当我浏览弹性搜索的网址时,当我尝试通过代码连接时,它会返回400错误的请求错误。当我请求www.google.com时,它会返回'200 OK'
。可能是什么问题?
这是我的代码:
url = new URL("http://192.168.1.68:9200");
httpCon = (HttpURLConnection) url.openConnection();
httpCon.setDoOutput(true);
httpCon.setRequestMethod("POST");
out = new OutputStreamWriter(
httpCon.getOutputStream());
int responseCode = httpCon.getResponseCode();
String responseMessage = httpCon.getResponseMessage();
System.out.println(responseCode);
System.out.println(responseMessage);
if (httpCon.getResponseCode()==HttpURLConnection.HTTP_OK){
JOptionPane.showMessageDialog(rootPane, "Connection Successful");
}
else {
JOptionPane.showMessageDialog(rootPane, "Connection Error!");
}
out.Close();
答案 0 :(得分:0)
您没有传递任何搜索参数...例如:
url = new URL("http://192.168.1.68:9200");
应该是
url = new URL("http://192.168.1.68:9200/_search?q=whatever+you+want");
或者你应该通过写入流出然后关闭它来发布你的JSON格式的查询(但是你仍然需要在你的URL上标记/ _search。)