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. — Tim Biegeleisen
– Tim Biegeleisen, Commented Mar 1, 2019 at 13:47

Marco Marchetti · Accepted Answer · 2019-03-01 14:05:50Z

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 1, 2019 at 14:05

answered Mar 1, 2019 at 13:45

Marco Marchetti

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2 Comments

Tyulpan Tyulpan Over a year ago

And I have to handle MailformedException as well in this case

Marco Marchetti Over a year ago

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.

Tim Biegeleisen · Accepted Answer · 2019-03-01 14:01:22Z

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 1, 2019 at 14:01

answered Mar 1, 2019 at 13:31

Tim Biegeleisen

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4 Comments

Tyulpan Tyulpan Over a year ago

where have you find this?) I need a reliable solution, better from a well-known and tested library.

Tim Biegeleisen Over a year ago

@TyulpanTyulpan My answer should work fine, but I will add an option using the Apache library.

Tyulpan Tyulpan Over a year ago

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

Tim Biegeleisen Over a year ago

I can't comment further. I have given you two valid solutions.

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