我目前正在尝试从 Android 应用程序发送一些数据到 php 服务器(两者都由我控制)。
应用程序中的表单上收集了大量数据,这些数据被写入数据库。这一切都有效。
在我的主代码中,首先我创建一个 JSONObject(在本例中我已将其删减):
JSONObject j = new JSONObject();
j.put("engineer", "me");
j.put("date", "today");
j.put("fuel", "full");
j.put("car", "mine");
j.put("distance", "miles");
接下来,我传递对象以进行发送,并接收响应:
String url = "http://www.server.com/thisfile.php";
HttpResponse re = HTTPPoster.doPost(url, j);
String temp = EntityUtils.toString(re.getEntity());
if (temp.compareTo("SUCCESS")==0)
{
Toast.makeText(this, "Sending complete!", Toast.LENGTH_LONG).show();
}
HTTPPoster 类:
public static HttpResponse doPost(String url, JSONObject c) throws ClientProtocolException, IOException
{
HttpClient httpclient = new DefaultHttpClient();
HttpPost request = new HttpPost(url);
HttpEntity entity;
StringEntity s = new StringEntity(c.toString());
s.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
entity = s;
request.setEntity(entity);
HttpResponse response;
response = httpclient.execute(request);
return response;
}
这会得到响应,但服务器返回 403 - Forbidden 响应。
我尝试稍微更改一下 doPost 函数(这实际上好一点,正如我所说,我有很多要发送,基本上是 3 个具有不同数据的相同表单 - 所以我创建了 3 个 JSONObject,每个表单条目一个 - 条目来自数据库而不是我正在使用的静态示例)。
首先我稍微改变了调用方式:
String url = "http://www.myserver.com/ServiceMatalan.php";
Map<String, String> kvPairs = new HashMap<String, String>();
kvPairs.put("vehicle", j.toString());
// Normally I would pass two more JSONObjects.....
HttpResponse re = HTTPPoster.doPost(url, kvPairs);
String temp = EntityUtils.toString(re.getEntity());
if (temp.compareTo("SUCCESS")==0)
{
Toast.makeText(this, "Sending complete!", Toast.LENGTH_LONG).show();
}
好的,对 doPost 函数的更改:
public static HttpResponse doPost(String url, Map<String, String> kvPairs) throws ClientProtocolException, IOException
{
HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(url);
if (kvPairs != null && kvPairs.isEmpty() == false)
{
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(kvPairs.size());
String k, v;
Iterator<String> itKeys = kvPairs.keySet().iterator();
while (itKeys.hasNext())
{
k = itKeys.next();
v = kvPairs.get(k);
nameValuePairs.add(new BasicNameValuePair(k, v));
}
httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
}
HttpResponse response;
response = httpclient.execute(httppost);
return response;
}
好的,这将返回响应 200
int statusCode = re.getStatusLine().getStatusCode();
然而,服务器上接收到的数据无法解析为 JSON 字符串。我认为它的格式很糟糕(这是我第一次使用 JSON):
如果在 php 文件中我对 $_POST['vehicle'] 进行回显,我会得到以下内容:
{\"date\":\"today\",\"engineer\":\"me\"}
谁能告诉我哪里出了问题,或者是否有更好的方法来实现我想要做的事情?希望以上说得有道理!