Question

我正在尝试在其正文中发送带有json对象的HTTP GET。有没有办法设置HttpClient HttpGet的主体？我正在寻找相当于HttpPost＃setEntity。

Answer 1

据我所知，你不能使用Apache库附带的默认HttpGet类。但是，您可以继承 HttpEntityEnclosingRequestBase 实体并将方法设置为GET。我没有对此进行测试，但我认为以下示例可能是您正在寻找的内容：

import org.apache.http.client.methods.HttpEntityEnclosingRequestBase;

public class HttpGetWithEntity extends HttpEntityEnclosingRequestBase {
    public final static String METHOD_NAME = "GET";

    @Override
    public String getMethod() {
        return METHOD_NAME;
    }
}

编辑：

然后您可以执行以下操作：

...
HttpGetWithEntity e = new HttpGetWithEntity();
...
e.setEntity(yourEntity);
...
response = httpclient.execute(e);

Answer 2

使用torbinsky的答案我创建了上面的课程。这让我可以使用与HttpPost相同的方法。

import java.net.URI;

import org.apache.http.client.methods.HttpPost;

public class HttpGetWithEntity extends HttpPost {

    public final static String METHOD_NAME = "GET";

    public HttpGetWithEntity(URI url) {
        super(url);
    }

    public HttpGetWithEntity(String url) {
        super(url);
    }

    @Override
    public String getMethod() {
        return METHOD_NAME;
    }
}

Answer 3

我们如何在这个例子中发送请求uri就像HttpGet＆amp; HttpPost ???

 public class HttpGetWithEntity extends HttpEntityEnclosingRequestBase
 {
    public final static String METHOD_NAME = "GET";
    @Override
     public String getMethod() {
         return METHOD_NAME;
     } 

        HttpGetWithEntity e = new HttpGetWithEntity(); 
        e.setEntity(yourEntity); 
        response = httpclient.execute(e); 
}

Answer 4

除了torbinsky的答案外，您还可以将这些构造函数添加到类中，以使设置uri更容易：

public HttpGetWithEntity(String uri) throws URISyntaxException{
    this.setURI(new URI(uri));
}

public HttpGetWithEntity(URI uri){
    this.setURI(uri);
}

setURI方法是从HttpEntityEnclosingRequestBase继承的，也可以在构造方法之外使用。

Apache HttpClient GET与body

4 个答案: