java 带有正文的 Apache HttpClient GET
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Apache HttpClient GET with body
提问by Scott Swank
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 HttpClientHttpGet? I am looking for the equivalent of HttpPost#setEntity.
我正在尝试发送一个包含 json 对象的 HTTP GET。有没有办法设置HttpClientHttpGet的主体？我正在寻找等效的 HttpPost#setEntity。
回答by torbinsky
From what I know, you can't do this with the default HttpGet class that comes with the Apache library. However, you can subclass the HttpEntityEnclosingRequestBaseentity and set the method to GET. I haven't tested this, but I think the following example might be what you're looking for:
据我所知，您无法使用 Apache 库附带的默认 HttpGet 类来执行此操作。但是，您可以继承 HttpEntityEnclosureRequestBase实体并将方法设置为 GET。我还没有测试过这个，但我认为下面的例子可能是你正在寻找的：
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;
}
}
Edit:
编辑：
You could then do the following:
然后，您可以执行以下操作：
...
HttpGetWithEntity e = new HttpGetWithEntity();
...
e.setEntity(yourEntity);
...
response = httpclient.execute(e);
回答by icapurro
Using torbinsky's answer I created the above class. This lets me use the same methods for HttpPost.
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;
}
}
回答by user2389517
How we can send request uri in this example just like HttpGet & HttpPost ???
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);
}
回答by Brandon Lyday
In addition torbinsky's answer, you can add these constructors to the class to make it easier to set the uri:
除了 torbinsky 的回答之外，您还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);
}
The setURImethod is inherited from HttpEntityEnclosingRequestBaseand can also be used outside the constructor.
该setURI方法是从HttpEntityEnclosingRequestBase构造函数继承而来的，也可以在构造函数之外使用。
