I have a very simple regex question. Suppose I have 2 conditions:
url =http://www.abc.com/cde/def
url =https://www.abc.com/sadfl/dsaf
How can I extract the baseUrl using regex?
Sample output:
http://www.abc.com
https://www.abc.com
Like this:
String baseUrl;
Pattern p = Pattern.compile("^(([a-zA-Z]+://)?[a-zA-Z0-9.-]+\\.[a-zA-Z]+(:\d+)?/");
Matcher m = p.matcher(str);
if (m.matches())
baseUrl = m.group(1);
However, you should use the URI class instead, like this:
URI uri = new URI(str);
A one liner without regexp:
String baseUrl = url.substring(0, url.indexOf('/', url.indexOf("//")+2));
/^(https?\:\/\/[^\/]+).*/$1/
This will capture ANYTHING that starts with http and $1 will contain everything from the beginning to the first / after the //
Except for write-and-throw-away scripts, you should always refrain from parsing complex syntaxes (e-mail addresses, urls, html pages, etc etc) using regexes.
believe me, you will get bitten eventually.
I'm pretty sure that there is a Java class that will allow path manipulations, but if it has to be a regex,
https?://[^/]+
would work. (s? included to also handle https:)
Looks like the simplest solution to your two specific examples would be the pattern:
[^/]_//[^/]+
i.e.: non-slash (0 or more times), two slashes, non-slash (0 or more times). You can be stricter than that if you wish, as the two existing answers are doing in different ways -- one will reject e.g. URLs starting with ftp:, the other will reject domains with underscores (but accept URLs without a leading protocol://, thereby being even broader than mine in that respect). This variety of answers (all correct wrt your scant specs;-) should suggest to you that your specs are too vague and should be tightened.
Here's a regex that should satisfy the problem as given.
https?://[^/]*
I'm assuming you're asking this partly to gain more knowledge of regexes. If, however, you're trying to pull the host from a URL, it's arguably much more correct to use Java's more robust parsing methods:
String urlStr = "https://www.abc.com/stuff";
URL url = new URL(urlStr);
String host = url.getHost();
String protocol = url.getProtocol();
URL baseUrl = new URL (protocol, host);
This is better, as it should catch more cases if your input URL isn't as strict as described above.
Old post.. thought I might as well put a simple answer to a simple regex Q:
(http|https):\/\/(www.)?(\w+)?\.(\w+)?
Related
Small question regarding how to use java to retain only the base URL of a rest API please.
As input, many strings, all valid rest APIs.
For instance, the inputs:
https://some-host.com/v1/someapi
https://another-host.fr/api/compute
https://somewhere.host.com/public/api/v3/getsomething
I would like to only retain the bold part, basically, the https, the : and the slashes, the host name. Everything that comes after the host, I would like to discard it.
Currently, I am trying some kind of string.split based on the / character, then trying to re-concat the arrays, but I have a feeling I am not going to the right direction.
What would be the most appropriate way please?
Thank you.
You could just try java.net.URL or java.net.URI. They behave pretty similar.
For example:
URL url = new URL("http://example.com/a/b/c");
url.getProtocol();
url.getHost();
url.getPath();
or:
URI uri = new URI("http://example.com/a/b/c");
uri.getScheme();
uri.getHost();
uri.getPath();
There are several methods in both classes to extract lot's of different parts.
Now I know about FilenameUtils.getExtension() from apache.
But in my case I'm processing extensions from http(s) urls, so in case I have something like
https://your_url/logo.svg?position=5
this method is gonna return svg?position=5
Is there the best way to handle this situation? I mean without writing this logic by myself.
You can use the URL library from JAVA. It has a lot of utility in this cases. You should do something like this:
String url = "https://your_url/logo.svg?position=5";
URL fileIneed = new URL(url);
Then, you have a lot of getter methods for the "fileIneed" variable. In your case the "getPath()" will retrieve this:
fileIneed.getPath() ---> "/logo.svg"
And then use the Apache library that you are using, and you will have the "svg" String.
