Question

我正在尝试将变量optionSelected传递给另一个Page get_usersjson.php。然后，此值将用于运行sql查询.get_usersjson.php返回将用于填充下拉菜单中的选项的值。但此代码似乎无法正常工作。

  $('#tabletomodify').on('change', '.selectname',
      function () {
          //alert( this.parentNode.parentNode.rowIndex );
          var row = ($(this).closest('tr').prop('rowIndex'));
          var optionSelected = $("option:selected", this);
          var valueSelected = this.value;
          $.ajax({
              type: "POST",
              url: "get_usersjson.php",
              data: valueSelected,
              function (data) {
                  var html = '';
                  var len = data.length;
                  for (var i = 0; i < len; i++) {
                      html += '<option value="' + data[i].monthId + '">' + data[i].month + '</option>';
                  }
                  $('select.Location').append(html);
              }
          });

      }
  );

以下是get_usersjson.php

        <?php

        $a = $_GET['a'];
        //$a=1;
        $connection_for_user_location= mysqli_connect('localhost', 'xxxx', 'xxxx', 'xxxx') or die ('Cannot connect to db');
        $sql_query_2 = "select * from user_info";
        $result_query_2=mysqli_query($connection_for_user_location,$sql_query_2) or die ("Fail");
        $result_query_2->data_seek($a);
        $row_query_2=$result_query_2->fetch_row();

        for ($i=3;$i<=21;$i++)
        {
            $locality[$i]=$row_query_2[$i];
            $city[$i] = $result_query_2->fetch_field_direct($i);
            if(($locality[$i] != null) && ($line_card[$i] != "Empty"))
            {
             $data[]=$city[$i]->name.'::'.$locality[$i];
             }
        }
        echo json_encode($data);

        mysqli_close($connection_for_user_location);
        ?>

以下是php页面的示例输出： -

["Select City::Select Locality","Bangalore::MG Road","Delhi::Shalimar Bagh","New Delhi::Barakhamba Road","Punjab::Ghuman Nagar"]

Answer 1

由于您获得了JSON，因此可以使用jQuery中的$.getJSON()函数。

$.getJSON("url");

这是完整版：

$.ajax({
  dataType: "json",
  url: url,
  data: data,
  success: success
});

您的代码

您的代码应该是这样的：

$.getJSON("get_usersjson.php", function(data){
  foreach(a in data)
     html += '<option value="' + a.monthId + '">' + a.month + '</option>';
  $('select.Location').append(html);
});

jquery相当于ajax xmlhttp.open

1 个答案: