HttpServletRequest完成URL

Question

我有一個HttpServletRequest對象。

如何獲得導致此調用到達我的servlet的完整URL和確切URL？或者至少儘可能準確，因爲有可能重新生成的東西（可能是參數的順序）。

Answer 1

的HttpServletRequest有以下方法：

• getRequestURL() - 返回完整的URL的查詢字符串分隔符之前的部分?
• getQueryString() - 查詢字符串分隔符?後返回完整的URL的一部分

所以，要得到完整的網址，只需要：

public static String getFullURL(HttpServletRequest request) { 
    StringBuffer requestURL = request.getRequestURL(); 
    String queryString = request.getQueryString(); 

    if (queryString == null) { 
     return requestURL.toString(); 
    } else { 
     return requestURL.append('?').append(queryString).toString(); 
    } 
} 

Answer 2

結合getRequestURL()和getQueryString()的結果應該會得到您想要的結果。

Answer 3

HttpUtil被棄用，這是正確的方法

StringBuffer url = req.getRequestURL(); 
String queryString = req.getQueryString(); 
if (queryString != null) { 
    url.append('?'); 
    url.append(queryString); 
} 
String requestURL = url.toString(); 

Answer 4

// http://hostname.com/mywebapp/servlet/MyServlet/a/b;c=123?d=789 

public static String getUrl(HttpServletRequest req) { 
    String reqUrl = req.getRequestURL().toString(); 
    String queryString = req.getQueryString(); // d=789 
    if (queryString != null) { 
     reqUrl += "?"+queryString; 
    } 
    return reqUrl; 
} 

Answer 5

我用這個方法：

public static String getURL(HttpServletRequest req) { 

    String scheme = req.getScheme();    // http 
    String serverName = req.getServerName();  // hostname.com 
    int serverPort = req.getServerPort();  // 80 
    String contextPath = req.getContextPath(); // /mywebapp 
    String servletPath = req.getServletPath(); // /servlet/MyServlet 
    String pathInfo = req.getPathInfo();   // /a/b;c=123 
    String queryString = req.getQueryString();   // d=789 

    // Reconstruct original requesting URL 
    StringBuilder url = new StringBuilder(); 
    url.append(scheme).append("://").append(serverName); 

    if (serverPort != 80 && serverPort != 443) { 
     url.append(":").append(serverPort); 
    } 

    url.append(contextPath).append(servletPath); 

    if (pathInfo != null) { 
     url.append(pathInfo); 
    } 
    if (queryString != null) { 
     url.append("?").append(queryString); 
    } 
    return url.toString(); 
} 

Answer 6

您可以使用過濾器。

@Override 
    public void doFilter(ServletRequest arg0, ServletResponse arg1, FilterChain arg2) throws IOException, ServletException { 
      HttpServletRequest test1= (HttpServletRequest) arg0; 

     test1.getRequestURL()); it gives http://localhost:8081/applicationName/menu/index.action 
     test1.getRequestURI()); it gives applicationName/menu/index.action 
     String pathname = test1.getServletPath()); it gives //menu/index.action 


     if(pathname.equals("//menu/index.action")){ 
      arg2.doFilter(arg0, arg1); // call to urs servlet or frameowrk managed controller method 


      // in resposne 
      HttpServletResponse httpResp = (HttpServletResponse) arg1; 
      RequestDispatcher rd = arg0.getRequestDispatcher("another.jsp");  
      rd.forward(arg0, arg1); 





    } 

DONOT忘記把<dispatcher>FORWARD</dispatcher>在過濾器映射在web.xml中

Answer 7

使用上的HttpServletRequest以下方法對象

java.lang.String中getRequestURI() - 返回該請求的URL從部分協議名稱直到HTTP請求的第一行中的查詢字符串。

java.lang.StringBuffer getRequestURL() - 重建客戶端用來發出請求的URL。

java.lang.String getQueryString() - 返回路徑後請求URL中包含的查詢字符串。

Answer 8

在Spring項目，你可以使用

UriComponentsBuilder.fromHttpRequest(new ServletServerHttpRequest(request)).build().toUriString() 

Answer 9

有點遲到了，但我包括這在我的MarkUtils-Web library在WebUtils - Checkstyle的認可和JUnit測試：

import javax.servlet.http.HttpServletRequest; 

public class GetRequestUrl{ 
    /** 
    * <p>A faster replacement for {@link HttpServletRequest#getRequestURL()} 
    * (returns a {@link String} instead of a {@link StringBuffer} - and internally uses a {@link StringBuilder}) 
    * that also includes the {@linkplain HttpServletRequest#getQueryString() query string}.</p> 
    * <p><a href="https://gist.github.com/ziesemer/700376d8da8c60585438" 
    * >https://gist.github.com/ziesemer/700376d8da8c60585438</a></p> 
    * @author Mark A. Ziesemer 
    * <a href="http://www.ziesemer.com.">&lt;www.ziesemer.com&gt;</a> 
    */ 
    public String getRequestUrl(final HttpServletRequest req){ 
     final String scheme = req.getScheme(); 
     final int port = req.getServerPort(); 
     final StringBuilder url = new StringBuilder(256); 
     url.append(scheme); 
     url.append("://"); 
     url.append(req.getServerName()); 
     if(!(("http".equals(scheme) && (port == 0 || port == 80)) 
       || ("https".equals(scheme) && port == 443))){ 
      url.append(':'); 
      url.append(port); 
     } 
     url.append(req.getRequestURI()); 
     final String qs = req.getQueryString(); 
     if(qs != null){ 
      url.append('?'); 
      url.append(qs); 
     } 
     final String result = url.toString(); 
     return result; 
    } 
} 

可能這是迄今爲止Mat Mat Banik的最快速和最健壯的答案 - 但即使他沒有考慮使用HTTP/HTTPS進行潛在的非標準端口配置。

參見：

• http://blogger.ziesemer.com/2017/08/httpservletrequestgetrequesturl.html
• https://gist.github.com/ziesemer/700376d8da8c60585438