Question

在我使用的大部分代码中都有可怕的东西，如：

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

或 - 更糟糕的是：

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

是否有正确的方法：

创建新网址（或者是URI）。
向其添加正确的转义参数。
在这些参数中添加格式正确的文件路径。
将其解析为字符串。

基本上 - 构建字符串要比正确执行它太容易了。有没有办法正确这就像构建字符串一样简单？

加

为了清楚起见 - 经过一番思考 - 我想我正在寻找类似的东西：

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

这样它就可以处理所有逃避和添加“？”/“＆amp;”的不愉快并将“\”更改为“/”，而不是将“\”更改为文件等。

如果我必须自己编写一个（即如果Apache不是一个选项），那么有真正的Java技术可以正确地转义各个部分。我的意思是在参数中将“”转换为“”。而逃避“”在其他地方“％20”。

Answer 1

您可以使用Apache URIBuilder

示例代码：Full Apache Example

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

输出：http://apache.org/shindig?hello+world=foo%26bar#foo

Answer 2

您还可以使用spring UriComponentsBuilder

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

Answer 3

我已经写完了，您可以在需要额外功能的地方进行更改。它不使用任何外部资源，如果我查看了某些内容，请告诉我！

它基本上是URI类的包装器，允许您更轻松地向URI添加子目录和参数。如果您对某些内容不感兴趣，可以设置默认值。

编辑：我添加了一个使用相对URI的选项（根据您的问题）。

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();
    }
}

输出：

绝对

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

相对

/ servlet的/ JSP / somesub / anothersub PARAM％20lol =未知＆安培; paramY =已知的

Answer 4

我喜欢@Jeroen的建议，但它并不是我想要的所有，使用他的想法收集部分然后使用URI来增长最终结果我把这个解决方案放在一起似乎做我想做的事：

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);
    }
  }

}

Answer 5

Recomendations

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

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

并在用于构建URL的代码中。

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

    return builder.toString();

}

有没有正确的方法来构建URL？

5 个答案: