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我会尝试从 mysql 数据库中检索一些数据到可能的 android 应用程序。这是我的php代码：

<?php
  $dbhost = "localhost";
  $dbuser = "root";
  $dbpass = "";
  $players = array();



  $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die("Could not connect to the database!");

  $db = mysql_select_db('arsenal') or die("Could not select DB!");

  $result = mysql_query("SELECT * FROM Player");

  $i = 0;

    while ($row = mysql_fetch_array($result)) {

    //array_push($data, $row["playerName"]);
    $players[] = $row['playerName']."  ".$row['surName']."          ";
    echo $players[$i];
    $i++;
    }
  print(json_encode($players));

  mysql_close($conn);  
?>

这是我的获取数据方法：

public ArrayList<String> getPlayers(){
        ArrayList<String> retRanks = new ArrayList<String>();
        InputStream is = null;
        String result = "";

        //http post
        try{
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://10.0.2.2/playerData.php");
            HttpResponse response = httpclient.execute(httppost);
            HttpEntity entity = response.getEntity();
            is = entity.getContent();
        } catch(Exception exc){
            retRanks.add("Error in http connection " + exc.toString());
            return retRanks;
        }

        //convert response to string
        try{
            BufferedReader reader = new BufferedReader(new InputStreamReader(is,"iso-8859-1"), 8);
            StringBuilder sb = new StringBuilder();
            String line = null;
            while ((line = reader.readLine()) != null){
                sb.append(line + "\n");
            }
            is.close();

            result=sb.toString();
        } catch(Exception exc){
            retRanks.add("Error converting result " + exc.toString());
            return retRanks;
        }

        //parse json data
        try{
            JSONArray jArray = new JSONArray(result);
            JSONObject json_data = null;
            for(int i=0;i<jArray.length();i++){
                retRanks.add(jArray.get(i).toString());

            }
            return retRanks;
        } catch(JSONException exc){
            retRanks.add("Error parsing data " + exc.toString());
            return retRanks;
        }
    }

我得到这个例外：

org.json.jsonexception value of type java.lang.string Santi cannot be converted to jsonarray

我不太确定我哪里出错了

当我为结果值敬酒时，它显示：

Santi Carzola Alex Chamberlin Theo Walcott ["Santi Carzola","Alex Chamberlin","Theo Walcott"]
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1 回答 1

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while在您的 php 代码中，您在循环中打印每个名称一次：

echo $players[$i];

然后在打印 json 数组时再次打印： 

print(json_encode($players));

因此，在 中result，您有以下内容：

Santi Carzola Alex Chamberlin Theo Walcott ["Santi Carzola","Alex Chamberlin","Theo Walcott"]

这不是一个JSONArray. 您需要摆脱 之前的内容[，因此echo $players[$i];从循环中删除 ，它应该可以工作。

于 2013-09-11T19:46:32.247 回答