Unable to json_decode output received from my php webservice?

Question

Webservice code:

function login($uname)
        {   

            $id=1;
            $link = mysql_pconnect("localhost", "root", "root") or die("Could not connect");
            mysql_select_db("sparq",$link) or die("Could not select database");
            $sql=mysql_query("select username,password from user_login where user_id=1");
            //$result = mysql_query($query);
            $arr = array();
            while($obj = mysql_fetch_object($sql)) 
                {
                    $arr[] = $obj;
                    }
           // $obj = mysql_fetch_object($sql);
           header("Content-type: application/json");
           echo json_encode($arr);
        }

Code from client:

$url="http://localhost/web.php";
if (isset($_POST['Login']))
    {

            $ch = curl_init($url); // Initialize a CURL session
            curl_setopt($ch, CURLOPT_HEADER, 0); // options for a CURL transfer
            curl_setopt($ch, CURLOPT_POST, 1);
            curl_setopt($ch, CURLOPT_POSTFIELDS,"username=".$username );
            curl_setopt($ch, CURLOPT_FOLLOWLOCATION, 0);
            curl_setopt($ch, CURLOPT_RETURNTRANSFER, 1);
            $data = curl_exec($ch); // Perform a CURL session
            curl_close($ch); 
            $arr =array();
            $arr=json_decode($data,true);
            echo 'I am here';    //echo1
            echo $data;   //echo2

            echo $arr[0]->username; //echo3

I am getting the below output:

I am here //echo1
[{"username":"[email protected]","password":"asdf123"}]  //echo2
Notice: Trying to get property of non-object in F:\xampp\htdocs\webtest\login.php on line 38  //echo3

Ariel · Accepted Answer · 2011-07-25 08:30:21Z

3

This:

echo $arr[0]->username;

Should be:

echo $arr[0]['username'];

And this:

$arr = array();
while($obj = mysql_fetch_object($sql)) {
  $arr[] = $obj;
}

Should be:

$arr = array();
while($row = mysql_fetch_assoc($sql)) {
  $arr[] = $row;
}

You can't send php objects over json. What javascript (and json) call an object is an associative array in php.

answered Jul 25, 2011 at 8:30

Ariel

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4 Comments

Akhil N K Over a year ago

But the below will return undefined constant username error "\n" echo $arr[0]['username'];

Ariel Over a year ago

@akhil-n-k If you are getting that error you forgot the quotes around 'username'.

Akhil N K Over a year ago

still it's not printing the value of username

Ariel Over a year ago

@akhil-n-k Do print_r($arr); and post the output.

Darm · Accepted Answer · 2011-07-25 10:50:39Z

0

Replace: echo $arr[0]->username; by $arr[0]['username'];

answered Jul 25, 2011 at 10:50

Darm

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4 Comments

Akhil N K Over a year ago

One more thing: The initial code I posted is working well in my friend's m/c and I am getting the correct output

Darm Over a year ago

Tell me what do you have by inserting var_dump($arr); before "echo3"

Darm Over a year ago

It's because you have to use UTF-8 encoding. Encode your file script in UTF-8 and your db table fields or use $arr=json_decode(utf8_encode($data),true);

Akhil N K Over a year ago

Still no hope. I am getting 4 as output from json_last_error(). I added that below the line $arr=json_decode($data,true);

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