使用cakephp在下拉列表中显示名称 [英] Getting names to show in dropdown list with cakephp

问题描述

我想在下拉列表中显示我们所有项目负责人的名字。

项目领导只是在公司工作的一些员工。 / p>

这是我的表：

  project_leaders 
， ---，----------------，
 | id | hr_employee_id | 
 | ---- | ---------------- | 
 | 1 | 18 | 
'----'----------------'
 
 projects 
，----，---- - ，-------------------，
 | id |名称| project_leader_id | 
 | ---- | ------ | ------------------- | 
 | 1 |蛋糕| 1 | 
'----' ------'-------------------'
 
 hr_employees 
 ，----，------，---------，
 | id |名称|姓氏
 | ---- | ------ | --------- 
 | 18 | joe |污垢| 
'----' ------'---------'
  

我的ProjectsController看起来像这样：

  public function add（）{
 if - > request-> is（'post'））{
 $ this-> Project-> create（）; 
 if（$ this-> Project-> save（$ this-> request-> data））{
 $ this-> _sendProjectRequestEmail（$ this-& getLastInsertID（））; 
 $ this-> Session-> setFlash（__（'项目已保存'））; 
 $ this-> redirect（array（'action'=>'index'））; 
} else {
 $ this-> Session-> setFlash（__（'无法保存项目，请重试。 
} 
} 
 $ this-> set（'projectLeaders'，$ this-> Project-> ProjectLeader-> find（'list'）; 
} 
  

这只返回项目领导者的id，而不是名字和姓氏。 ，它返回1。

我试过做$ this-> Project-> ProjectLeader-> HrEmployee-> find（'list'），但列出所有的员工。

我也试过指定字段，但返回未知的字段错误。

我做错了？

解决方案

  $ this-> set（'projectLeaders'，$ this-> Project-> ProjectLeader-> find（'list'））; 
  

将只列出 project_leaders 表中的记录，很可能是因为表本身不包含名称/标题字段（该蛋糕将自动作为 displayField ）如下所示：

  ; 1 
）
  

使用适当的查找

要获得有意义的项目负责人名单，您需要确保您在 project_leaders 和 hr_employees 一种方法是使用可包含的行为，并简单地指定

  $ projectLeaders = $ this-> Project-> ProjectLeader-> find list'，array（
'contains'=> array（
'HrEmployee'
），
'fields'=> array（
'ProjectLeader.id'，
'HrEmployee.name'，
）
））; 
 $ this-> set（Compact（'projectLeaders'））;您的数据库结构是否适合您的需要？ 
 
  
  
有一个表格相当于说这个用户是管理员可能不是最好的主意 - 这将更容易避免数据问题，并给你更简单的查询，如果表 project_leaders 在 hr_employees 表和 project_leader_id 中指向一个布尔字段 hr_employee 表，不是一些其他数据抽象。
 
 
 
当然，我不知道你的整个模式，非常有可能是有一个单独的 project_leaders 表的好理由。
 
 
 
或者 -  denormalize < h2> 
 
 
如果您向 project_leaders 添加一个名称字段 - 您不需要加入就可以知道项目领导或任何funkiness：
  alter table project_leaders添加名称varchar（255） 
 update project_leaders set name =（select name from hr_employees where hr_employees.id = hr_employee_id）; 
  
通过这种方式，您可以通过一个查询/联接知道谁是相关项目潜在客户，做两个连接以获得员工姓名。
 
I want to show the names of all our project leaders in a dropdown list.

The project leaders are only some of the employees that work in the company.

Here are my tables:
project_leaders
,----,----------------,
| id | hr_employee_id |
|----|----------------|
| 1  |  18            |
'----'----------------'

projects
,----,------,-------------------,
| id | name | project_leader_id |
|----|------|-------------------|
| 1  | cake |         1         |
'----'------'-------------------'

hr_employees
,----,------,---------,
| id | name | surname |
|----|------|---------|
| 18 | joe  | dirt    |
'----'------'---------'
My ProjectsController looks like this:
public function add() {
    if ($this->request->is('post')) {
        $this->Project->create();
        if ($this->Project->save($this->request->data)) {
            $this->_sendProjectRequestEmail($this->Project->getLastInsertID() );
            $this->Session->setFlash(__('The project has been saved'));
            $this->redirect(array('action' => 'index'));
        } else {
            $this->Session->setFlash(__('The project could not be saved. Please, try again.'));
        }
    }
    $this->set('projectLeaders',$this->Project->ProjectLeader->find('list');
}
This only returns the id of the Project Leaders, not the name and surname. So instead of Joe Dirt, it returns 1.

I've tried doing $this->Project->ProjectLeader->HrEmployee->find('list') but that lists all the employees.

I've also tried specifying the fields, but it returns a unknown field error.

What am I doing wrong?
解决方案
$this->set('projectLeaders',$this->Project->ProjectLeader->find('list'));
This will just list the records from the project_leaders table and most likely, as the table doesn't itself contain a name/title field (which cake would pick up automatically as the displayField) be like so:
array(
    1 => 1
)


Use an appropriate find

To get a meaningful list of project leaders, you need to ensure you get a join between project_leaders and hr_employees one way to do that is using the containable behavior and simply specifying which fields to use in the list:
$projectLeaders = $this->Project->ProjectLeader->find('list', array(
    'contain' => array(
        'HrEmployee'
    ),
    'fields' => array(
        'ProjectLeader.id',
        'HrEmployee.name',
    )
));
$this->set(compact('projectLeaders'));


Is your db structure appropriate for your needs?

Having a table equivalent to saying "this user is admin" may not be the best idea - it would be easier to avoid data problems, and give you simpler queries, if the table project_leaders was (only) a boolean field on your hr_employees table and project_leader_id pointed at the hr_employee table, not some other data-abstraction.

Of course I don't know your whole schema, there very well may be a good reason for having a seperate project_leaders table.

Alternatively - denormalize

If you add a name field to project_leaders - you don't need a join to know the name of a project leader or any funkiness:
alter table project_leaders add name varchar(255) after id;
update project_leaders set name = (select name from hr_employees where hr_employees.id = hr_employee_id);
In this way you can know who is the relevant project lead with one query/join instead of needing to do two joins to get an employee's name.

                        
这篇关于使用cakephp在下拉列表中显示名称的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！