在Codeigniter中显示联接中的数据 [英] Displaying Data from a Join in Codeigniter

问题描述

我正在使用一个简单的联接从两个数据库中提取数据.这是模型中的联接:

I am using a simple join to pull data from two databases. This is the join in the model:

function com_control(){
    $this->db->select('*');
    $this->db->from('comments');
    $this->db->join('posts', 'comments.entry_id = posts.id');
    $query = $this->db->get();
    return $query->result;
}

我想要的显示方法将在表格中，所以我开始像这样使用:

My desired method of display is going to be in a table so I am starting out to use like this:

foreach($comm_control as $row){

   $this->table->add_row(
   $row->entry_id,
   $row->comments.id,
   $row->comment,
   $row->title             
   );      
}//end of foreach

我的问题是显示comment.id中的数据.什么是将comment.id添加到表行中的正确格式?我需要两个表中的ID才能在表中进一步显示，编辑和删除.我目前唯一显示的"comment.id"是单词id.

My problem is the display of data from comments.id. What is the proper format to add the comment.id into the table rows? I need the ID from both tables for display, edit and delete further on in the table. The only display I get at this time for "comment.id" is the word id.

任何帮助将不胜感激.

推荐答案

好的，我想您需要做的是为注释表中的id字段设置一个别名:

Okay, I think what you need to do is set an alias for the id field in the comments table:

function com_control() {
    $this->db->select('entry_id, comments.id AS comment_id, comment, title');
    $this->db->from('comments');
    $this->db->join('posts', 'comments.entry_id = posts.id');
    $query = $this->db->get();

    return $query->result;
}

然后，您可以简单地通过$ row-> comment_id来引用comments.id字段:

Then you can reference the comments.id field as simply $row->comment_id:

$this->table->set_heading('Entry ID', 'Comment ID', 'Comment', 'Title');

foreach ($comm_control as $row ) {
    $this->table->add_row(
        $row->entry_id,
        $row->comment_id,
        $row->comment,
        $row->title             
    );      
}

echo $this->table->generate();

实际上，如果'id'列对于注释表是唯一的，则可以只使用$ row-> id;.当您联接具有相同名称的列的表时，会出现问题；它变得模棱两可，并且计算机将不知道您在引用什么.

Actually if the column 'id' is unique to the comments table, then you could just use $row->id; the problem arises when you are joining tables that both have a column named the same; it becomes ambiguous and the computer won't know what you're referencing.

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