. The square brackets are a set matching operator, it's essentially, match any character in this set of characters, the ^ at the start makes it an inverse match, so match anything not in this set.
None of the proposed answers did work for me. (e.g. in notepad++)
But
^.*?(?=\;)
did.
Try /[^;]*/
Google regex character classes for details.
sample text:
"this is a test sentence; to prove this regex; that is g;iven below"
If for example we have the sample text above, the regex /(.*?\;)/ will give you everything until the first occurence of semicolon (;), including the semicolon: "this is a test sentence;"
Try /[^;]*/
That's a negating character class.
This was very helpful for me as I was trying to figure out how to match all the characters in an xml tag including attributes. I was running into the "matches everything to the end" problem with:
/<simpleChoice.*>/
but was able to resolve the issue with:
/<simpleChoice[^>]*>/
after reading this post. Thanks all.
this is not a regex solution, but something simple enough for your problem description. Just split your string and get the first item from your array.
$str = "match everything until first ; blah ; blah end ";
$s = explode(";",$str,2);
print $s[0];
output
$ php test.php
match everything until first
This will match up to the first occurrence only in each string and will ignore subsequent occurrences.
/^([^;]*);*/
"/^([^\/]*)\/$/" worked for me, to get only top "folders" from an array like:
a/ <- this
a/b/
c/ <- this
c/d/
/d/e/
f/ <- this
Really kinda sad that no one has given you the correct answer....
In regex, ? makes it non greedy. By default regex will match as much as it can (greedy)
Simply add a ? and it will be non-greedy and match as little as possible!
Good luck, hope that helps.
This works for getting the content from the beginning of a line till the first word,
/^.*?([^\s]+)/gm
I faced a similar problem including all the characters until the first comma after the word entity_id. The solution that worked was this in Bigquery:
SELECT regexp_extract(line_items,r'entity_id*[^,]*')
This is the text sample:
$text = "asd dasjfd fdsfsd http://11111.com/asdasd/?s=423%423%2F gfsdf http://22222.com/asdasd/?s=423%423%2F
asdfggasd http://3333333.com/asdasd/?s=423%423%2F";
This is my regex pattern:
preg_match_all( "#http:\/\/(.*?)[\s|\n]#is", $text, $m );
That match the first two urls, but how do I match the last one? I tried adding [\s|\n|$] but that will also only match the first two urls.
Don't try to match \n (there's no line break after all!) and instead use $ (which will match to the end of the string).
Edit:
I'd love to hear why my initial idea doesn't work, so in case you know it, let me know. I'd guess because [] tries to match one character, while end of line isn't one? :)
This one will work:
preg_match_all('#http://(\S+)#is', $text, $m);
Note that you don't have to escape the / due to them not being the delimiting character, but you'd have to escape the \ as you're using double quotes (so the string is parsed). Instead I used single quotes for this.
I'm not familar with PHP, so I don't have the exact syntax, but maybe this will give you something to try. the [] means a character class so |$ will literally look for a $. I think what you'll need is another look ahead so something like this:
#http:\/\/(.*)(?=(\s|$))
I apologize if this is way off, but maybe it will give you another angle to try.
See What is the best regular expression to check if a string is a valid URL?
It has some very long regular expressions that will match all urls.
I'm building this regex with a positive look ahead in it. Basically it must select all text in the line up to last period that precedes a ":" and add a "|" to the end to delimit it. Some sample text below. I am testing this in gskinner and editpadpro which has full grep regex support apparently so if I could get the answers in that for I'd appreciate it.
The regex below works to a degree but I am unsure if it is correct. Also it falls down if the text contains brackets.
Finally I would like to add another ignore rule like the one that ignores but includes "Co." in the selection. This second ignore rule would ignore but include periods that have a single Capital letter before them. Sample text below too. Thanks for all the help.
^(?:[^|]+\|){3}(.*?)[^(?:Co)]\.(?=[^:]*?\:)
121| Ryan, T.N. |2001. |I like regex. But does it like me (2) 2: 615-631.
122| O' Toole, H.Y. |2004. |(Note on the regex). Pages 90-91 In: Ryan, A. & Toole, B.L. (Editors) Guide to the regex functionality in php. Timmy, Tommy& Stewie, Quohog. * Produced for Family Guy in Quohog.
I don't think I understand what you want to do. But this part [^(?:Co)] is definitely not correct.
With the square brackets you are creating a character class, because of the ^ it is a negated class. That means at this place you don't want to match one of those characters (?:Co), in other words it will match any other character than "?)(:Co".
Update:
I don't think its possible. How should I distinguish between L. Co. or something similar and the end of the sentence?
But I found another error in your regex. The last part (?=[^:]*?\:) should be (?=[^.]*?\:) if you want to match the last dot before the : with your expression it will match on the first dot.
See it here on Regexr
This seems to do what you want.
(.*\.)(?=[^:]*?:)
It quite simply matches all text up to the last full stop that occurs before the colon.
I'm trying to pull a word out of an email subject line to use as a category for attached email. Preg_match works great as long as it's not just a single word (which is what I'd like to do anyway). If there is only one word in the subject line, I just get an empty array. I've tried to treat $matches as just a variable in that case, but that doesn't work either. Can anyone tell me if preg_match will work on a single word, or what the better way to do this would be?
Thanks very much
Assuming \b(?:word1|word2|word3)\b
The reason it wont match "word1" is because you included a word separator, the \b.
What you can do is just simply always inject the word separator:
preg_match("\b(?:word1|word2|word3)\b", "." . $subject . ".", $matches);
Crude but effective.
preg_match will work on a string one character long. I think that the issue here is probably your regex. My guess is that you're testing for whitespace and because it isn't finding any it says that there is no match. Try appending '^([^\s]*)$|' to your regex and I wager it will start picking up those one word values. ([^\s] means give me anything which has no spaces in it, | means 'or'. By adding it to the front of your regex, it will include things without whitespace or whatever you already had)
I'm looking for a simple replacement of [[wiki:Title]] into Title.
So far, I have:
$text = preg_replace("/\[\[wiki:(\w+)\]\]/","\\1", $text);
The above works for single words, but I'm trying to include spaces and on occasion special characters.
I get the \w+, but \w\s+ and/or \.+ aren't doing anything.
Could someone improve my understanding of basic regex? And I don't mean for anyone to simply point me to a webpage.
\w\s+ means "a word-character, followed by 1 or more spaces". You probably meant (\w|\s)+ ("1 or more of a word character or a space character").
\.+ means "one or more dots". You probably meant .+ (1 or more of any character - except newlines, unless in single-line mode).
The more robust way is to use
\[wiki:(.+?)\]
This means "1 or more of any character, but stop at first position where the rest matches", i.e. stop at first right bracket in this case. Without ? it would look for the longest available match - i.e. past the first bracket.
You need to use \[\[wiki:([\w\s]+)\]\]. Notice square brackets around \w\s.
If you are learning regular expressions, you will find this site useful for testing: http://rexv.org/
You're definitely getting there, but you've got a couple syntax errors.
When you're using multiple character classes like \w and \s, in order to match within that group, you have to put them in [square brackets] like so... ([\w\s]+) this basically means one or more of words or white space.
Putting a backslash in front of the period escapes it, meaning the regex is searching for a period.
As for matching special characters, that's more of a pain. I tried to come up with something quickly, but hopefully someone else can help you with that.
(Great cheat sheet here, I keep a copy on my desk at all times: http://www.addedbytes.com/cheat-sheets/regular-expressions-cheat-sheet/ )