Php JSON Response Array

Posted by Nick Kl on Stack Overflow See other posts from Stack Overflow or by Nick Kl
Published on 2012-06-08T16:28:58Z Indexed on 2012/06/08 16:40 UTC
Read the original article Hit count: 224

Filed under:
|
|
|

I have this php code. As you can see i query a mysql database through a function showallevents. I return a the $result to the $event variable. I try to return all rows of the data that i take with the msql_fetch_assoc. I don't get response even when i encode the $response variable. It returns null to all fields. Can anyone help me on what i am doing wrong. I had a valid code but it was returning only 1 row of data so i tried to make an associative array but seems i am failing.

    if ($tag == 'showallevents') 
{
    // Request type is show all events
    // show all events
    $event = $db->showallevents();
    if ($event != false)
    {
    $data = array();
    while($row = mysql_fetch_assoc($event)) 
        {
        $data[] = array(
        $response["success"] = 1,
        $response["uid"] = $event["uid"],
        $response["event"]["date"] = $event["date"],
        $response["event"]["hours"] = $event["hours"],
        $response["event"]["store_name"] = $event["store_name"],
        $response["event"]["event_information"] = $event["event_information"],
        $response["event"]["event_type"] = $event["event_type"],
        $response["event"]["Phone"] = $event["Phone"],
        $response["event"]["address"] = $event["address"],
        $response["event"]["created_at"] = $event["created_at"],
        $response["event"]["updated_at"] = $event["updated_at"]);
        }
        echo json_encode($data);            
    }
    else 
    {
        // event not found
        // echo json with error = 1
        $response["error"] = 1;
        $response["error_msg"] = "Events not found";
        echo json_encode($response);
    }
}

     else
    {
echo "Access Denied";
   }
   }
  ?>

© Stack Overflow or respective owner

Related posts about php

Related posts about android