使用ajax jquery的位置地址

Question

我正在尝试在格式化的表格中打印位置地址和last_seen时间。从res.location[i].lat, res.location[i].long正确获取格式化地址。但是当我将last_seen附加到字符串时，只获取最后一个记录的值。我的意思是为last_seen获取相同的值，这是最后一条记录的值。

请帮我找出错误的地方？

function getAddress(res)
{
    var html = '';
    var latitude = '';
    var longitude = '';
    var location_id = '';
    for (i = 0; i < res.location.length; i++){ 
        latitude = res.location[i].lat;
        longitude = res.location[i].long;
        location_id = res.location[i].id;
        var last_seen = moment(res.location[i].created_at).calendar();
        console.log(location_id);
        console.log(last_seen);
        url = 'https://maps.googleapis.com/maps/api/geocode/json?latlng='+parseFloat(latitude)+','+ parseFloat(longitude)+'&key=AIzaSyCJeRq6PHsq6DOwC3YDeIR1357I3w5BPLI';
        $.ajaxq ('getAddress', {
            url: url,
            type: 'post',
            data: 
            {
                delay: 5
            },
            success: function (response) 
            {
                console.log(last_seen);
                html = '<tr><td>' + response.results[0].formatted_address+'<br />Last Seen : '+ last_seen +'</td><td><input type="checkbox" checked="true" name="selected-location" id="'+location_id +'" value="'+location_id +'"></td></tr>'; 
                $('#single-user tr:last').after(html);

             }
         }); 
    }

}