我正在尝试在 Java 中复制以下 C# 代码。这段代码是一个帮助类，它发送一个包含 xml 的请求，并读取一个响应。
internal static String Send(String url, String body)
{
HttpWebRequest request = (HttpWebRequest)HttpWebRequest.Create(url);
try
{
// create the new httpwebrequest with the uri
request.ContentLength = 0;
// set the method to POST
request.Method = "POST";
if (!String.IsNullOrEmpty(body))
{
request.ContentType = "application/xml; charset=utf-8";
byte[] postData = Encoding.Default.GetBytes(body);
request.ContentLength = postData.Length;
using (Stream s = request.GetRequestStream())
{
s.Write(postData, 0, postData.Length);
}
}
using (HttpWebResponse response = (HttpWebResponse)request.GetResponse())
{
String responseString = new StreamReader(response.GetResponseStream()).ReadToEnd();
if (response.StatusCode != HttpStatusCode.OK)
{
throw new ResponseException(((int)response.StatusCode),
response.StatusCode.ToString(), request.RequestUri.ToString(),
responseString);
}
return responseString;
}
}
catch (WebException e)
{
using (WebResponse response = e.Response)
{
HttpWebResponse httpResponse = response as HttpWebResponse;
if (httpResponse != null)
{
using (Stream data = response.GetResponseStream())
{
data.Position = 0;
throw new ResponseException(((int)httpResponse.StatusCode),
httpResponse.StatusCode.ToString(), request.RequestUri.ToString(),
new StreamReader(data).ReadToEnd()
);
}
}
else
{
throw;
}
在阅读了其他线程后，我确定 Apache HttpComponents 库将是我获得相同功能的最佳选择。阅读文档并按照此处的示例进行操作后：
http://hc.apache.org/httpcomponents-client-ga/quickstart.html
我无法弄清楚如何将正文字符串作为 xml 发送。当我尝试为请求设置实体时，它要求我声明一个 BasicNameValuePair，但我不明白这是什么，或者我将如何格式化正文字符串以满足此规范。以下是我目前所做的。
protected static String Send(String url, String body)
{
HttpPost request = new HttpPost(url);
try
{
request.setHeader("ContentType", "application/xml; charset=utf=8");
// Encode the body if needed
request.setEntity(new UrlEncodedFormEntity());
//get the response
// if the response code is not valid throw a ResponseException
// else return the response string.
} finally {
request.releaseConnection();
}
return null;
}
编辑：或者我应该使用 StringEntity 并执行以下操作
protected static String SendToJetstream(String url, String body)
{
HttpPost request = new HttpPost(url);
try
{
StringEntity myEntity = new StringEntity(body,
ContentType.create("application/xml", "UTF-8"));
// Encode the body if needed
request.setEntity(myEntity);
//get the response
// if the response code is not valid throw a ResponseException
// else return the response string.
} finally {
request.releaseConnection();
}
return null;
}