Question

这是Curl命令

curl -H "Content-Type: application/json" -X POST --data '[{"a": 4, "b": 5}\]' 
\
-u person@email.com:111sdreiwoiewur_y \
http://mysite.website.com/api/v1/A/B/P

我尝试将上面的curl请求转换为scala postRequest。 curl请求成功运行，但代码遇到400错误。

import java.util
import java.util.ArrayList

import com.google.gson.Gson
import org.apache.catalina.util.Base64
import org.apache.http.client.entity.UrlEncodedFormEntity
import org.apache.http.client.methods.HttpPost
import org.apache.http.entity.StringEntity
import org.apache.http.impl.client.HttpClientBuilder
import org.apache.http.message.BasicNameValuePair
import org.apache.http.{HttpHeaders, NameValuePair}

 case class dummy_information( a:Int, b:Int)

object postTest extends App {

  val restUrl="http://mysite.website.com/api/v1/A/B/P"
  val TOKEN= "111sdreiwoiewur_y"
  val username="person@email.com"

  val dummyCaseObj =  dummy_information(4,5)
  val dummyAsJson: String = new Gson().toJson(dummyCaseObj)


  val httpPost: HttpPost = new HttpPost(restUrl)

  val base64=username+":"+ TOKEN
  val encoding: String =Base64.encode(base64.getBytes("UTF-8"))
  httpPost.setHeader(HttpHeaders.AUTHORIZATION,"Basic "+new String(encoding))
  httpPost.setHeader(HttpHeaders.ACCEPT,"application/json")
  httpPost.setHeader(HttpHeaders.CONTENT_TYPE, "application/json")

  //passing payload
  httpPost.setEntity(new StringEntity(dummyAsJson))

  //POST response
  val client = HttpClientBuilder.create().build()
  var response = client.execute(httpPost)

  println("Response Code : " +response.getStatusLine().getStatusCode())

}

Responsecode即将400。我如何更改代码以获得200响应？

Answer 1

dummyAsJson在下面的行是JSONObject，curl接受为JSONArray。 httpPost.setEntity（new StringEntity（dummyAsJson））

因此改为上线 httpPost.setEntity（new StringEntity（＆＃34; [＆＃34; + dummyAsJson +＆＃34;]＆＃34;））

使用Basic64授权，Scala HttpPost请求引发400错误

这是Curl命令

1 个答案: