Removing the end [ and ] from php Json result for Android Java code

Question

My PHP code is this:

$userdetails = mysqli_query($con, "SELECT *FROM aircraft_status"); 

#$row = mysql_fetch_row($userdetails) ;

while($rows=mysqli_fetch_array($userdetails)){
$status[]= array($rows['Aircraft']=>$rows['Status']);
}
#Output the JSON data
echo json_encode($status);

and gives this:

[{"A70_870":"1"},{"A70_871":"1"},{"A70_872":"1"},{"A70_873":"1"},{"A70_874":"1"},{"A70_875":"1"},{"A70_876":"2"},{"A70_877":"1"},{"A70_878":"2"},{"A70_879":"2"},{"A70_880":"2"},{"A70_881":"0"},{"A70_882":"0"},{"A70_883":"0"},{"A70_884":"0"},{"A70_885":"0"}]

The java code that reads it is this:

// Create a JSON object from the request response
    JSONObject jsonObject = new JSONObject(result);

    //Retrieve the data from the JSON object
        n870 = jsonObject.getInt("A70_870");
        n871 = jsonObject.getInt("A70_871");
        n872 = jsonObject.getInt("A70_872");
        n873 = jsonObject.getInt("A70_873");
        n874 = jsonObject.getInt("A70_874");
        n875 = jsonObject.getInt("A70_875");
        n876 = jsonObject.getInt("A70_876");
        n877 = jsonObject.getInt("A70_877");
        n878 = jsonObject.getInt("A70_878");
        n879 = jsonObject.getInt("A70_879");
        n880 = jsonObject.getInt("A70_880");
        n881 = jsonObject.getInt("A70_881");
        n882 = jsonObject.getInt("A70_882");
        n883 = jsonObject.getInt("A70_883");
        n884 = jsonObject.getInt("A70_884");
        n885 = jsonObject.getInt("A70_885");

When i run my android app I seem to keep getting the error:

"of type org.json.JSONArray cannot be converted into Json object"

However when I send the app dummy code without the square brackets, it seems to work fine! How do I get rid of those [ and ] brackets on the ends???

Alternatively is there a way to accept the json as it is and adapt the java to read it?

removing the brackets will break the json. You've got an array of objects. taking off the brackets "deletes" the array, and now you've got a bunch of comma-separated objects, which is NOT valid javascript/json. — Marc B
– Marc B, Commented Oct 10, 2013 at 21:17

Peter · Accepted Answer · 2013-10-10 21:16:46Z

1

echo json_encode($status, JSON_FORCE_OBJECT);

Demo: http://codepad.viper-7.com/lrYKv6

or

echo json_encode((Object) $status);

Demo; http://codepad.viper-7.com/RPtchU

answered Oct 10, 2013 at 21:16

Peter

17k9 gold badges55 silver badges77 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Zeus · Accepted Answer · 2013-10-10 21:18:39Z

1

Instead of using JSonobject, use JSONArray

   JSONArray array = new JSONArray(sourceString);

Later loop through the array and do the business logic.

http://www.json.org/javadoc/org/json/JSONArray.html

answered Oct 10, 2013 at 21:18

Zeus

6,6347 gold badges60 silver badges94 bronze badges

Comments

Lajos Veres · Accepted Answer · 2013-10-10 21:19:20Z

0

Maybe with this kind of json structure?

$status[$rows['Aircraft']]= $rows['Status'];

answered Oct 10, 2013 at 21:19

Lajos Veres

13.7k7 gold badges47 silver badges57 bronze badges

1 Comment

Riggs78 Over a year ago

Thanks to everyone else who answered also, but this worked perfectly first go!!!! Cheers Lajos! you are the man.

Community · Accepted Answer · 2017-05-23 12:28:04Z

0

You get an JSONArray, Not Object, you could create an Object holding an array, or parsing the array. Refering to this post

edited May 23, 2017 at 12:28

CommunityBot

11 silver badge

answered Oct 10, 2013 at 21:19

MemLeak

4,6705 gold badges49 silver badges85 bronze badges

Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

Solution #1 (Java)

How about a helper method like this:

private int getProp(String name, JSONArray arr) throws Exception {
    for (int i = 0; i < arr.length(); ++i) {
        JSONObject obj = arr.getJSONObject(i);
        if (obj.has(name))
            return obj.getInt(name);
    }
    throw new Exception("Key not found");
}

Then you could use it like:

JSONArray jsonArray = new JSONArray(result); // note the *JSONArray* vs your *JSONObject*
n870 = getProp("A70_870", jsonArray);
n871 = getProp("A70_871", jsonArray);
...

Note I haven't tested this code, so you may need to make some changes...

Alternate solution (PHP)

It's been awhile since I've worked with PHP, but you might be able to leave your Java code intact and change your PHP int the while-loop body to:

$status[$rows['Aircraft']] = $rows['Status'];

Collectives™ on Stack Overflow

Removing the end [ and ] from php Json result for Android Java code

5 Answers 5

Comments

Comments

1 Comment

Comments

Solution #1 (Java)

Alternate solution (PHP)

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

5 Answers 5

Comments

Comments

1 Comment

Comments

Solution #1 (Java)

Alternate solution (PHP)

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related