java - Apache HttpClient GET with body

Question

I am trying to send an HTTP GET with a json object in its body. Is there a way to set the body of an HttpClient HttpGet? I am looking for the equivalent of HttpPost#setEntity.

Answer 1

据我所知，您不能使用 Apache 库附带的默认 HttpGet 类来执行此操作。但是，您可以子类化HttpEntityEnclosureRequestBase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：

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

使用托宾斯基的回答我创建了上面的类。这让我可以对 HttpPost 使用相同的方法。

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

在这个例子中，我们如何像 HttpGet 和 HttpPost 一样发送请求 uri ？？？

 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 的回答，您可以将这些构造函数添加到类中，以便更轻松地设置 uri：

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

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

该setURI方法继承自HttpEntityEnclosingRequestBase构造函数，也可以在构造函数之外使用。