请建议我,如何使用http post方法。
我想将数据发送回PHP服务器,如下所示:
{
"customer_detail": {
"customer_phone": "6774980865",
"customer_comments": "Enter Your comments here...gvhhxjkjs",
"customer_email": "prasanth@dunamis.com",
"customer_name": "prasanth"
},
"feedBack": [
{
"qu_id": "1",
"feedans": "Good"
},
{
"qu_id": "2",
"feedans": "Good"
},
{
"qu_id": "3",
"feedans": "Bad"
},
{
"qu_id": "4",
"feedans": "Bad"
},
{
"qu_id": "5",
"feedans": "Excellent"
}
]
}
请给我解决方案。
答案 0 :(得分:0)
使用此方法发出POST请求:
public String makePOSTRequest(String url, List<NameValuePair> nameValuePairs) {
String response = "";
try {
HttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(url);
httpPost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
HttpResponse httpResponse = httpClient.execute(httpPost);
HttpEntity httpEntity = httpResponse.getEntity();
response = EntityUtils.toString(httpEntity);
} catch (UnsupportedEncodingException e) {
e.printStackTrace();
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
Log.d(LOGTAG, "POST Response >>> " + response);
return response;
}
用法:
在Java中:
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();
nameValuePairs.add(new BasicNameValuePair("json",jsonObject.toString()));
String response = makePOSTRequest(String url, nameValuePairs );
服务器端Php:
$jsonInput = $_POST['json'];
json_decode($jsonInput);