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up vote 0 down vote favorite

Im trying to use the address from a form input that is in the POST and insert that adress into this javascript for geocoding it into a pin point on the map. I put the POST into a javscript variable but i cant seem to get that into the function that geodoces it. here is it live on a test site http://nickshanekearney.com/geo/

<?php $address = $_POST["address"]; ?>
<script> 
var address = "<?php echo $_POST["address"]; ?>";
var geocoder;
var map;

function codeAddress() {
var address = document.getElementById("address").value;
geocoder.geocode( { 'address': address}, function(results, status) {
  if (status == google.maps.GeocoderStatus.OK) {
    map.setCenter(results[0].geometry.location);
    var marker = new google.maps.Marker({
        map: map,
        position: results[0].geometry.location
    });

    infowindow = new google.maps.InfoWindow();
    var service = new google.maps.places.PlacesService(map);
    service.nearbySearch(request, callback);
  } else {
    alert("Geocode was not successful for the following reason: " + status);
  }
});
}
</script>

<div id="map"></div>
share|improve this question
 
see the double quotes (") here var address = "<?php echo $_POST["address"]; ?>";. It's an error writing –  Oki Erie Rinaldi 2 days ago
 
Try to replace with var address = "<?php echo $_POST['address']; ?>"; –  Oki Erie Rinaldi 2 days ago
 
with double quotes , it works –  Charaf jra 2 days ago
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4 Answers

up vote 2 down vote

I think you have an error in quotes writing.
See your line below: 

var address = "<?php echo $_POST["address"]; ?>";

it must be :

 var address = "<?php echo $_POST['address']; ?>";

or

var address = "<?php echo $address; ?>";// because you already declared variable $address at the top of your source code

And the variable address that you declared is overwritten when you call codeAddress() function. So clear this line too : 

var address = document.getElementById("address").value;

Or if you are using it,you have to check if the posted variable exist or not :

function codeAddress(){
<?php 
if (isset($_POST['address'])){
echo "var address = $_POST['address'];"; 
}
else{
echo "var address = document.getElementById('address').value;";
}

?>

...
share|improve this answer
 
var address = "<?php echo $_POST["address"]; ?>"; this works too –  Charaf jra 2 days ago
add comment
up vote 1 down vote

Just remove this line:

var address = document.getElementById("address").value;

Because it overwrites the address you already declared on the third line.

share|improve this answer
add comment
up vote 1 down vote

It is so strange that you echo the value to JS, but your JS gets the address from element with ID = address, overwritten your PHP-echoed result.

You probably want: 

function codeAddress() {
  var address = "<?php echo $_POST['address']; ?>";
  geocoder.geocode( { 'address': address}, function(results, status) {
    if (status == google.maps.GeocoderStatus.OK) {
      map.setCenter(results[0].geometry.location);
      var marker = new google.maps.Marker({
        map: map,
        position: results[0].geometry.location
      });

      infowindow = new google.maps.InfoWindow();
      var service = new google.maps.places.PlacesService(map);
      service.nearbySearch(request, callback);
    } else {
      alert("Geocode was not successful for the following reason: " + status);
    }
  });
}
</script>
share|improve this answer
add comment
up vote 0 down vote

You can pass the address as argument to your function like this : 

<?php $address = $_POST["address"]; ?>
<script> 
var address = "<?php echo $_POST["address"]; ?>";
var geocoder;
var map;

function codeAddress(address) {

geocoder.geocode( { 'address': address}, function(results, status) {
  if (status == google.maps.GeocoderStatus.OK) {
    map.setCenter(results[0].geometry.location);
    var marker = new google.maps.Marker({
        map: map,
        position: results[0].geometry.location
    });

    infowindow = new google.maps.InfoWindow();
    var service = new google.maps.places.PlacesService(map);
    service.nearbySearch(request, callback);
  } else {
    alert("Geocode was not successful for the following reason: " + status);
  }
});
}
</script>

<div id="map"></div>
share|improve this answer
 
This is all good advice im so close you can see here my progress nickshanekearney.com/geo It works now but not on load I have to press the submit button twice. is it because I need to start the function on load instead of on submit of form? –  user3029877 2 days ago
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