Codeigniter和AJAX从数据库错误中检索/获取数据 [英] Codeigniter and AJAX retrieving/getting data from database error

问题描述

我总是进入ajax的错误函数，这是什么问题?

I always land into the error function in ajax, what is the problem here?

我想检索存储在数据库中的img URL，并使用jquery附加它以显示我的img画廊.但是，在执行任何操作之前，我必须确保它始终位于成功函数上，但我始终位于错误函数上……所以我无法继续进行，因此我被困在这里，

I want to retrieve a img url stored in the database and append it using jquery to show my img gallery. But before I can do any of this, I have to make sure it always land on the success function, but I always land on error function... so I can not proceed and I am stuck here,

更新:

检查控制台，它在 success:function(result)

我的脚本:

    function load_contents(track_page)
    {
        $('#loading').show();

        $.ajax({
            url:'<?php echo base_url('gallery/load_design');?>',
            type:'GET',
            dataType:'json',
            success:function(result)
            {
                alert("success");
            },
            error:function()
            {
                alert("failed"); //it always show the alert failed.

            }
        });
    }

我的控制器:

function load_design()
{
    $this->load->model('design');
    $this->load-model('profile');

    $user_id = $this->profile->retrieve_userid();

    $json = $this->design->load_gallery($user_id->id);
    echo json_encode($json);
}

我的模特:

    function load_gallery($user_id)
    {
        $query = $this->db->query("SELECT * from designs WHERE user_id = '".$user_id."' LIMIT 9");
        return $query->result_array();
    }

推荐答案

您应该在页脚中创建第一个全局变量

You should create first global variable in footer

var base_url = "<?= base_url(); ?>";

现在您需要像这样使用

function load_contents(track_page)
    {
        $('#loading').show();
        var ref = new Date(); 
        var url = base_url + "gallery/load_design?=" + ref.getTime();
        $.ajax({
            url:url,
            type:'GET',
            cache: false,
            success:function(result)
            {
                alert("success");
            },
            error:function()
            {
                alert("failed"); //it always show the alert failed.

            }
        });
    }

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