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Javascript 尝试用jquery和ajax填充下拉列表_Javascript_Php_Jquery - Fatal编程技术网

Javascript 尝试用jquery和ajax填充下拉列表

Javascript 尝试用jquery和ajax填充下拉列表,javascript,php,jquery,Javascript,Php,Jquery,这是我的密码:- <script> $(document).ready(function(){ //#This script uses jquery and ajax it is used to set the values in $("#day").change(function(){ //# the time field whenever a day

这是我的密码:-

  <script>
           $(document).ready(function(){                               //#This script uses jquery and ajax it is used to set the values in
           $("#day").change(function(){             //# the time field whenever a day is selected.

           var day=$("#day").val();
           var doctor=$("#doctor").val();

           $.ajax({
                 type:"post",
                 url:"time.php",
                 data:"day="+day+"&doctor="+doctor,
                 dataType : 'json', 
                 success:function(data){
                            var option = '';
            $.each(data.d, function(index, value) {
                option += '<option>' + value.res + '</option>';
                });
            $('#timing').html(option);
                             }

                  });

                  });

                 });
   </script>
下面是php脚本

  <?
    $con=mysqli_connect("localhost","clinic","myclinic","myclinic");
    // Check connection

    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    $doctor = $_POST['doctor'];

    $day = $_POST['day'];

    $query="SELECT * FROM schedule WHERE doctor='" .$doctor."'AND day='" .$day. "'";

    $result = mysqli_query($con, $query);

    $i = 0;                                 //Initialize the variable which passes over the array key values

    $row = mysqli_fetch_assoc($result);    //Fetches an associative array of the row
    $index = array_keys($row);             // Fetches an array of keys for the row.

    while($row[$index[$i]] != NULL)
    {

        if($row[$index[$i]] == 1) {
            $res = $index[$i];              
            echo json_encode($res);

        }
        $i++;
    }       



  ?>
我希望在我的html页面上的select中插入带有时间值的选项,该页面如下所示:-

  <select id="timing" name="timing"></select>

我的java脚本代码正在将值发布到php脚本中,但代码仍然不起作用。在我看来,我的javascript中没有任何错误。请帮帮我,希望对你有帮助

   success:function(data){
           var select= '<select>';
           var option = '';
            $.each(data.d, function(index, value) {
                option += '<option>' + value.res + '</option>';
            });
           select = select+option'</select>';
           $('#timing').html(select);
      }
HTML:

<div id="timing"> </div>

我想你已经把所有的代码都放在JavaScription中了。你能告诉我们你在成功回调中得到的数据结构吗?我知道变量正在进入php脚本。当我在主页上更改“day”下拉列表中的值时,它会在控制台中弹出。@SilverBlade如何查看数据结构?您能告诉我吗?我是新手:使用console.logdata;
    var day=$("#day option:selected").val();
    var doctor=$("#doctor option:selected").val();

    data:"{day:'"+day+"', doctor: '" + doctor + "'}" ,  
      var postUrl = "time.php";
      $.ajax({
            type: "POST",
            url: postUrl,
            data: {day: day,doctor: doctor},
            dataType: "json",
            success: function (data) {
                $.each(data, function (key, value) {
                    $('#timing').append('<option value="' + key + '">' + value + '</option>');
                });
            }
        });