Question

大家好我有一个在codeigniter中开发的网站。在某些函数中，我必须从mysql中检索数据并在javascript中打印结果，因为它是一个ajax调用。

这是我检索数据的php函数：

public function getCityByNameOnly($name) {
      $query = $this->db->query('SELECT *  FROM city WHERE name LIKE "%'.$name.'%" ');
      $city = array();
      foreach ($query->result() as $row)
     array_push($city, $row);
     return json_encode($city);
      }
}

这是我的ajax电话：

$('#ricerca-ajax').keyup(function(){
        var val = $(this).val();
        if (val.length>2){
            var site_url_city ="<?php echo(site_url('/city/get_city_by_text')); ?>";
            $.ajax({   
                url: site_url_city, 
                type: "POST", 
                data: {search: val},
                dataType: "text",             
                success: function(data) {
                    console.log(data);
                    for(var i=0;i<data.length;i++)
                    {
                          $('#msgid').append(data[i].name_en  + '<br> ');

                    }
                }
            });
        }
    })

我将结果附加到div中，但始终未定义。这是我创建的json的console.log：

{"id":"125","name_it":"L&egrave;sina (Hvar)","name_en":"Hvar","nation_id":"23","region_id":"0","active":"1"},{"id":"127","name_it":"Trogir (Tra&ugrave;)","name_en":"Trogir","nation_id":"23","region_id":"0","active":"1"},{"id":"1088","name_it":"Citt&agrave; del Capo","name_en":"Cape Town","nation_id":"101","region_id":"0","active":"1"}]

如何在javascript中将此json打印到我的成功函数中？感谢

Answer 1

更改ajax中的datatype:"json"将解决问题

$('#ricerca-ajax').keyup(function(){
        var val = $(this).val();
        if (val.length>2){
            var site_url_city ="<?php echo(site_url('/city/get_city_by_text')); ?>";
            $.ajax({   
                url: site_url_city, 
                type: "POST", 
                data: {search: val},
                dataType: "json",             
                success: function(data) {
                    console.log(data);
                    for(var i=0;i<data.length;i++)
                    {
                          $('#msgid').append(data[i].name  + '<br> ');

                    }
                }
            });
        }
    })

从mysql打印json

1 个答案: