我有以下 php 脚本，用户可以在其中通过他的 Android 手机对玩家进行评分并将该数据发布到我的服务器中。
所以:
<?php
session_start();
require "init.php";
header('Content-type: application/json');
error_reporting(E_ALL);
ini_set("display_errors", 1);
$id = isset($_POST['player_id']);
$user_id = isset($_POST['User_Id']);
$best_player = isset($_POST['player']);
$rate = isset($_POST['rating']);
if($id && $user_id && $best_player && $rate){
$sql_query = "insert into rating_players_table values('$id','$best_player','$rate','$user_id');";
if(mysqli_query($con,$sql_query)){
$don = array('result' =>"success","message"=>"Επιτυχής πρόσθεση παίχτη");
}
}else if(!$best_player){
$don = array('result' =>"fail","message"=>"Insert player name");
}else if(!$rate){
$don = array('result' =>"fail","message"=>"Rate player");
}
echo json_encode($don);
?>
但是我收到一条消息说 echo 中的 $don 变量未被识别。
发送数据的Android代码是:
private void ratingPlayer(final String player, final int rating,final String UserId) {
String tag_string_req = "req_register";
StringRequest strReq = new StringRequest(Request.Method.POST,
URL.URL_BEST_PLAYERS, new Response.Listener<String>() {
@Override
public void onResponse(String response) {
Log.d("Response", "Best player response: " + response.toString());
try {
JSONObject jsonObject = new JSONObject(response);
if (jsonObject.getString("result").equals("success")) {
Toast.makeText(getApplicationContext(), jsonObject.getString("message"), Toast.LENGTH_LONG).show();
} else if (jsonObject.getString("result").equals("fail")) {
Toast.makeText(getApplicationContext(), jsonObject.getString("message"), Toast.LENGTH_LONG).show();
}
} catch (JSONException e) {
e.printStackTrace();
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.e("Error", "Registration Error: " + error.getMessage());
Toast.makeText(getApplicationContext(),
error.getMessage(), Toast.LENGTH_LONG).show();
}
}) {
@Override
protected Map<String, String> getParams() {
// Posting params to register urls
Map<String, String> params = new HashMap<String, String>();
params.put("player_id", "");
params.put("User_Id", UserId);
params.put("player", player);
params.put("rating", String.valueOf(rating));
return params;
}
};
// Adding request to request queue
AppController.getInstance().addToRequestQueue(strReq, tag_string_req);
}
在哪里
final String player, final int rating,final String UserId
我在 Android 手机的编辑文本字段中输入的值。
有什么问题吗？
谢谢，
西奥。
最佳答案
这样更准确
$id_isSet = isset($_POST['player_id']);
$user_id_isSet = isset($_POST['User_Id']);
$best_player_isSet = isset($_POST['player']);
$rate_isSet = isset($_POST['rating']);
if($id_isSet && $user_id_isSet && $best_player_isSet && $rate_isSet){
$id = $_POST["player_id"];
$user_id = $_POST['User_Id'];
$best_player = $_POST['player'];
$rate = $_POST['rating'];
$sql_query = "INSERT INTO rating_players_table VALUES('$id','$best_player','$rate','$user_id');";
if(mysqli_query($con,$sql_query)){
$don = array('result' =>"success","message"=>"Επιτυχής πρόσθεση παίχτη");
}
}else if(!$best_player){
$don = array('result' =>"fail","message"=>"Insert player name");
}else if(!$rate){
$don = array('result' =>"fail","message"=>"Rate player");
}
echo json_encode($don);
?>
如果你回显 isset($_POST["id"]) 并且是真的你将打印“1”而不是实际的 post 值
关于无法识别 php 变量。，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/38414491/