I've tried and tried again to find a regex for this pattern.
I have a string like this picked from HTML source.
<!-- TAG=Something / Something else -->
And sometimes it's just:
<!-- TAG=Something -->
In both cases I want the regex to just match "Something", i.e. everything between TAG= and an optional /.
My first attempt was:
TAG=(.*)[/]?(.*) -->
But the first parenthesis matches everything between TAG= and --> no matter what. So what is the correct way here?
Try this:
TAG=([^/]*)(?:/(.*))?-->
Group 1 will contain "Something".
Group 2 will contain "Something else" or null.
Test it.
<!--.*?=(.*?)(-->|/)
It matches everything you need.
Use a non-greedy modifier ?:
TAG=(.*?)[/]?.* -->
Also your usage of [/] seems unusual - you don't need a character class to write a single character. The most likely explanation for this unusual syntax is probably because you are using / as the regular expression delimiter, meaning that / is treated as a special character. In many (not all) regex dialects it is possible solve this issue by using a different delimiter, such as #. This prevents you from needing to escape the slashes.
Related
I need some help with regex with this test string:
Kershing_User ID/Electronic Delivery_6ZZ138429_ 3142-999999__1
I want one match to select everything except the forward slash, so this would be acceptable:
Kershing_User IDElectronic Delivery_6ZZ138429_ 3142-999999__1
Even better would be to return this with a substitution of a _.
Kershing_User ID_Electronic Delivery_6ZZ138429_ 3142-999999__1
I know how to do lookarounds and can individually match the part before and after the /, but not all in one match. Anything else I have tried has come up with two separate matches. I am using this with an application called Laserfiche, so as far as I know there is not the ability to do find & replace or to extract a group, just doing it with one match. My regrets if I don't have the terminology correct. I am not even sure if this is possible. I tried for a while and come up with these below, but can't get it in one match.
This does the before: .*(?=\/)
This does the after: (?<=\/).*
Even better would be to return this with a substitution of a _.
import re
a = "Kershing_User ID/Electronic Delivery_6ZZ138429_ 3142-999999__1"
print(re.sub('\/', '_', a))
This will replace / with _. Read more on this here
I'm creating a javascript regex to match queries in a search engine string. I am having a problem with alternation. I have the following regex:
.*baidu.com.*[/?].*wd{1}=
I want to be able to match strings that have the string 'word' or 'qw' in addition to 'wd', but everything I try is unsuccessful. I thought I would be able to do something like the following:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
but it does not seem to work.
replace [wd|word|qw] with (wd|word|qw) or (?:wd|word|qw).
[] denotes character sets, () denotes logical groupings.
Your expression:
.*baidu.com.*[/?].*[wd|word|qw]{1}=
does need a few changes, including [wd|word|qw] to (wd|word|qw) and getting rid of the redundant {1}, like so:
.*baidu.com.*[/?].*(wd|word|qw)=
But you also need to understand that the first part of your expression (.*baidu.com.*[/?].*) will match baidu.com hello what spelling/handle????????? or hbaidu-com/ or even something like lkas----jhdf lkja$##!3hdsfbaidugcomlaksjhdf.[($?lakshf, because the dot (.) matches any character except newlines... to match a literal dot, you have to escape it with a backslash (like \.)
There are several approaches you could take to match things in a URL, but we could help you more if you tell us what you are trying to do or accomplish - perhaps regex is not the best solution or (EDIT) only part of the best solution?
For clarity, I have created this:
http://rubular.com/r/ejYgKSufD4
My strings:
http://blablalba.com/foo/bar_soap/foo/dir2
http://blablalba.com/foo/bar_soap/dir
http://blablalba.com/foo/bar_soap
My Regular expression:
\/foo\/(.*)
This returns:
/foo/bar_soap/dir/dir2
/foo/bar_soap/dir
/foo/bar_soap
But I only want
/foo/bar_soap
Any ideas how I can achieve this? As illustrated above, I want everything after foo up until the first forward slash.
Thanks in advance.
Edit. I only want the text after foo until until the next forward slash after. Some directories may also be named as foo and this would render incorrect results. Thanks
. will match anything, so you should change it to [^/] (not slash) instead:
\/foo\/([^\/]*)
Some of the other answers use + instead of *. That might be correct depending on what you want to do. Using + forces the regex to match at least one non-slash character, so this URL would not match since there isn't a trailing character after the slash:
http://blablalba.com/foo/
Using * instead would allow that to match since it matches "zero or more" non-slash characters. So, whether you should use + or * depends on what matches you want to allow.
Update
If you want to filter out query strings too, you could also filter against ?, which must come at the front of all query strings. (I think the examples you posted below are actually missing the leading ?):
\/foo\/([^?\/]*)
However, rather than rolling out your own solution, it might be better to just use split from the URI module. You could use URI::split to get the path part of the URL, and then use String#split split it up by /, and grab the first one. This would handle all the weird cases for URLs. One that you probably haven't though of yet is a URL with a specified fragment, e.g.:
http://blablalba.com/foo#bar
You would need to add # to your filtered-character class to handle those as well.
