Question

我有这个js做反向地理编码，效果很好。我现在需要使用AJAX将txt变量发送到PHP，但什么都没有，这是我的js，它是通过按钮点击调用的。我知道js工作，因为它将结果返回到我的html页面中的div。

<?php
// Start XML file, create parent node

$dom = new DOMDocument("1.0");
$node = $dom->createElement("markers");
$parnode = $dom->appendChild($node);
$lat = $lng = $address = "";

$servername = "localhost";
$username = "";
$password = "";
$dbname = "";

// Gelocation - get current location

$currPostCode = $_POST['post_code'];

// Opens a connection to a MySQL server

$connection=mysqli_connect($servername, $username, $password, $dbname);
if (!$connection) {  die('Not connected : ' . mysqli_error());}

// Select all the rows in the markers table

$query = "SELECT * FROM tblProperties WHERE propertyPostCode = '$currPostCode';";

$result = mysqli_query($connection, $query);
if (!$result) {
  die('Invalid query: ' . mysqli_error($connection));
}

header("Content-type: text/xml");

// Iterate through the rows, adding XML nodes for each

while ($row = @mysqli_fetch_assoc($result)){
  $lat = $lng = $address = "";
  $address = $row['propertyNumber'];
  $address .= " ";
  $address .= $row['propertyStreet'];
  $address .= ", ";
  $address .= $row['propertyPostCode'];

  // ADD TO XML DOCUMENT NODE
  $node = $dom->createElement("marker");
  $newnode = $parnode->appendChild($node);
  $newnode->setAttribute("name",$row['customerName']);
  $newnode->setAttribute("address", $address);
  $newnode->setAttribute("lat", $lat);
  $newnode->setAttribute("lng", $lng);
  $newnode->setAttribute("ID_Status", $row['ID_Status']);
}
echo $dom->saveXML();
echo $currPostCode;

?>

我的php文件没有从POST获得任何内容，或者查询中至少没有返回任何内容，我做错了什么？...

std::move

Answer 1

首先尝试检查一下是否收到数据：

var d1 = new Date();
var cache = d1.getTime();

$.ajax({
    url: "genMapXML.php?cache=" + cache,
    type: "post",
    data: {post_code: txt},
    success: function (data) {
        alert(data);
    },
    error: function () {
        alert('failure');
    }
});

Answer 2

以下是代码段：

<!DOCTYPE html>
<html>
<head>
    <title></title>
     <script src="https://code.jquery.com/jquery-1.10.2.js"></script>
     <script type="text/javascript">
    function getLocation()
  {
  if (navigator.geolocation)
    {
       navigator.geolocation.getCurrentPosition(reverseGeoLookup);
    }
  else
    {
       x.innerHTML="Geolocation is not supported by this browser.";
    }
  }

function reverseGeoLookup(position) {
    var lat = position.coords.latitude;
    var lon = position.coords.longitude;
    var txt;
  var req = new XMLHttpRequest()
  req.open("GET", "https://maps.googleapis.com/maps/api/geocode/json?latlng="+lat+","+lon+"&sensor=true", true)
  req.onreadystatechange = function() {
      if(req.readyState == 4) {

          var result = JSON.parse(req.response).results
          for(var i = 0, length = result.length; i < length; i++) {
              for(var j = 0; j < result[i].address_components.length; j++) {
                  var component = result[i].address_components[j]

                  if(~component.types.indexOf("postal_code")) {
                    txt = component.long_name;            

                $.ajax({
                 url: "genMapXML.php",
                 type: "post",
                 data: { post_code: txt },
                 success: function(){
                 alert('success');
                 },
                 error: function(){
                 alert('failure');
                 }
                 }); 
                 document.getElementById("trace_location").innerHTML = txt;
                 return false;                  
                  }
              }
          }
      }

  }
  req.send()
}
getLocation();
</script>
</head>

<body>
<div id="trace_location"></div>
</body>
</html>

在你的php页面中首先有请求变量：

 <?php
    echo $_REQUEST['post_code'];
// write the required code after having it as a response and do the necessary tasks after that.
    ?>

希望这能解决您的问题。感谢。

AJAX POST到PHP没有返回任何内容

2 个答案: