我在PHP代码中设置AJAX变量时遇到问题
当我更改区域的值时,将调用以下函数:
function find_map(){
var value1 = $("#area").val();
var value2 = $("#city").val();
$.ajax({
url :"find_my_map.php", // json datasource
type: "post", // method , by default get
dataType:"json",
data:
{
area_id:value1,
city_id:value2
},
success: function(data)
{
console.log(data);
$.each(data, function(index, element) {
alert("area_name:" + element.area_name);
alert("city_name:" + element.city_name);
var map_area_name= element.area_name;
var map_city_name= element.city_name;
$("#map_area_name").val(map_area_name);
$("#map_city_name").val(map_city_name);
});
},
error:function(data){
console.log(data);
}
});
}
find_my_map.php文件
include_once("config.php");
if(isset($_POST['area_id']) && isset($_POST['city_id']))
{
$area_id = $_POST['area_id'];
$city_id = $_POST['city_id'];
$sql = "SELECT * FROM area a, city c WHERE a.city_id=c.city_id AND c.city_id='$city_id' AND a.area_id='$area_id'";
$qry = mysql_query($sql);
while($fetch = mysql_fetch_array($qry))
{
$area_name=$fetch['area_name'];
$city_name=$fetch['city_name'];
$data[]=array(
'area_name' => $area_name,
'city_name' => $city_name
);
}
$json_data = array($data);
echo json_encode($data);
}
这是我的PHP代码
<?php
echo $address ="here i want to use **map_area_name AND map_city_name**"; // Google HQ
$prepAddr = str_replace(' ','+',$address);
$geocode=file_get_contents('https://maps.google.com/maps/api/geocode/json?address='.$prepAddr.'&sensor=false');
$output= json_decode($geocode);
$area_latitude = $output->results[0]->geometry->location->lat;
$area_longitude = $output->results[0]->geometry->location->lng;
echo "<br><input type='text' name='area_latitude' id='area_latitude'value='".$area_latitude."'> <br>";
echo "<input type='text' name='area_longitude' id='area_longitude' value='".$area_longitude."'>";
?>