How do i get multiple table rows from the JSON Object in to the Application

Question

How can i get multiple table rows from the MySQL database into the application by a JSON object.

Getting one row from the table works well for me! But i cant get multiple rows, it throws an error saying "SyntaxError: Unexpected token { in JSON at position..."

How do i achieve this please help me!!

This is my angular Code..(here im passing a variable and get the matching rows and get them to the console.)

  $scope.companySelected = function(emp_Comp){

    console.log('Selected Item is : '+ emp_Comp);

    $http({
      method: 'GET',
      url: 'http://localhost/companyPopulate.php',
      params: {employeeCompany : emp_Comp}

    }).then(function successCallback(response) {
      $scope.data = response.records;
      console.log('success :'+ JSON.stringify(response));
      //console.log(testdata.data);

    }, function errorCallback(response) {
        //console.log('error',response);
      // called asynchronously if an error occurs
      // or server returns response with an error status.
    });

  }

This is my PhP Script...

<?php
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");

    //http://stackoverflow.com/questions/18382740/cors-not-working-php
    if (isset($_SERVER['HTTP_ORIGIN'])) {
        header("Access-Control-Allow-Origin: {$_SERVER['HTTP_ORIGIN']}");
        header('Access-Control-Allow-Credentials: true');
        header('Access-Control-Max-Age: 86400');    // cache for 1 day
    }

    // Access-Control headers are received during OPTIONS requests
    if ($_SERVER['REQUEST_METHOD'] == 'OPTIONS') {

        if (isset($_SERVER['HTTP_ACCESS_CONTROL_REQUEST_METHOD']))
            header("Access-Control-Allow-Methods: GET, POST, OPTIONS");

        if (isset($_SERVER['HTTP_ACCESS_CONTROL_REQUEST_HEADERS']))
            header("Access-Control-Allow-Headers:        {$_SERVER['HTTP_ACCESS_CONTROL_REQUEST_HEADERS']}");

        exit(0);
    }

    include_once 'config.php';

    //create connection
    $conn = new mysqli($mysql_host, $mysql_user, $mysql_password,$mysql_database);

    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    }


    $fields = array();

    if(isset($_GET['employeeNo']))
        $fields['employeeNo'] = $_GET['employeeNo'];

    if(isset($_GET['employeeName']))
        $fields['employeeName'] = $_GET['employeeName'];

    if(isset($_GET['employeeCompany']))
        $fields['employeeCompany'] = $_GET['employeeCompany'];


    $sql = "SELECT employeeName,employeeNo FROM employee_info WHERE ";
    $count = 0;

    foreach($fields as $key => $value){
        if($count == 0)
            $sql .= $key . "='" . $value . "' ";

        else
            $sql .= "AND " . $key . "='" . $value . "' ";

        $count++;
    }

    $fields = array();

        $result = $conn->query($sql);

            $outp = "";

            while($rs = $result->fetch_array(MYSQLI_ASSOC)) 
            {
                if ($outp != "") 
                {
                    $outp .= ",";
                } 

                $outp .= '{"employeeName":"' . $rs["employeeName"] . '"}'; 



            }

            $outp ='{"records":['.$outp.']}';


            $conn->close();

        echo $outp;
    /* }else{
        echo 'parameters expected from the client are missing! :) ';
    } */

?>

I'm getting the correct results when I'm running the script alone in the browser.

Please Help me!! Thanks in Advance!!

use Joins to fetch the data from Multiple tables – Karthi Aug 18 '16 at 07:01 — Karthi, Aug 18 '16 at 07:01
what is the response you log on success? – AranS Aug 18 '16 at 07:01 — AranS, Aug 18 '16 at 07:01

score 0 · Answer 1 · answered Aug 18 '16 at 07:01

0

I think you should be able to simplify the json creation like this

$output=array();

while($rs = $result->fetch_array(MYSQLI_ASSOC)){
    $output[]=$rs;
}
json_encode( $output );


header( 'Content-Type: application/json' );
echo $output;

answered Aug 18 '16 at 07:01

Professor Abronsius

33,063
5
32
46

my jason format was wrong!! json_encode worked fine!! thanks!! – Nisal Malinda Livera Aug 18 '16 at 08:07

score 0 · Accepted Answer · answered Aug 18 '16 at 07:15

0

try echo json_encode($outp); instead of echo $outp; because. it need to be pass as a json string. then only you can parse it in to array.

answered Aug 18 '16 at 07:15

Abhinav

388
4
12

my jason format was wrong!! json_encode worked fine!! thanks!! – Nisal Malinda Livera Aug 18 '16 at 08:07

How do i get multiple table rows from the JSON Object in to the Application

2 Answers2