Question

我在sub_regions的基础上提取region_id名称。通过ajax完成它，所以在加入region和sub_region表之后，我将获得谷歌Chrome控制台的 network 中的结果。这意味着查询和其他一切都运行良好。

现在的问题是：

数据未在指定的Div标签上打印

我得到的数据是sub_regions，而非ID特定sub_region

JavaScript的：

<script>
 $("li a").click(function() {
var x = $(this).attr('id'); // this is getting the id of the region from region links
if (x.length > 0) {
  $.ajax({
    type: "POST",
    data: "id=" + x,
    url: "http://localhost/sitename/autocomplete/GetRegionName",
    dataType: "json",
    success: function(result) {
        $.each(result, function(i, v) {

          // For each record in the returned array
          $('#result').append(v.name); // specified name as to display only the name of sub_region, not everything else
          alert(v.name);
        });
      } // end success
  }); // end ajax
} // if condition ends
return false;
 });
</script>

PHP：

    <?php
    // This is the controller
    public function GetRegionName(){
            $keyword=$this->input->post('keyword');
            $data=$this->regioncomplete->GetRow($keyword);        
            echo json_encode($data);
        }   
    // This is the Model where i have joined 2 tables "regions" and "subregions" and declared that region ID = Sub region ID
            class Regioncomplete extends CI_Model{
            public function GetRow($keyword) {
            $this->db->select('*');
            $this->db->from('regions');
            $this->db->join('sub_regions', 'regions.id = sub_regions.region_id');        
            $this->db->order_by('sub_regions.id', 'DESC');
            $this->db->like("sub_regions.name", $keyword);
            return $this->db->get()->result_array();
        }
    }
    ?>

我正在网络中获得结果

{id: "2", code: "CF", name: "California", country_id: "1", region_id: "2"}
{id: "1", code: "ST", name: "Seattle", country_id: "1", region_id: "1"}

Answer 1

success: function (result) {
$.each(result, function(key, item) {
    var get_result = "<p>"+item.name+"</p><br/>";
    $(get_result).appendTo('#result');
});
}

OR

success: function (result) {
for (var i = 0; i < data.result; i++) {
 var get_result = "<p>"+data[i].name+"</p> <br/>";
     $(get_result).appendTo('#result');
}

}

检查出来

无法通过json在codeigniter中通过id获取数据

1 个答案: