I have a regular expression I am trying to use to rewrite an incoming REST url and am getting stuck on one use case when one section of the URL is excluded.
Here is the regex I'm currently using:
^(/[^/]+/(?:books))/([^/]+?)(?:/(?:(?!page).+?))?(?:/page/(\\d+))?$
As example I'm using "$1 - $2 - $3" as parts to use in writing new URL.
Here are the examples that are working correctly...
"/mySite/books/topic1/page/2" results in "/mySite/books - topic1 - 2"
"/mySite/books/topic1/subtopic1/page/2" results in "/mySite/books - topic1 - 2"
All the above work as intended. The problem is when the URL excludes the "topic1" part of the URL then the results are not what I need. Example:
"/mySite/books/page/2" results in "/mySite/books - page - "
What I need is the $2 to be blank, because there is no topic, and the page number still as $3. What I need as output...
"/mySite/books/page/2" results in "/mySite/books - - 2"
What can I change in my regex to satisfy that scenario without disrupting the existing ones that work correctly? This is being done in Java.
You might try to use regex pattern
^(/[^/]+/books)/(?:(?!page/)([^/]+)/)?page/(\\d+)$
It should suffice to make your second group ungreedy. Then the engine will first try to find a match without using it (trying only /page/\\d+ instead). And if that fails it tries to include the second group:
^(/[^/]+/(?:books))/([^/]+?)(?:/(?:(?!page).+?))??(?:/page/(\\d+))?$
Prepending any kind of quantifier (+, *, ? and {..} with ?) makes it ungreedy.
Related
i want to extract query part(name,find,sort,limit - split by dot(.)) from mongo query via regex
input - >
db.metrics.find(
{
"brand_name":"Apple",
"job_status.status":"SUCCESS",
'host.user':'root',
"current_time":{$gt:new Date(Date.now() - 3*60*60 * 1000)}
}
).sort({"current_time" : -1}).limit(10)
with help of 2-3 stackoverflow answer i have build below regex
regex = `\.(?=(([^']*'){2})*[^']*$)(?=(([^\"]*\"){2})*[^\"]*$)(?![^()]*\\)`
which solves my use case till certain extent
i am not able to ignore dot(.) char group inside curly braces (Date.now())
regExr.com matched screen shot
i need regex which should ignore .now() part from above query
I have Strings like the following:
"parameter: param0=true, param1=401230 param2=asset client: desktop"
"parameter: param0=false, param1=15230 user: user213 client: desktop"
"parameter: param0=false, param1=51235 param2=asset result: ERROR"
The pattern is parameter:, then the param's, and after the params either client: and/or user: and/or result.
I want to match the stuff between parameter: and the first occurrence of either client:, user: or result:
So for the 2nd String it should match param0=false, param1=15230.
My regex is:
parameter:\s+(.*)\s+(result|client|user):
But now if I match the 2nd String it captures param0=false, param1=15230 user: user213 (looks like regex is matching greedy)
How to fix this? parameter:\s+(.*)\s+(result|client|user)+?: won't fix it
With this regex tester I can add the modifier U to the regex to make regex lazy by default, is this possible in Java too?
Try putting the ? character inside the first captured group (the subpattern you intend to extract):
parameter:\\s+(.*?)\\s+(result|client|user):
No. There is no ungreedy modifier in Java. You have to use ? behind modifiers to make the quantifiers as lazy capture.
This means you should denote all quantifiers with a ?, see the following pattern:
"parameter:\\s+?(.*?)\\s+?(result|client|user):"
Specified by:http://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html
I am new to regex. I would like to retrieve the Hostname from postgreSQL jdbc URL using regex.
Assume the postgreSQL url will be jdbc:postgresql://production:5432/dbname. I need to retrieve "production", which is the hostname. I want to try with regex and not with Java split function. I tried with
Pattern PortFinderPattern = Pattern.compile("[//](.*):*");
final Matcher match = PortFinderPattern.matcher(url);
if (match.find()) {
System.out.println(match.group(1));
}
But it's matching all the string from hostname till the end.
Pattern PortFinderPattern = Pattern.compile(".*:\/\/([^:]+).*");
regex without grouping :
"(?<=//)[^:]*"
[//]([\\w\\d\\-\\.]+)\:
Should be enough to find it reliably. Though this is probably a better regex:
The Hostname Regex
There are some errors in your regex:
[//] - This is only one character, because the [] marks a character class, so it will not fully match //. To match it, you need to write it like this: [/][/] or \/\/.
(.*) - This will match all characters to the end of line. You need to be more specific if you want to go till a certain character. For example you could go to the colon by fetching all characters, which are not colons, like this: ([^:]*).
:* - This makes the colon optional. I guess you forgot to put a dot( every character ) after the colon, like this: :.*.
So here is your regex corrected: \/\/([^:]*):.*.
Hope this helps.
BTW. If the port number is optional after production (:5432), then I suggest the following regex:
\/\/([^/]*)(?::\d+)?\/
To capture also Oracle and MySQL JDBC URL variants with their quirks (e.g. Oracle allowing to use # instead of // or even #//), I use this regexp to get the hostname: [/#]+([^:/#]+)([:/]+|$) Then the hostname is in group 1.
