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由于我在某些业余时间通过我的Android学习进展,我遇到了HttpPost
请求的奇怪行为。发送来自Android的HttpPost请求,请阅读Post Post from PHP
我想要实现: 请从Android应用程序到Apache Web服务器我的开发PC上运行一个简单的POST请求,并显示从PHP脚本发布的数据,其形式发送。
我的Android应用程序的Java代码驻留在Activity
内为AsyncTask
如下:
private class DoSampleHttpPostRequest extends AsyncTask<Void, Void, CharSequence> {
@Override
protected CharSequence doInBackground(Void... params) {
BufferedReader in = null;
String baseUrl = "http://10.0.2.2:8080/android";
try {
HttpClient httpClient = new DefaultHttpClient();
HttpPost request = new HttpPost(baseUrl);
List<NameValuePair> postParameters = new ArrayList<NameValuePair>();
postParameters.add(new BasicNameValuePair("login", "someuser"));
postParameters.add(new BasicNameValuePair("data", "somedata"));
UrlEncodedFormEntity form = new UrlEncodedFormEntity(postParameters);
request.setEntity(form);
Log.v("log", "making POST request to: " + baseUrl);
HttpResponse response = httpClient.execute(request);
in = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
StringBuffer sb = new StringBuffer("");
String line = "";
String NL = System.getProperty("line.separator");
while ((line = in.readLine()) != null) {
sb.append(line + NL);
}
in.close();
return sb.toString();
} catch (Exception e) {
return "Exception happened: " + e.getMessage();
} finally {
if (in != null) {
try {
in.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
@Override
protected void onPostExecute(CharSequence result) {
// this refers to a TextView defined as a private field in the parent Activity
textView.setText(result);
}
}
我PHP代码如下:
<?php
echo "Hello<br />";
var_dump($_SERVER);
if ($_SERVER["REQUEST_METHOD"] == "POST") {
echo "Page was posted:<br />";
foreach($_POST as $key=>$var) {
echo "[$key] => $var<br />";
}
}
?>
而且最后的问题 : 正如你所见,$_SERVER
内容被转储,并且在输出$_SERVER["REQUEST_METHOD"]
的值为GET
,尽管事实上我实际上正在提出POST
请求。即使我尝试转储$_POST
的内容,也是空的。
我在做什么错?提前致谢。
如果您访问PC上的指定URL,它是否会执行302重定向?您可能需要在URL末尾指定一个尾部斜线。 – Jon
令人惊叹!这是问题所在。并立即解决它,当我在网址 – Trogvar
@Jon后添加尾部斜杠作为答案; Trogvar接受回答:) – verbumSapienti