You can try this regular expression
/\/foo\/([^\/]+)/
\/foo\/([^\/]+)
[^\/]+ gives you a series of characters that are not a forward slash.
the parentheses cause the regex engine to store the matched contents in a group ([^\/]+), so you can get bar_soap out of the entire match of /foo/bar_soap
For example, in javascript you would get the matched group as follows:
regexp = /\/foo\/([^\/]+)/ ;
match = regexp.exec("/foo/bar_soap/dir");
console.log(match[1]); // prints bar_soap
im looking to use a regular expression to parse a URL to get a specific section of the url and nothing if I cannot find the pattern.
A url example is
/te/file/value/jifle?uil=testing-cdas-feaw:jilk:&jklfe=https://value-value.jifels/temp.html/topic?id=e997aad4-92e0-j30e-a3c8-jfkaliejs5#c452fds-634d-f424fds-cdsa&bf_action=jildape
I wish to get the bolded text in it.
Currently im using the regex "d=([^#]*)" but the problem is im also running across urls of this pattern:
and im getting the bold section of it
/te/file/value/jifle?uil=testing-cdas-feaw:jilk:&jklfe=https://value-value.jifels/temp.html/topic?id=e997aad4-92e0-j30e-a3c8-jfkaliejs5&bf_action=jildape
I would prefer it have no matches of this url because it doesnt contain the #
Regexes are not a magic tool that you should always use just because the problem involves a string. In this case, your language probably has a tool to break apart URLs for you. In PHP, this is parse_url(). In Perl, it's the URI::URL module.
You should almost always prefer an existing, well-tested solution to a common problem like this rather than writing your own.
So you want to match the value of the id parameter, but only if it has a trailing section containing a '#' symbol (without matching the '#' or what's after it)?
Not knowing the specifics of what style of regexes you're using, how about something like:
id=([^#&]*)#
regex = "id=([\\w-])+?#"
This will grab everything that is character class[a-zA-Z_0-9-] between 'id=' and '#' assuming everything between 'id=' and '#' is in that character class(i.e. if an '&' is in there, the regex will fail).
id=
-Self explanatory, this looks for the exact match of 'id='
([\\w-])
-This defines and character class and groups it. The \w is an escaped \w. '\w' is a predefined character class from java that is equal to [a-zA-Z_0-9]. I added '-' to this class because of the assumed pattern from your examples.
+?
-This is a reluctant quantifier that looks for the shortest possible match of the regex.
#
-The end of the regex, the last character we are looking for to match the pattern.
If you are looking to grab every character between 'id=' and the first '#' following it, the following will work and it uses the same logic as above, but replaces the character class [\\w-] with ., which matches anything.
regex = "id=(.+?)#"
how can I write regular expression that dose not contain some string at the end.
in my project,all classes that their names dont end with some string such as "controller" and "map" should inherit from a base class. how can I do this using regular expression ?
but using both
public*.class[a-zA-Z]*(?<!controller|map)$
public*.class*.(?<!controller)$
there isnt any match case!!!
Do a search for all filenames matching this:
(?<!controller|map|anythingelse)$
(Remove the |anythingelse if no other keywords, or append other keywords similarly.)
If you can't use negative lookbehinds (the (?<!..) bit), do a search for filenames that do not match this:
(?:controller|map)$
And if that still doesn't work (might not in some IDEs), remove the ?: part and it probably will - that just makes it a non-capturing group, but the difference here is fairly insignificant.
If you're using something where the full string must match, then you can just prefix either of the above with ^.* to do that.
Update:
In response to this:
but using both
public*.class[a-zA-Z]*(?<!controller|map)$
public*.class*.(?<!controller)$
there isnt any match case!!!
Not quite sure what you're attempting with the public/class stuff there, so try this:
public.*class.*(?<!controller|map)$`
The . is a regex char that means "anything except newline", and the * means zero or more times.
If this isn't what you're after, edit the question with more details.
Depending on your regex implementation, you might be able to use a lookbehind for this task. This would look like
(?<!SomeText)$
This matches any lines NOT having "SomeText" at their end. If you cannot use that, the expression
^(?!.*SomeText$).*$
matches any non-empty lines not ending with "SomeText" as well.
You could write a regex that contains two groups, one consists of one or more characters before controller or map, the other contains controller or map and is optional.
^(.+)(controller|map)?$
With that you may match your string and if there is a group() method in the regex API you use, if group(2) is empty, the string does not contain controller or map.
Check if the name does not match [a-zA-Z]*controller or [a-zA-Z]*map.
finally I did it in this way
public.*class.*[^(controller|map|spec)]$
it worked