IAm*_*aja 45 java http apache-httpclient-4.x
我试图让Apache HttpClient触发HTTP请求,然后显示HTTP响应代码(200,404,500等)以及HTTP响应正文(文本字符串).重要的是要注意我正在使用,v4.2.2因为大多数HttpClient示例来自v.3.x.x,并且API从版本3变为版本4.
不幸的是,我只能让HttpClient返回状态代码或响应体(但不是两者).
这就是我所拥有的:
// Getting the status code.
HttpClient client = new DefaultHttpClient();
HttpGet httpGet = new HttpGet("http://whatever.blah.com");
HttpResponse resp = client.execute(httpGet);
int statusCode = resp.getStatusLine().getStatusCode();
// Getting the response body.
HttpClient client = new DefaultHttpClient();
HttpGet httpGet = new HttpGet("http://whatever.blah.com");
ResponseHandler<String> handler = new BasicResponseHandler();
String body = client.execute(httpGet, handler);
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所以我问:使用该v4.2.2库,如何从同一个client.execute(...)调用中获取状态代码和响应正文?提前致谢!
Avi*_*gal 49
不要提供处理程序execute.
获取HttpResponse对象,使用处理程序获取正文并直接从中获取状态代码
HttpResponse response = client.execute(httpGet);
String body = handler.handleResponse(response);
int code = response.getStatusLine().getStatusCode();
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lka*_*mal 32
您可以避免使用BasicResponseHandler,但使用HttpResponse本身将状态和响应都作为String获取.
HttpResponse response = httpClient.execute(get);
// Getting the status code.
int statusCode = response.getStatusLine().getStatusCode();
// Getting the response body.
String responseBody = EntityUtils.toString(response.getEntity());
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esk*_*tos 13
如果状态不是2xx,则抛出BasicResponseHandler.看它的javadoc.
我将如何做到这一点:
HttpResponse response = client.execute( get );
int code = response.getStatusLine().getStatusCode();
InputStream body = response.getEntity().getContent();
// Read the body stream
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或者您也可以从BasicResponseHandler源开始编写一个ResponseHandler,当状态不是2xx时,它不会抛出.
流畅的外观 API:
Response response = Request.Get(uri)
.connectTimeout(MILLIS_ONE_SECOND)
.socketTimeout(MILLIS_ONE_SECOND)
.execute();
HttpResponse httpResponse = response.returnResponse();
StatusLine statusLine = httpResponse.getStatusLine();
if (statusLine.getStatusCode() == HttpStatus.SC_OK) {
// ??????????(???????)
String responseContent = EntityUtils.toString(
httpResponse.getEntity(), StandardCharsets.UTF_8.name());
}
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