Is there a right way to build a URL?

Question

In much of the code I work with there is horrible stuff like:

String url = "../Somewhere/SomeServlet?method=AMethod&id="+object.getSomething()+ "&aParam="+object.getSomethingElse());

or - even worse:

String url = "Somewhere/Here/Something.jsp?path="+aFile.toString().replace("\\","/")+ "&aParam="+object.getSomethingElse());

Is there a right way to:

Create a new URL (or is it a URI).
Add correctly escaped parameters to it.
Add well-formed file paths in those params.
Resolve it to a String.

Essentially - it is too easy to just build the string than it is to do it properly. Is there a way to do it properly that is as easy as just building the string?

Added

For clarity - and after a little thought - I suppose I am looking for something like:

String s = new MyThing()
    .setPlace("Somewhere/Something.jsp")
    .addParameter(aName,aValue)
    .addParameter(aName,aFile)
    .toString();

so that it will deal with all of the unpleasantness of escaping and adding "?"/"&" and changing "\" to "/" instead of using "\" for files etc.

If I have to write one myself (i.e. if Apache is not an option) are there real Java techniques for correctly escaping the various parts. I mean things like escaping " " in parameters as "." while escaping " " in other places a "%20".

I think this seems similar. http://stackoverflow.com/questions/1861620/is-there-a-java-package-to-handle-building-urls — peter.petrov, Oct 23 '13 at 09:59
Also, http://stackoverflow.com/questions/883136/is-there-a-good-url-builder-for-java — siledh, Oct 23 '13 at 10:02

score 67 · Answer 1 · edited Jul 29 '20 at 17:14

67

You can use Apache URIBuilder

Sample code: Full Apache Example

URIBuilder builder = new URIBuilder()
    .setScheme("http")
    .setHost("apache.org")
    .setPath("/shindig")
    .addParameter("helloWorld", "foo&bar")
    .setFragment("foo");
builder.toString();

Output: http://apache.org/shindig?helloWorld=foo%26bar#foo

edited Jul 29 '20 at 17:14

calvin

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answered Oct 23 '13 at 11:10

Barun

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It's `URIBuilder`, not `UriBuilder`. Last URL is not valid, and I'd remove link to that example. – ooi Jan 29 '18 at 14:39
This also works with only query parameter, in case you already have a predefined base URL – Markus Dec 15 '20 at 06:49

score 33 · Answer 2 · answered Jun 17 '17 at 16:04

33

You can also use spring UriComponentsBuilder

UriComponentsBuilder
    .fromUriString(baseUrl)
    .queryParam("name", name)
    .queryParam("surname", surname)
    .build().toUriString();

answered Jun 17 '17 at 16:04

Marcin Szymczak

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Be aware that this doesn't encode each query param so you need to do manually https://stackoverflow.com/a/21259193/524588 – Sanghyun Lee Sep 28 '18 at 13:34
4

@SanghyunLee: instead you can follow https://www.baeldung.com/spring-uricomponentsbuilder and use .build().encode().toUriString(). – Dan Locks Oct 24 '19 at 20:57

Jeroen Vannevel · Accepted Answer · 2013-10-23T10:57:47.387

I have written this up, you can change it where you want extra functionality. It doesn't use any external resources, let me know if I have looked over something!

It's basically a wrapper for the URI class that allows you to more easily add subdirectories and parameters to the URI. You can set default values if you're not interested in some things.

Edit: I have added an option to use a relative URI (per your question).

public class Test {
    public static void main(String[] args) throws URISyntaxException,
            MalformedURLException {
        URLBuilder urlb = new URLBuilder("www.example.com");
        urlb.setConnectionType("http");
        urlb.addSubfolder("somesub");
        urlb.addSubfolder("anothersub");
        urlb.addParameter("param lol", "unknown");
        urlb.addParameter("paramY", "known");
        String url = urlb.getURL();
        System.out.println(url);


        urlb = new URLBuilder();
        urlb.addSubfolder("servlet");
        urlb.addSubfolder("jsp");
        urlb.addSubfolder("somesub");
        urlb.addSubfolder("anothersub");
        urlb.addParameter("param lol", "unknown");
        urlb.addParameter("paramY", "known");
        String relUrl = urlb.getRelativeURL();
        System.out.println(relUrl);
    }
}

class URLBuilder {
    private StringBuilder folders, params;
    private String connType, host;

    void setConnectionType(String conn) {
        connType = conn;
    }

    URLBuilder(){
        folders = new StringBuilder();
        params = new StringBuilder();
    }

    URLBuilder(String host) {
        this();
        this.host = host;
    }

    void addSubfolder(String folder) {
        folders.append("/");
        folders.append(folder);
    }

    void addParameter(String parameter, String value) {
        if(params.toString().length() > 0){params.append("&");}
        params.append(parameter);
        params.append("=");
        params.append(value);
    }

    String getURL() throws URISyntaxException, MalformedURLException {
        URI uri = new URI(connType, host, folders.toString(),
                params.toString(), null);
        return uri.toURL().toString();
    }

    String getRelativeURL() throws URISyntaxException, MalformedURLException{
        URI uri = new URI(null, null, folders.toString(), params.toString(), null);
        return uri.toString();
    }
}

Output:

Absolute

http://www.example.com/somesub/anothersub?param%20lol=unknown&paramY=known

Relative

/servlet/jsp/somesub/anothersub?param%20lol=unknown&paramY=known

I eventually used a variant of this technique [like this](http://stackoverflow.com/a/19541585/823393). — OldCurmudgeon, Oct 31 '13 at 15:58
This is NOT correct as it does not escape parameters or paths! — Alexander Klimetschek, Mar 15 '19 at 23:17
I would follow the answers pointing to the Apache and Spring URI builders. That code will be well used and covered by tests! — pmckeown, May 24 '20 at 20:49

score 1 · Answer 4 · edited May 17 '14 at 22:03

I like @Jeroen's suggestion but it didn't quite do all I wanted so, using his idea of gathering the parts together and then using a URI to grow the final result I put together this solution which seems to do what I want:

public class URLBuilder {
  // The scheme - http
  private String scheme = null;
  // The user - user
  private String user = null;
  // The host - example.com
  private String host = null;
  // The port - 8080
  private int port = -1;
  // The paths - /Path/To/Somewhere/index.jsp
  private final ArrayList<String> paths = new ArrayList<String>();
  // The parameters - ?a=b&c=d
  private final ArrayList<Pair<String, String>> queries = new ArrayList<Pair<String, String>>();
  // The fragment - #n
  private String fragment = null;

  public URLBuilder addQuery(String name, String value) {
    queries.add(new Pair(name, value));
    return this;
  }

  public URLBuilder addQuery(String name, long value) {
    addQuery(name, String.valueOf(value));
    return this;
  }

  public URLBuilder addQuery(String name, File file) {
    addQuery(name, file.toURI().getPath());
    return this;
  }

  public URLBuilder addPath(String path) {
    paths.add(path);
    return this;
  }

  @Override
  public String toString() {
    // Build the path.
    StringBuilder path = new StringBuilder();
    for (String p : paths) {
      path.append("/").append(p);
    }
    // Build the query.
    StringBuilder query = new StringBuilder();
    String sep = "";
    for (Pair<String, String> p : queries) {
      query.append(sep).append(p.p).append("=").append(p.q);
      sep = "&";
    }
    String url = null;
    try {
      URI uri = new URI(
              scheme,
              user,
              host,
              port,
              path.length() > 0 ? path.toString() : null,
              query.length() > 0 ? query.toString() : null,
              fragment);
      url = uri.toString();
    } catch (URISyntaxException ex) {
      Logger.getLogger(URLBuilder.class.getName()).log(Level.SEVERE, null, ex);
    }

    return url;
  }

  /**
   * @param host the host to set
   * @return this
   */
  public URLBuilder setHost(String host) {
    this.host = host;
    return this;
  }

  /**
   * @param scheme the scheme to set
   * @return this
   */
  public URLBuilder setScheme(String scheme) {
    this.scheme = scheme;
    return this;
  }

  /**
   * @param user the user to set
   * @return this
   */
  public URLBuilder setUser(String user) {
    this.user = user;
    return this;
  }

  /**
   * @param port the port to set
   * @return this
   */
  public URLBuilder setPort(int port) {
    this.port = port;
    return this;
  }

  /**
   * @param fragment the fragment to set
   * @return this
   */
  public URLBuilder setFragment(String fragment) {
    this.fragment = fragment;
    return this;
  }

  public static void main(String args[]) {
    try {
      URLBuilder url = new URLBuilder();
      System.out.println(url.toString());
      url.setFragment("fragment");
      System.out.println(url.toString());
      url.setHost("host.com");
      System.out.println(url.toString());
      url.addPath("APath");
      System.out.println(url.toString());
      url.addPath("AnotherPath");
      System.out.println(url.toString());
      url.addQuery("query1", "param1");
      System.out.println(url.toString());
      url.addQuery("query 2", "param 2");
      System.out.println(url.toString());
      url.addQuery("file", new File("Hello World.txt"));
      System.out.println(url.toString());
    } catch (Throwable t) {
      t.printStackTrace(System.err);
    }
  }

}

Could you clarify what functionality you have added? Glad I could help :) — Jeroen Vannevel, Oct 23 '13 at 16:02
@JeroenVannevel - I added the handling of `File` as queries and a more complete use of the `URI` class. I plan to mark your answer as correct as it inspired my builder. I am waiting for any late but inspiring suggestions. — OldCurmudgeon, Oct 23 '13 at 23:51

score -6 · Answer 5 · answered Oct 23 '13 at 09:57

-6

Recomendations

private final String BASE_URL = Properties.getProperty("base-url");

private Map propertiesMap; // = new HashMap<String,String>();

and in the code to build the URL.

public String buildURL(){
    StringBuilder builder = new StringBuilder();
    builder.append(BASE_URL);
    //for each property, append it

    return builder.toString();

}

answered Oct 23 '13 at 09:57

RamonBoza

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Downvoting, because this is only half-baked code that re-invents the wheel. Also it doesn't address all requirements of the OP. – Hubert Grzeskowiak Jan 05 '17 at 13:17

Is there a right way to build a URL?

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