Question

I am passing back a JSON object (consisting of an array of strings) from php. I am trying to convert the object into a Java jArray but the string I get back from my php file isn’t formed correctly:

String from php (I have 2 entries in my db tips table)

[{"0":"2","id":"2","1":"2","household_id":"2","2":"3","stepgreen_id":"3","3":"tip 1","tip":"tip 1","4":"2011-08-05","dateOfTip":"2011-08-05","5":"3","likes":"3"}]

[{"0":"2","id":"2","1":"2","household_id":"2","2":"3","stepgreen_id":"3","3":"tip 1","tip":"tip 1","4":"2011-08-05","dateOfTip":"2011-08-05","5":"3","likes":"3"},{"0":"91","id":"91","1":"1","household_id":"1","2":"1","stepgreen_id":"1","3":"tip 2","tip":"tip 2","4":"2011-08-04","dateOfTip":"2011-08-04","5":"1","likes":"1"}]

Here is my php code

    <?php
    // mysql connection, etc....
    $query  = "SELECT * FROM Tips";
    $result = mysql_query($query);
    while($row = mysql_fetch_array($result))
    {
    $output[]=$row;
    print(json_encode($output));
    } 
    mysql_close($con);
    ?>

In Java, I do the following. What happens is that I get back a jArray with only one entry. I expect 2. I’m not sure why the php json object is returning an array of 1 string and a second array of 2 strings. I only expect to receive an array of 2 strings.

     try {
      InputStream responseData;
      responseData = httpEntity.getContent();
      js = convertStreamToString(responseData);
      Log.v(LOG_TAG, js);
      jArray = new JSONArray(js);

      JSONObject json_data = null;

      Log.v(LOG_TAG, "Arraysize: " + jArray.length());
         for (int i = 0; i < jArray.length(); i++) {
        Log.v(LOG_TAG, "entering loop");
        json_data = jArray.getJSONObject(i);
        tip = json_data.getString("tip");
        stepgreenId = json_data.getString("stepgreen_id");
      dateOfTip = json_data.getString("dateOfTip");
      householdId = json_data.getString("household_id");
      likes = json_data.getString("likes");
      Log.v(LOG_TAG, "Json tip= " + tip + " stepgreen id: " + stepgreenId +  " household_id: " + householdId + " likes: " + likes);

    }
} catch (IllegalStateException e1) {
    // TODO Auto-generated catch block
    e1.printStackTrace();
    } catch (IOException e1) {
    // TODO Auto-generated catch block
    e1.printStackTrace();
    }catch (JSONException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

Answer 1

If you do the following:

$first_array = array('first array');
$second_array = array('second array');

echo json_encode($first_array);
echo json_encode($second_array);

The JSON that you will get is invalid if you want to deserialize it. You have to create a common array for them and then print it out. In your case:

<?php
// mysql connection, etc....
$query  = "SELECT * FROM Tips";
$result = mysql_query($query);
$output = array();
while($row = mysql_fetch_array($result))
    $output[]=$row;

print(json_encode($output));
mysql_close($con);

This should work.