Question

当我试图通过php mysql在android中将图像显示到imageview中时，我的代码出错了。我正在使用JSON来解析图像..

所以我有一张名为movie的表..

movie
id_movie   |   Tittle   |    link_poster
-------------------------------------------------------------------
 MP01      |    Frozen  |http://10.0.2.2/cinemainfo/image/movie1.jpg

我将图像保存到htdocs中名为cinemainfo / image的文件夹中。

所以这是我的detail.php代码：

$response = array();

$id_movie = $_REQUEST['id_movie'];

$sql="select tittle, link_poster from movie where id_movie = '".$id_movie."'";

$result = mysql_query($sql);

 if (mysql_num_rows($result) > 0)
      {
     $response["detail"] = array();

     while ($row = mysql_fetch_array($result))
        {
        $detail = array();
        $detail["tittle"] = stripslashes($row["tittle"]);
        $detail["link_poster"] = base64_encode($row["link_poster"]);

        array_push($response["detail"], $detail);
        }

     $response["success"] = 1;

     echo json_encode($response);
      }
      else {

      $response["success"] = 0;
      $response["message"] = "No data";

      echo json_encode($response);
        }

这是我的detail.java代码：

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.detail);

    Bundle b = this.getIntent().getExtras(); 
    kode = b.getString("kode_intent");

    tittle=(TextView)findViewById(R.id.tv_tittle);
    poster=(ImageView)findViewById(R.id.img_poster);

    ArrayList<NameValuePair> postParameters = new ArrayList<NameValuePair>();
    postParameters.add(new BasicNameValuePair("id_movie", kode));

    String response = null;

    try {
        response = CustomHttpClient.executeHttpPost("http://10.0.2.2/cinemainfo/detail.php", postParameters);
        String result = response.toString();

            try {
                JSONArray jArray = new JSONArray(result);
                JSONObject json_data=null;
                for(int i=0;i<jArray.length();i++){

                     json_data = jArray.getJSONObject(i);

                     ttl =json_data.getString("tittle");

                     pstr = json_data.getString("link_poster"); 
                }

                byte[] rawImage = Base64.decode(pstr, Base64.DEFAULT);
                bmp = BitmapFactory.decodeByteArray(rawImage, 0, rawImage.length); 

            }

            catch(JSONException e){
                Log.e("log_tag", "Error parsing data "+e.toString());
                }

            try {
                tittle.setText(ttl);
                poster.setImageBitmap(bmp);
                 }

            catch(Exception e){
                Log.e("log_tag","Error in Display!" + e.toString());;          
               }
        }

        catch(Exception e){
            Log.e("log_tag", "Error in http connection"+e.toString());
        }
}
  }

这是日志猫：

E/log_tag(338): Error parsing data org.json.JSONException: Value {"success":1,"detail":[{"link_poster":"aHR0cDovLzEwLjAuMi4yL2NpbmVtYWluZm8vaW1hZ2UvZmlsbTEuanBn","tittle":"Frozen"}]} of type org.json.JSONObject cannot be converted to JSONArray

当我运行这个网站时："http://10.0.2.2/cinemainfo/detail.php"我想确保我的PHP代码很好......它是.. 这是我的PHP代码的结果：

{"detail":[
    {"tittle":"Frozen",
     "link_poster":"aHR0cDovLzEwLjAuMi4yL2NpbmVtYWluZm8vaW1hZ2UvZmlsbTEuanBn"
     }
          ],"success":1
 }

有谁知道如何解决我的问题？我陷入困境以掩盖这个问题..任何帮助都会非常有帮助谢谢你:)）

Answer 1

您在json解析时遇到错误请尝试此代码。

 ArrayList<String> title= new ArrayList<String>();
        ArrayList<String> imgurl=new ArrayList<String>();
        String status="";

        try { 

 JSONObject result=new JSONObject(jsonouput);
 status    = result.getJSONObject("success");
 JSONArray jArray = new JSONArray(result.getString("detail"));

 for(int i=0;i<jArray.length();i++){

json_data = jArray.getJSONObject(i).toString();

title.add(json_data.getString("tittle"));

imgurl.add(json_data.getString("link_poster")); 

          }

使用Json将PHP中的图像显示到Android中的imageview（错误解析数据）

1 个答案: