What is the right way to safely concatenate an URL without forgetting a separator?

Question

lets assume I have some method like that:

void getMagicUrl(Sting urlBase, String addedPath) {
    urlBase = urlBase.endsWith("/") ? urlBase : urlBase + "/";      // *1*
    urlBase = urlBase + "constantPath/";                            // *2*
    urlBase = urlBase + addedPath;                                  // *3*
    return new URL(urlBase);                                        // *4*
}

Is there with Java (8) no better way to concatenate the URL together and not to check manual for the ending / as path separator to not forget it? (Like in line 1)
and have it in the path names always? (like in line 2)

or does the JDK have something in there?

Are you trying to duplicate this constructor that's already in URL? https://docs.oracle.com/javase/7/docs/api/java/net/URL.html#URL(java.net.URL,%20java.lang.String) — ControlAltDel, Jul 26 '18 at 17:29
@Wow: I always have to check to not forget the "/" and if an url is passed to a method check if it is there. always! — PaulEdison, Jul 26 '18 at 17:44
@ControlAltDel: no, this is just an example - I always have to do this not only on construction. — PaulEdison, Jul 26 '18 at 17:45
Does this answer your question? [Creating a URL Using Java - What's the Best Practice?](https://stackoverflow.com/questions/50989746/creating-a-url-using-java-whats-the-best-practice) — naXa stands with Ukraine, Dec 03 '19 at 13:24

score 3 · Answer 1 · answered Jul 26 '18 at 17:29

3

One of the most efficient ways to build url is to use UriComponentsBuilder

Something like this

 UriComponents uriComponents = UriComponentsBuilder.newInstance()
      .scheme("http").host("www.blahblah.com).path("/somepath").build().encode();

 uriComponents.toUriString());

This avoids all the manual concatenation using +

There are many methods to add path variables and query params and much more. Here is a nice article on it.

answered Jul 26 '18 at 17:29

pvpkiran

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But it is not part of JDK :( I don't add just for that another lib (especially as the app has to be downloaded) – PaulEdison Jul 26 '18 at 17:43

Víctor Cabrera · Accepted Answer · 2018-07-26T18:39:42.727

There are more alternatives to build a url in both type-safety and fluent way. They involve adding new dependencies but some of them are very common and useful.

Apache commons (JavaDoc)

URI uri = new URIBuilder()
        .setHost("your_host")
        .setPath("/you_path")
        .setParameter("param1", "value1")
        .build();

uri.toString();

JAX-RS (JavaDoc)

URI uri= UriBuilder
    .fromPath("your_domain")
    .scheme("http")
    .path("your_path/")
    .queryParam("param1", "value1")
    .build();

URI uri = builder.build();

OkHttp (JavaDoc) (GitHub)

URL url = new HttpUrl.Builder()
        .scheme("http")
        .host("your_domain")
        .addPathSegments("your_path")
        .addQueryParameter("param1", "value1")
        .build()
        .url();

If you want to use Java 8 without additional dependencies, you could use URI class' constructor directly but it's not so friendly as the previous alternatives.

URI class (JavaDoc)

URI uri = new URI("your_host", "your_path", "param1=value1");

What is the right way to safely concatenate an URL without forgetting a separator?

2 Answers2