Getting file extension from http url using Java

Question

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.

Possible duplicate of [How to determine the file extension of a file from a uri](https://stackoverflow.com/questions/3234090/how-to-determine-the-file-extension-of-a-file-from-a-uri). — Tim Biegeleisen, Mar 01 '19 at 13:47

Marco Marchetti · Accepted Answer · 2019-03-01T14:05:50.447

3

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

edited Mar 01 '19 at 14:05

answered Mar 01 '19 at 13:45

Marco Marchetti

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And I have to handle MailformedException as well in this case – Tyulpan Tyulpan Mar 01 '19 at 14:06
Yep, I forgot to mention that, but I think that is the only downside of this solution. At least in my case, I don't like the use of regular expressions. – Marco Marchetti Mar 01 '19 at 14:08

Tim Biegeleisen · Answer 2 · 2019-03-01T14:01:22.103

0

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

edited Mar 01 '19 at 14:01

answered Mar 01 '19 at 13:31

Tim Biegeleisen

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where have you find this?) I need a reliable solution, better from a well-known and tested library. – Tyulpan Tyulpan Mar 01 '19 at 13:41
@TyulpanTyulpan My answer should work fine, but I will add an option using the Apache library. – Tim Biegeleisen Mar 01 '19 at 13:54
I was actually considering something like FilenameUtils.getExtension(new URL(url).getPath()); - but looks not so great and I need to handle exception coz of url creation – Tyulpan Tyulpan Mar 01 '19 at 14:00
I can't comment further. I have given you two valid solutions. – Tim Biegeleisen Mar 01 '19 at 14:02

Getting file extension from http url using Java

2 Answers2