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Home/ Questions/Q 8034803
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T02:07:50+00:00 2026-06-05T02:07:50+00:00

I have a database table whose fields contain latitude and longitude coordinates of positions.

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I have a database table whose fields contain latitude and longitude coordinates of positions. I want to create markers for a Google Map view using the information from the database.

I’ve implmented the query function as

function getCords(){
  $link = connectDB();
  $query = "SELECT * FROM tour";
  $results = mysqli_query($link, $query);

  $jsonArray = array();
  while ($row = mysqli_fetch_assoc($results)){
    $jsonArray[] = array('fileName' => $row['FileName'], 'lat' => $row['Lat'], 'lon' => $row['Lon']);

  }

return json_encode($jsonArray);
}

When I call this function from a php page, it returns the usual JSON format.

My issue is executing an ajax query. I have the query function above in a php scripts file containing six or so utility functions controlling login, logout, registration and the like. To query the database via jquery, I tried

var request = $.ajax({
  type:"GET",
  url: "includes/phpscripts.php?action=cords",
  type: "json"
});

var response = request.responseText;

My problem is the response is always empty. Is this due to the formation of the URL or for some other reason?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-05T02:07:51+00:00Added an answer on June 5, 2026 at 2:07 am
       $.ajax({
          type:"GET",
          url: "includes/phpscripts.php?action=cords",
          dataType: 'json', // necessary, because you're sending json from server
          success: function(response) {  // response will catch within success function
            console.log(response);
          }
        });
    

    or

       var request = $.ajax({
          type:"GET",
          url: "includes/phpscripts.php?action=cords",
          dataType: 'json', // necessary, because you're sending json from server
        }).done(function(response) {
           console.log(response);
        });
    

    NOTE

    Instead of return json_encode($jsonArray);, use echo json_encode($jsonArray);

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