1

No value for {"username":"sara"}{"username":"john"} JSONException当我尝试使用 android 应用程序从 MySql 数据库中查看数据时,我得到了一个。

  08-20 04:26:39.396: W/System.err(4732): org.json.JSONException: No value for {"username":"sara"}{"username":"john"}
  08-20 04:26:39.497: W/System.err(4732):   at org.json.JSONObject.get(JSONObject.java:354)
  08-20 04:26:39.497: W/System.err(4732):   at org.json.JSONObject.getJSONArray(JSONObject.java:544)
  08-20 04:26:39.517: W/System.err(4732):   at com.example.phpapp.ViewData$MyTask.doInBackground(ViewData.java:69)
  08-20 04:26:39.577: W/System.err(4732):   at com.example.phpapp.ViewData$MyTask.doInBackground(ViewData.java:1)
  08-20 04:26:39.606: W/System.err(4732):   at android.os.AsyncTask$2.call(AsyncTask.java:287)
  08-20 04:26:39.606: W/System.err(4732):   at java.util.concurrent.FutureTask.run(FutureTask.java:234)
  08-20 04:26:39.631: W/System.err(4732):   at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
  08-20 04:26:39.636: W/System.err(4732):   at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1080)
  08-20 04:26:39.636: W/System.err(4732):   at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:573)
  08-20 04:26:39.636: W/System.err(4732):   at java.lang.Thread.run(Thread.java:856)

PHP 代码

$con = mysql_connect($dbhost, $dbuser, $dbpass) or die("can't connect");
$db = mysql_select_db($dbname) or die("no such db name");
$query = "select username from users";
$result = mysql_query($query);
while($row=  mysql_fetch_array($result)){
    echo json_encode($row);
}

json数组

{"username":"sara"}{"username":"john"}

安卓代码

HttpResponse response = client.execute(post);
HttpEntity entity = response.getEntity();
InputStream is = entity.getContent();
BufferedReader br = new BufferedReader(new InputStreamReader(is));

String str = "";
while((str = br.readLine()) != null){
    s.append(str+"\n");
    mystr = s.toString();
}   

JSONObject object = new JSONObject(mystr);
JSONArray Jarray = object.getJSONArray(mystr);
for (int i = 0; i < Jarray.length(); i++) {
    JSONObject Jasonobject = Jarray.getJSONObject(i);
    st = Jasonobject.getString("username");
    alist.add(str);
}

如何解决这个问题?谢谢

4

2 回答 2

2

问题出在这一行

 JSONArray Jarray = object.getJSONArray(mystr);

哪里mystr={"username":"sara"}{"username":"john"} (来自输入流的数据)

所以在这里你试图mystr从 JSONObject访问一个带有索引的 JSON 数组object

从技术上讲,它永远不存在,所以它给出了 JSONException: No value for {"username":"sara"}{"username":"john"}

解决方案

然后使用 JJPA 的答案更正 PHP 代码

利用

JSONArray Jarray=new JSONArray(mystr);
于 2013-08-20T04:45:06.873 回答
1

这个

{"username":"sara"}{"username":"john"}

不是 json 数组,甚至不是有效的 json 字符串。

如果这是一个 json 数组,那么它将如下所示,

[{"username":"sara"},{"username":"john"}]

所以你需要在这些对象和 FYI json 对象之间有一个逗号,{}并且数组将在[]. 因此,您需要更正 json 字符串,然后为此编写 json 解析器。

您的 php 代码应修改如下,打印 json 数组,

$con = mysql_connect($dbhost, $dbuser, $dbpass) or die("can't connect");
$db = mysql_select_db($dbname) or die("no such db name");
$query = mysql_query("select username from users");
$return_arr = array();

while ($row = mysql_fetch_array($query, MYSQL_ASSOC)) {
    $row_array['id'] = $row['id'];
    $row_array['col1'] = $row['col1'];// and your respective cols
    array_push($return_arr,$row_array);
}

echo json_encode($return_arr);
于 2013-08-20T04:45:53.460 回答