FilenameUtils.getExtension(fileIneed.getPath()) ---> "svg"
JAVA URL library docs >>>
https://docs.oracle.com/javase/7/docs/api/java/net/URL.html
If you want a brandname® solution, then consider using the Apache method after stripping off the query string, if it exists:
String url = "https://your_url/logo.svg?position=5";
url = url.replaceAll("\\?.*$", "");
String ext = FilenameUtils.getExtension(url);
System.out.println(ext);
If you want a one-liner which does not even require an external library, then consider this option using String#replaceAll:
String url = "https://your_url/logo.svg?position=5";
String ext = url.replaceAll(".*/[^.]+\\.([^?]+)\\??.*", "$1");
System.out.println(ext);
svg
Here is an explanation of the regex pattern used above:
.*/ match everything up to, and including, the LAST path separator
[^.]+ then match any number of non dots, i.e. match the filename
\. match a dot
([^?]+) match AND capture any non ? character, which is the extension
\??.* match an optional ? followed by the rest of the query string, if present
I would like to use Java regex to match a domain of a url, for example,
for www.table.google.com, I would like to get 'google' out of the url, namely, the second last word in this URL string.
Any help will be appreciated !!!
It really depends on the complexity of your inputs...
Here is a pretty simple regex:
.+\\.(.+)\\..+
It fetches something that is inside dots \\..
And here are some examples for that pattern: https://regex101.com/r/L52oz6/1.
As you can see, it works for simple inputs but not for complex urls.
But why reinventing the wheel, there are plenty of really good libraries that correctly parse any complex url. But sure, for simple inputs a small regex is easily build. So if that does not solve the problem for your inputs then please callback, I will adjust the regex pattern then.
Note that you can also just use simple splitting like:
String[] elements = input.split("\\.");
String secondToLastElement = elements[elements.length - 2];
But don't forget the index-bound checking.
Or if you search for a very quick solution than walk through the input starting from the last position. Work your way through until you found the first dot, continue until the second dot was found. Then extract that part with input.substring(index1, index2);.
There is also already a delegate method for exactly that purpose, namely String#lastIndexOf (see the documentation).
Take a look at this code snippet:
String input = ...
int indexLastDot = input.lastIndexOf('.');
int indexSecondToLastDot = input.lastIndexOf('.', indexLastDot);
String secondToLastWord = input.substring(indexLastDot, indexSecondToLastDot);
Maybe the bounds are off by 1, haven't tested the code, but you get the idea. Also don't forget bound checking.
The advantage of this approach is that it is really fast, it can directly work on the internal structures of Strings without creating copies.
My attempt:
(?<scheme>https?:\/\/)?(?<subdomain>\S*?)(?<domainword>[^.\s]+)(?<tld>\.[a-z]+|\.[a-z]{2,3}\.[a-z]{2,3})(?=\/|$)
Demo. Works correctly for:
http://www.foo.stackoverflow.com
http://www.stackoverflow.com
http://www.stackoverflow.com/
http://stackoverflow.com
https://www.stackoverflow.com
www.stackoverflow.com
stackoverflow.com
http://www.stackoverflow.com
http://www.stackoverflow.co.uk
foo.www.stackoverflow.com
foo.www.stackoverflow.co.uk
foo.www.stackoverflow.co.uk/a/b/c
private static final Pattern URL_MATCH_GET_SECOND_AND_LAST =
Pattern.compile("www.(.*)//.google.(.*)", Pattern.CASE_INSENSITIVE);
String sURL = "www.table.google.com";
if (URL_MATCH_GET_SECOND_AND_LAST.matcher(sURL).find()){
Matcher matchURL = URL_MATCH_GET_SECOND_AND_LAST .matcher(sURL);
if (matchURL .find()) {
String sFirst = matchURL.group(1);
String sSecond= matchURL.group(2);
}
}
what is the regular expression to find the path param from the url?
http://localhost:8080/domain/v1/809pA8
https://localhost:8080/domain/v1/809pA8
Want to retrieve the value(809pA8) from the above URL using regular expression, java is preferable.
I would suggest you do something like
url.substring(url.lastIndexOf('/') + 1);
If you really prefer regexps, you could do
Matcher m = Pattern.compile("/([^/]+)$").matcher(url);
if (m.find())
value = m.group(1);
I would try:
String url = "http://localhost:8080/domain/v1/809pA8";
String value = String.valueOf(url.subSequence(url.lastIndexOf('/'), url.length()-1));
No need for regex here, I think.
EDIT: I'm sorry I made a mistake:
String url = "http://localhost:8080/domain/v1/809pA8";
String value = String.valueOf(url.subSequence(url.lastIndexOf('/')+1, url.length()));
See this code working here: https://ideone.com/E30ddC
For your simple case, regex is an overkill, as others noted. But, if you have more cases and this is why you prefer regex, give Spring's AntPathMatcher#extractUriTemplateVariables a look, if you're using Spring. It's actually better equipped for extracting path variables than regex directly. Here are some good examples.
For the given url like "http://google.com//view/All/builds", i want to replace the double slash with single slash. For example the above url should display as "http://google.com/view/All/builds"
I dint know regular expressions. Can any one help me, how can i achieve this using regular expressions.
To avoid replacing the first // in http:// use the following regex :
String to = from.replaceAll("(?<!http:)//", "/");
PS: if you want to handle https use (?<!(http:|https:))// instead.
Is Regex the right approach?
In case you wanted this solution as part of an exercise to improve your regex skills, then fine. But what is it that you're really trying to achieve? You're probably trying to normalize a URL. Replacing // with / is one aspect of normalizing a URL. But what about other aspects, like removing redundant ./ and collapsing ../ with their parent directories? What about different protocols? What about ///? What about the // at the start? What about /// at the start in case of file:///?
If you want to write a generic, reusable piece of code, using a regular expression is probably not the best appraoch. And it's reinventing the wheel. Instead, consider java.net.URI.normalize().
java.net.URI.normalize()
java.lang.String
String inputUrl = "http://localhost:1234//foo//bar//buzz";
String normalizedUrl = new URI(inputUrl).normalize().toString();
java.net.URL
URL inputUrl = new URL("http://localhost:1234//foo//bar//buzz");
URL normalizedUrl = inputUrl.toURI().normalize().toURL();
java.net.URI
URI inputUri = new URI("http://localhost:1234//foo//bar//buzz");
URI normalizedUri = inputUri.normalize();
Regex
In case you do want to use a regular expression, think of all possibilities. What if, in future, this should also process other protocols, like https, file, ftp, fish, and so on? So, think again, and probably use URI.normalize(). But if you insist on a regular expression, maybe use this one:
String noramlizedUri = uri.replaceAll("(?<!\\w+:/?)//+", "/");
Compared to other solutions, this works with all URLs that look similar to HTTP URLs just with different protocols instead of http, like https, file, ftp and so on, and it will keep the triple-slash /// in case of file:///. But, unlike java.net.URI.normalize(), this does not remove redundant ./, it does not collapse ../ with their parent directories, it does not other aspects of URL normalization that you and I might have forgotten about, and it will not be updated automatically with newer RFCs about URLs, URIs, and such.
String to = from.replaceAll("(?<!(http:|https:))[//]+", "/");
will match two or more slashes.
Here is the regexp:
/(?<=[^:\s])(\/+\/)/g
It finds multiple slashes in url preserving ones after protocol regardless of it.
Handles also protocol relative urls which start from //.
#Test
public void shouldReplaceMultipleSlashes() {
assertEquals("http://google.com/?q=hi", replaceMultipleSlashes("http://google.com///?q=hi"));
assertEquals("https://google.com/?q=hi", replaceMultipleSlashes("https:////google.com//?q=hi"));
assertEquals("//somecdn.com/foo/", replaceMultipleSlashes("//somecdn.com/foo///"));
}
private static String replaceMultipleSlashes(String url) {
return url.replaceAll("(?<=[^:\\s])(\\/+\\/)", "/");
}
Literally means:
(\/+\/) - find group: /+ one or more slashes followed by / slash
(?<=[^:\s]) - which follows the group (*posiive lookbehind) of this (*negated set) [^:\s] that excludes : colon and \s whitespace
g - global search flag
I suggest you simply use String.replace which documentation is http://docs.oracle.com/javase/7/docs/api/java/lang/String.html#replace(java.lang.CharSequence, java.lang.CharSequence)
Something like
`myString.replace("//", "/");
If you want to remove the first occurence:
String[] parts = str.split("//", 2);
str = parts[0] + "//" + parts[1].replaceAll("//", "/");
Which is the simplest way (without regular expression). I don't know the regular expression corresponding, if there is an expert looking at the thread.... ;)