Php Android HttpPost没有';不要发送数据
我正在使用HttpPost从我的服务器请求代码,但在添加其他参数之前,它工作得很好。这是我的代码:Php Android HttpPost没有';不要发送数据,php,android,post,http-post,Php,Android,Post,Http Post,我正在使用HttpPost从我的服务器请求代码,但在添加其他参数之前,它工作得很好。这是我的代码: private String getShareCode(String t, String d, String lat, String lng) { String respTxt; try{ HttpClient dbClient = new DefaultHttpClient(); HttpPost actionPost = new HttpPos
private String getShareCode(String t, String d, String lat, String lng) {
String respTxt;
try{
HttpClient dbClient = new DefaultHttpClient();
HttpPost actionPost = new HttpPost(Config.URL_CODE);
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("t",t));
params.add(new BasicNameValuePair("d",d));
params.add(new BasicNameValuePair("lat",lat));
params.add(new BasicNameValuePair("lng",lng));
actionPost.setEntity(new UrlEncodedFormEntity(params, "UTF-8"));
HttpResponse dbResp = dbClient.execute(actionPost);
HttpEntity ent = dbResp.getEntity();
respTxt = EntityUtils.toString(ent).trim();
Log.i("ShareCode", t + ", " + d + ", " + lat + ", " + lng);
Log.i("ShareCode",respTxt);
}catch(Exception e){
e.printStackTrace();
Log.i("ShareCode", String.valueOf(e));
respTxt = null;
}
return respTxt;
}
我希望你能帮我,我有点紧张…试试这个:
它将向请求对象发送数据
<? //Receive post data
echo $t = $_REQUEST['t'];
$d = $_REQUEST['d'];
$l = $_REQUEST['lat'];
$lo= $_REQUEST['lng'];
显示脚本的第17到20行。最后添加了哪个参数?第17到20行是上面PHP代码中的行。我添加的最后一个参数是“d”。您已经在没有添加最后一个参数的情况下再次测试了吗?为什么只回显一个参数?使用isset($_POST['d'])检查是否已设置,如果未设置,则回显消息。Log.i(“ShareCode”,String.valueOf(e))代码>最好将其更改为Log.i(“Exception”,即getMessage())代码>
Notice: Undefined index: t in /home/iplaces/public_html/iplaces_url.php on line 17
Notice: Undefined index: d in /home/iplaces/public_html/iplaces_url.php on line 18
Notice: Undefined index: lat in /home/iplaces/public_html/iplaces_url.php on line 19
Notice: Undefined index: lng in /home/iplaces/public_html/iplaces_url.php on line 20
<? //Receive post data
echo $t = $_REQUEST['t'];
$d = $_REQUEST['d'];
$l = $_REQUEST['lat'];
$lo= $_REQUEST['lng'];