CodeIgniter 表类：从生成的单元添加链接

发布于 2024-07-11 15:53:11 字数 1047 浏览 4 评论 0原文

我正在使用表类，该类根据从数据库中提取的数据数组自动生成一个表。

模型：

function get_reports_by_user_id($userid)
{
    return $this->db->get_where('ss2_report',array('userid' => $userid))->result_array();
}

控制器：

function index()
{
    echo $this->table->generate($this->mymodel->get_reports_by_user_id('1234'));
}

当我让它工作时，控制器最终将被移动到视图。这会很好地生成表格，但我想添加指向字段的链接。例如，id 列允许我链接到仅包含该报告 ID 的数据页面。我知道我可以手动以老式方式输出表格。然后我可以添加我想要的任何链接，但我希望能够尽可能地使用自动生成。必须有一种方法来完成像链接表格单元格这样常见的事情。有人有什么想法吗？

编辑：

用户Java PHP大部分都在下面。这是使其工作的代码：

function get_reports_by_user_id($userid)
{
    $rows = $this->db->get_where('ss2_report',array('userid' => $userid))->result_array();

    foreach ($rows as $count => $row)
    {
        $rows[$count]['id'] = anchor('report/'.$row['id'],$row['id']);
    }
    return $rows;
}

我只需要用锚文本版本替换原始数组中的值。

原文

I'm using the table class that auto-generates a table for me from an array of data pulled from my database.

Model:

function get_reports_by_user_id($userid)
{
    return $this->db->get_where('ss2_report',array('userid' => $userid))->result_array();
}

Controller:

function index()
{
    echo $this->table->generate($this->mymodel->get_reports_by_user_id('1234'));
}

The controller will eventually be moved to a view when I have it working. This generates the table just fine, but I'd like to add a link to a field. For example, the id column that would allow me to link to a page of data for just that report's id. I know I can just output the table the old fashioned way by hand. I can then add whatever links I want, but I'd love to be able to use the auto-generation as much as possible. There's got to be a way to do something as common as linking a table cell. Does anyone have any ideas?

EDIT:

User Java PHP has it mostly right below. Here's the code that makes it work:

function get_reports_by_user_id($userid)
{
    $rows = $this->db->get_where('ss2_report',array('userid' => $userid))->result_array();

    foreach ($rows as $count => $row)
    {
        $rows[$count]['id'] = anchor('report/'.$row['id'],$row['id']);
    }
    return $rows;
}

I just needed to replace the value in the original array with the anchor text version.

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欢你一世 2024-07-18 15:53:11

唯一的方法是，在函数 get_reports_by_user_id() 中，您将循环遍历所有结果并将标记添加到 ids 中。像这样的事情：

function get_reports_by_user_id($userid)
{
   $rows=$this->db->get_where('ss2_report',array('userid' => $userid))->result_array();
   foreach ($rows as $row)
   {
     $row->id=anchor('site.com/some_controller/some_function/'.$row->id,$row->id);
   }
   return $rows;
}

我不使用 CodeIgniter 的数据库库，所以我不确定它返回 $rows 的格式，但上面的代码应该让您大致了解需要做什么。

The only way is, in the function get_reports_by_user_id() , you would loop through all the results and add the <a href> tag to the ids. Something like this:

function get_reports_by_user_id($userid)
{
   $rows=$this->db->get_where('ss2_report',array('userid' => $userid))->result_array();
   foreach ($rows as $row)
   {
     $row->id=anchor('site.com/some_controller/some_function/'.$row->id,$row->id);
   }
   return $rows;
}

I don't use CodeIgniter's database library so I'm not sure of what format it returns $rows in, but the code above should give you the general idea of what you need to do.

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圈圈圆圆圈圈 2024-07-18 15:53:11

一个想法可能是做类似的事情..

foreach ($row in $this->mymodel->get_reports_by_user_id('1234'))
{
    $row->id = anchor(site_url(array('report', 'user', $row->id)), $row->id);
    $this->table->add_row($row);
}
$this->table->generate();

One idea might be to do something like..

foreach ($row in $this->mymodel->get_reports_by_user_id('1234'))
{
    $row->id = anchor(site_url(array('report', 'user', $row->id)), $row->id);
    $this->table->add_row($row);
}
$this->table->generate();

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