Code e.g.
String jdbcURL = "jdbc:oracle:thin:#//hostname:1521/service.domain.local";
Pattern hostFinderPattern = Pattern.compile("[/#]+([^:/#]+)([:/]+|$)");
final Matcher match = hostFinderPattern.matcher(jdbcURL);
if (match.find()) {
System.out.println(match.group(1));
}
This works for all these URLs (and other variants):
jdbc:oracle:thin:#//hostname:1521/service.domain.local
jdbc:oracle:thin:#hostname:1521/service.domain.local
jdbc:oracle:thin:#hostname/service.domain.local
jdbc:mysql://localhost:3306/sakila?profileSQL=true
jdbc:postgresql://production:5432/dbname
jdbc:postgresql://production/
jdbc:postgresql://production
This assumes that
The hostname is after // or # or a combination thereof (single / would also work, but I don't think JDBC allows that).
After the hostname either : or / or the end of the string follows.
Note that the the + are greedy, this is especially important for the middle one.
[introduction][position]Lead Researcher and Research Manager[/position] in the [affiliation]Web Search and Mining Group, Microsoft Research[/affiliation]</b>.
I am a [position]lead researcher[/position] at [affiliation]Microsoft Research[/affiliation]. I am also [position]adjunct professor[/position] of [affiliation]Peking University[/affiliation], [affiliation]Xian Jiaotong University[/affiliation] and [affiliation]Nankai University[/affiliation].
I joined [affiliation]Microsoft Research[/affiliation] in June 2001. Prior to that, I worked at the Research Laboratories of NEC Corporation.
I obtained a [bsdegree]B.S.[/bsdegree] in [bsmajor]Electrical Engineering[/bsmajor] from [bsuniv]Kyoto University[/bsuniv] in [bsdate]1988[/bsdate] and a [msdegree]M.S.[/msdegree] in [msmajor]Computer Science[/msmajor] from [msuniv]Kyoto University[/msuniv] in [msdate]1990[/msdate]. I earned my [phddegree]Ph.D.[/phddegree] in [phdmajor]Computer Science[/phdmajor] from the [phduniv]University of Tokyo[/phduniv] in [phddate]1998[/phddate].
I am interested in [interests]statistical learning[/interests], [interests]natural language processing[/interests], [interests]data mining, and information retrieval[/interests].[/introduction]
I'm able to strip all tags from the paragraph above with:
String stripped = html.replaceAll("\\[.*?\\]", "");
But I'd like to keep three pairs of tags in the paragraph, which are [bsuniv][/bsuniv],[msuniv][/msuniv] and [phduniv][/phduniv]. In other words, I don't want to strip those tags containing the keyword "univ". I can't find a convenient way to rewrite the regular expression. Anyone help me?
You can use a negative-look ahead assertion here: -
str = str.replaceAll("\\[(.(?!univ))*?\\]", "");
or: -
str = str.replaceAll("\\[((?!univ).)*?\\]", "");
Both of them will give you the desired output. There is only one difference -
The first one does a negative look-ahead, against the current character, and if it is not followed by univ, it moves to the next character.
The second one does a negative look-ahead against an empty string before every character, and if it is not followed by univ, it goes ahead to match a single character.
need to find an expression for the following problem:
String given = "{ \"questionID\" :\"4\", \"question\":\"What is your favourite hobby?\",\"answer\" :\"answer 4\"},{ \"questionID\" :\"5\", \"question\" :\"What was the name of the first company you worked at?\",\"answer\" :\"answer 5\"}";
What I want to get: "{ \"questionID\" :\"4\", \"question\":\"What is your favourite hobby?\",\"answer\" :\"*******\"},{ \"questionID\" :\"5\", \"question\" :\"What was the name of the first company you worked at?\",\"answer\" :\"******\"}";
What I am trying:
String regex = "(.*answer\"\\s:\"){1}(.*)(\"[\\s}]?)";
String rep = "$1*****$3";
System.out.println(test.replaceAll(regex, rep));
What I am getting:
"{ \"questionID\" :\"4\", \"question\":\"What is your favourite hobby?\",\"answer\" :\"answer 4\"},{ \"questionID\" :\"5\", \"question\" :\"What was the name of the first company you worked at?\",\"answer\" :\"******\"}";
Because of the greedy behaviour, the first group catches both "answer" parts, whereas I want it to stop after finding enough, perform replacement, and then keep looking further.
The pattern
("answer"\s*:\s*")(.*?)(")
Seems to do what you want. Here's the escaped version for Java:
(\"answer\"\\s*:\\s*\")(.*?)(\")
The key here is to use (.*?) to match the answer and not (.*). The latter matches as many characters as possible, the former will stop as soon as possible.
The above pattern won't work if there are double quotes in the answer. Here's a more complex version that will allow them:
("answer"\s*:\s*")((.*?)[^\\])?(")
You'll have to use $4 instead of $3 in the replacement pattern.
The following regex works for me :
regex = "(?<=answer\"\\s:\")(answer.*?)(?=\"})";
rep = "*****";
replaceALL(regex,rep);
The \ and " might be incorrectly escaped since I tested without java.
http://regexr.com?303mm