How can I create a json object in php

Question

I am working in android and php.

I want to return a json object to android program from php program.

If these is a entry in a database then it is working properly. But when there is no record in database then it goes wrong.

I would welcome suggestions

I want to make json object like this ([{"id":"5"}])

This is my php program:-

    $server = "localhost";
    $username = "root";
    $password = "";
    $database = "mymusic";

    $con = mysql_connect($server, $username, $password) or die ("Could not connect: " . mysql_error());
    mysql_select_db($database, $con);
     $id=$_GET["id"];
     $pass=$_GET["password"];
    //$id='ram';
    //$pass='ram';
    $sql = "SELECT id FROM login where userid='$id' and password='$pass'";
    $result = mysql_query($sql) or die ("Query error: " . mysql_error());

    $records = array();
    if (mysql_num_rows($result)==0)
     {
      //what should i right here to make jsonobject like this:- ([{"id":"5"}])
       echo myjsono;
     }
     else
     {
       while($row = mysql_fetch_assoc($result)) 
       {
         $records[] = $row;
       }
       echo $_GET['jsoncallback'] . '(' . json_encode($records) . ');';
     }

    ?>

You're asking for each row to be encoded as an object, so start fetching them as objects from mysql rather than arrays. Replace mysql_fetch_assoc with mysql_fetch_object. — El Yobo
– El Yobo, Commented Sep 8, 2011 at 1:19
Thank you Dezigo and other persons who helped me... Thank you all... — Pushpendra Kuntal
– Pushpendra Kuntal, Commented Sep 8, 2011 at 4:40

Metalmini · Accepted Answer · 2011-09-07 10:11:48Z

3

How about something like this: (replace with your own variables)

if (empty($row)){
$arr = array('success' => 'false');
} else {
$arr = array('success' => 'true');
}
echo json_encode($arr);

answered Sep 7, 2011 at 10:11

Metalmini

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Comments

davidethell · Accepted Answer · 2011-09-07 10:07:32Z

0

If you want your android app to receive an object with a special id in the case of a not found condition I would return this:

{"id":"0"}

Then in your android app check if the id == 0 and that will tell you no records were found.

answered Sep 7, 2011 at 10:07

davidethell

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Comments

Pushpendra Kuntal · Accepted Answer · 2011-09-08 07:40:50Z

This is very correct solution for my question:-

$server = "localhost";
$username = "root";
$password = "";
$database = "mymusic";

$con = mysql_connect($server, $username, $password) or die ("Could not connect: " . mysql_error());
mysql_select_db($database, $con);
 $id=$_GET["id"];
 $pass=$_GET["password"];
//$id='ram';
//$pass='ram';
$sql = "SELECT id FROM login where userid='$id' and password='$pass'";
$result = mysql_query($sql) or die ("Query error: " . mysql_error());

$records = array();
if (mysql_num_rows($result)==0)
 {
  //  {"messages":{"message":[{"id":  "17","user": "Ryan Smith","text": "This is an example of JSON","time": "04:41"}]};}

  **echo '('.'['.json_encode(array('id' => 0)).']'.')';** //**note this**
 }
 else
 {
   while($row = mysql_fetch_assoc($result)) 
   {

     $records[] = $row;
   }
   mysql_close($con);
   echo $_GET['jsoncallback'] . '(' . json_encode($records) . ');';
 }

//mysql_close($con);

//echo $_GET['jsoncallback'] . '(' . json_encode($records) . ');';
?>

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