Question

我尝试使用POST中的表单输入中的地址，并将该地址插入到此javascript中，以便将其地理编码到地图上的引脚点。我把POST放到一个javscript变量中，但我似乎无法将其转换为geodoces它的函数。这是在测试网站上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>

Answer 1

我认为quotes写作时出错请参阅以下行：

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

<小时/> 必须：

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

或

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

当您调用address函数时，您声明的变量codeAddress()会被覆盖。所以也清楚这一行：

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

或如果您正在使用，则必须检查发布的变量是否存在：

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

Answer 2

只需删除此行：

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

因为它会覆盖你已经在第三行声明的address。

Answer 3

您回复JS的值是很奇怪的，但是您的JS从ID = address的元素获取地址，覆盖了您的PHP回显结果。

你可能想要：

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>

Answer 4

您可以将地址作为参数传递给您的函数，如下所示：

<?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>

php POST插入到javascript中

4 个答案: