如何从php中提取值并传递给jquery函数

Question

我想计算两个机场之间的距离，并将此值返回给jquery函数，并在进一步的计算中使用此值。 我不知道哪里出问题了？

html：dept和dest输入从数据库中检索到机场信息

  <form id="airform1" class="on">
    <div>
        From
        <div class="textinput">
            <input type="text" id="dept" name="departure" placeholder="City name or aiport code" >
        </div>
    </div>

    <div>
        To
        <div>
            <input type="text" id="dest" placeholder="City name or airport code" >
        </div>
    </div>

    <div>
        Via
        <div>
            <input type="text" id="via" placeholder="City name or airport code" >
        </div>
    </div>

    <div>
        <div>
            <input type="radio" name="trip" value="2.0">Round-trip <br>
            <input type="radio" name="trip" value="1.0">One-way
        </div>
    </div>

    <div>
        Number of flights/passengers
        <input type="text" id="numpass" value="1">
    </div>

    <div>
        Class
        <select style="width: 82px">
            <option selected="selected" value="1.0">Economy</option>
            <option value="2.04">Business</option>
            <option value="2.69">First</option>
        </select>
    </div>

    <div id="line">

    </div>
    </form>

php代码：

    <?php


    if(isset($_POST['dept'], $_POST['dest'])){
    $dept=$_POST['dept'];
    $dest=$_POST['dest'];
   }



        mysql_connect("localhost","abv","asdasda") or die (mysql_error());
        mysql_select_db("asda") or die(mysql_error());


       $strSQL1 = "SELECT display, lat, longi FROM airport WHERE display = '$dept'";
       $rs1 = mysql_query($strSQL1);
       $row1 = mysql_fetch_array($rs1);
       $lat1= $row1['lat'];
       $long1= $row['longi'];

       $strSQL2 = "SELECT display, lat, longi FROM airport WHERE display = '$dest'";
       $rs2 = mysql_query($strSQL2);
       $row2 = mysql_fetch_array($rs2);
       $lat2= $row1['lat'];
       $long2= $row['longi'];



$earthradius = 6366.707;    
    $km_to_miles = 1/1.609344;

    $radlat1 = M_PI * (lat1)/180;
    $radlat2 = M_PI * (lat2)/180;

    $radlng1 = M_PI * (long1)/180;
    $radlng2 = M_PI * (long2)/180;

    $radlng1 += M_PI / 2;
    $radlng2 += M_PI / 2;

    if($radlng1<0)
        radlng1 += M_PI * 2;
    if($radlng2<0)
        radlng2 += M_PI * 2;

    $x1 = $earthradius * cos($radlng1) * sin($radlat1);
    $y1 = $earthradius * sin($radlng1) * sin($radlat1);
    $z1 = $earthradius * cos($radlat1);

    $x2 = $earthradius * cos($radlng2) * sin($radlat2);
    $y2 = $earthradius * sin($radlng2) * sin($radlat2);
    $z2 = $earthradius * cos($radlat2);

    $d = sqrt(($x1-$x2)*($x1-$x2)+($y1-$y2)*($y1-$y2)+($z1-$z2)*($z1-$z2));

    $theta = acos(($earthradius*$earthradius + $earthradius * $earthradius - $d * $d)/(2 * $earthradius * $earthradius));

    $distance = $theta * $earthradius;
    $miles = $distance * $km_to_miles;

    $co2 = (($miles / 41.986) * 20.88 * 2.7) / 2204.6;

    echo $co2; 

 ?>

javascript：airanswer没有任何显示，我不知道哪里出了问题？

     $("#dept").change(function(){
            $("#dest").change(function(){
                var dept = $("#dept").val();
                var dest = $("#dest").val();
                $.post('airtravel.php',{dept: dept, dest: dest}, function(data){
                    $("#aircalc").click(function(){
                        $("#airanswer").html(data);
                    });

                });
            });
        });

Answer 1

来自服务器端脚本的数据是HTML。 因为您在Ajax中使用它，所以返回JSON响应可能是一个更好的主意：

header('Content-Type: application/json');
echo json_encode(array('co2' => $co2));

然后在JavaScript中：

$.post('airtravel.php',{dept: dept, dest:dest},function(data){
    var dist = data.co2,
    passengers = $("#numpass"),
    carbon = dist * passengers;

    $("#airanswer").text(carbon);
});