Question

我有这个，

@POST
@Path("/login")
public Response postLogin() throws IOException, SQLException,NamingException {
    return Response.temporaryRedirect(URI.create("localdir")).build();
}

它返回POST。但是我想获得相同帖子的GET方法。

我该怎么做？

Answer 1

试试这个，

@POST
@Path("/login")
public Response postLogin() throws IOException, SQLException,NamingException {

    URL obj; 

    obj = new URL(url);
    HttpURLConnection con = (HttpURLConnection) obj.openConnection();
    con.setRequestMethod("GET");
    int responseCode = con.getResponseCode();

    BufferedReader in = new BufferedReader(
            new InputStreamReader(con.getInputStream()));
    String inputLine;
    StringBuffer response = new StringBuffer();

    while ((inputLine = in.readLine()) != null) {
        response.append(inputLine);
    }
    in.close();         
    return response;
}

或者，

@POST
@Path("/login")
public Response postLogin() throws IOException, SQLException,NamingException {

    Request.Get("your url").execute().returnContent();
}

或者，

@POST
@Path("/login")
public Response postLogin() throws IOException, SQLException,NamingException {
    String uri = "your url";
    HttpParams httpParams = new BasicHttpParams();
    HttpClient client = new DefaultHttpClient(httpParams);
    HttpGet request = new HttpGet(uri);
    HttpResponse response = client.execute(request);
    String responseStr = buildResponseFromEntity(response.getEntity());

    private String buildResponseFromEntity(HttpEntity entity)throws IllegalStateException, IOException {

        BufferedReader r = new BufferedReader(new InputStreamReader(entity.getContent()));
        StringBuilder total = new StringBuilder();
        String line;
        while ((line = r.readLine()) != null) {
            total.append(line);
        }
        return total.toString();
    }
    // check if error
    if (response.getStatusLine().getStatusCode() != 200) {
        JSONObject jsonObj = null;
        try {
            jsonObj = new JSONObject(responseStr);
        } catch (Exception e) {
            // do your code
        }
    }
}

Answer 2

尽管答案是对的，并且因为我将它用于HTML，所以这是一种简单的黑客方法。

    String s = "<script>window.location.href="%s"</script>";
    s = String.format(s, "localdir");
    return Response.status(Response.Status.OK)
            .entity(s)
            .build();

如何在JAX-RS中从@POST结束返回GET响应

2 个答案: