Unable to return latitude/longitude result to Javascript JSON from PHP MySQL query

Question

I am trying to return database values for latitude and longitude from a PHP/MySQL query to my Javascript function in order to populate a google maps JSON request.

My PHP/MySQL query script, phpsearch2.php is:

<?php
include "base.php";
$name = $_GET["name"];
$query = "SELECT lat, lng FROM markers WHERE name = '".$name."'";
$result = mysql_query($query);
while($row = mysql_fetch_array ($result))     
{
     echo '{';
           echo '"latitude":"'.$row['lat'].'",';
           echo '"longitude":"'.$row['lng'].'",';
     echo '}';    
}
?>

It returns a value in this format:

{"latitude":"37.730267","longitude":"-122.498589",}

Here is my calculate root function, when I run the program I get an error saying 'origin is not defined' even though I think I am setting it to have the same value as the return result from the JSON request to phpsearch, I just don't know where the mistake lies.

 function calcRoute() {
    var startname = document.getElementById('start').value;
    var endname = document.getElementById('end').value;
    var waypts = [];
    var checkboxArray = document.getElementById('waypoints');
    for (var i = 0; i < checkboxArray.length; i++) {
      if (checkboxArray.options[i].selected == true) {
        waypts.push({
            location:checkboxArray[i].value,
            stopover:true});
      }
    }

$.getJSON("phpsearch2.php", {name : startname}, function (result) {
origin = google.maps.LatLng('result');
});

var end = new google.maps.LatLng('37.738029', '-122.499481');
     var request = {
        origin: origin,
        destination: end,
        waypoints: waypts,
        optimizeWaypoints: true,
        travelMode: google.maps.DirectionsTravelMode.WALKING
    };
        directionsService.route(request, function(response, status) {
    //document.write('<b>'+ start + end + '</b>');
      if (status == google.maps.DirectionsStatus.OK) {
        directionsDisplay.setDirections(response);
        var route = response.routes[0];
        var summaryPanel = document.getElementById('directions_panel');
        summaryPanel.innerHTML = '';
        // For each route, display summary information.
        for (var i = 0; i < route.legs.length; i++) {
          var routeSegment = i + 1;
          summaryPanel.innerHTML += '<b>Time for a Walkabout </b><br>';
          summaryPanel.innerHTML += '<b>From ' + startname + '   </b>';
          summaryPanel.innerHTML += '<b>to ' + endname + '('+ route.legs[i].distance.text +')</b><br>';

      } 
      }
    });
 }

OT: Always use `json_encode` to generate JSON in PHP. Have a look at this answer to learn the difference between synchronous and asynchronous code: http://stackoverflow.com/a/14220323/218196. Ajax is asynchronous. — Felix Kling, Apr 07 '13 at 00:25

jszobody · Accepted Answer · 2013-04-07T00:23:14.657

1

You are passing 'result' in as a string. You need to pass the result object, without the quotes:

var origin = google.maps.LatLng(result);

Keep in mind that $.getJSON is asynchronous. You probably need to wrap the rest of your gmaps code in a function declaration, and call that function in the $.getJSON callback after origin is set. Otherwise you're still trying to use origin before it has been fetched from the server.

Something like:

$.getJSON("phpsearch2.php", {name : startname}, function (result) {
    var origin = google.maps.LatLng(result);

    // Now that we have our origin, we can initialize the rest
    init(origin);
});

function init(origin) {
   var end = new google.maps.LatLng('37.738029', '-122.499481');
   var request = {
        origin: origin,
   <snip>
}

edited Apr 07 '13 at 00:23

answered Apr 06 '13 at 23:41

jszobody

28,495
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Thanks, that was definitely an issue, I am still seeing an error after I make that change though Error: Invalid value for property : undefined ...y)/d);f&&f.io&&(qa(k,f.io[0]),Ka(k,f.io[1]));return{ya:f,W:e,df:g,anchorOffset:k... – Mark Harrison Apr 06 '13 at 23:48
Look at the updated answer. You need to wait for $.getJSON to return before you move on. – jszobody Apr 07 '13 at 00:07

Unable to return latitude/longitude result to Javascript JSON from PHP MySQL query

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