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Database 如何在DataMapper Code Igniter中显示get_by_user_id()返回的数据?_Database_Codeigniter_Datamapper - Fatal编程技术网

Database 如何在DataMapper Code Igniter中显示get_by_user_id()返回的数据?

Database 如何在DataMapper Code Igniter中显示get_by_user_id()返回的数据?,database,codeigniter,datamapper,Database,Codeigniter,Datamapper,我是新的代码点火器数据映射器。我有一个名为user的表,我正在尝试从数据库表中检索数据并将其显示给用户。 以下是我在模型中得到的: $u=new User(); $results=$u->get_by_user_id($id); //$results here will be set to huge bunch of none sense data( which also includes the row that I am looking for as well

我是新的代码点火器数据映射器。我有一个名为user的表,我正在尝试从数据库表中检索数据并将其显示给用户。 以下是我在模型中得到的:

    $u=new User();
$results=$u->get_by_user_id($id); 
    //$results here will be set to huge bunch of none sense data( which also includes the     row that I am looking for as well) 

if ($u->exists())
{ 

 foreach ($results->all as $row){
      $data['user']['first_name']=($row->user_first); //this  where I am stuck ..
  $data['user']['last_name']=($row->user_last);//this is also where I am stuck..

}
我不知道如何处理结果,以获得所需的字段,并将它们存储在传递给用户查看的$data中。
谢谢

在模型上调用
get_by_x()
时,字段将填充数据,您可以这样访问它们:

$u = new User();
$u->get_by_user_id($id);
if($u->exists())
{
    // you can access the table columns as object fields
    $data['user']['first'] = $u->first;
    $data['user']['last'] = $u->last;
}
else
{
    $data['error'] = 'No such user!';
}
看一看真正有用的文档:查看并通过


此外,DataMapper希望所有表都有一个
id
列:请参阅。如果您的列名为
id
,那么您应该调用
$u->get\u by\u id($id)
,而不是
$u->get\u by\u user\u id($id)

当您在模型上调用
get\u by\u x()
时,字段将填充数据,您可以像这样访问它们:

$u = new User();
$u->get_by_user_id($id);
if($u->exists())
{
    // you can access the table columns as object fields
    $data['user']['first'] = $u->first;
    $data['user']['last'] = $u->last;
}
else
{
    $data['error'] = 'No such user!';
}
看一看真正有用的文档:查看并通过

此外,DataMapper希望所有表都有一个
id
列:请参阅。如果列名为
id
,则应调用
$u->get\u by\u id($id)
,而不是
$u->get\u by\u user\u id($id)