Php 从数据库Codeigniter创建下拉菜单
我正在尝试创建下拉菜单。 数据库表:Php 从数据库Codeigniter创建下拉菜单,php,codeigniter,Php,Codeigniter,我正在尝试创建下拉菜单。 数据库表: *cat* catid, catname, catsex *menus* id,parrentid,name,icon,slug,number 视图中的菜单代码 echo $this->multi_menu->render(array( 'nav_tag_open' => '<ul class="nav nav-pills">', 'parentl1_tag_op
*cat*
catid, catname, catsex
*menus*
id,parrentid,name,icon,slug,number
视图中的菜单代码
echo $this->multi_menu->render(array(
'nav_tag_open' => '<ul class="nav nav-pills">',
'parentl1_tag_open' => '<li class="dropdown">',
'parentl1_anchor' => '<a tabindex="0" data-toggle="dropdown" href="%s">%s<span class="caret"></span></a>',
'parent_tag_open' => '<li class="dropdown-submenu">',
'parent_anchor' => '<a href="%s" data-toggle="dropdown">%s</a>',
'children_tag_open' => '<ul class="dropdown-menu">',
'item_active' => 'Gallery/foto'
));
模型库中的菜单代码\u模型
public function every()
{
return $this->db->get("menus")
->result_array();
}
现在我将试着解释我的问题是什么。
我想创建一个下拉菜单项,其中显示所有猫的名字(我想从数据库表中获取这些名字),若我在数据库中更改了任何猫的名字记录,那个么它也应该在菜单项中更改。诸如此类
Cats Female
Adorada Amor
Dani Vitala
Cats Male
Glant Diamo
Lorris Diatore
Paol Duece
其中名称来自数据库表cat。
有什么帮助吗?你可以这样试试
在控制器中:
public function abc()
{
$data['plan_data']=$this->plan_model->view_plan();
$this->load->view('templates/header',$header);
$this->load->view('xyz',$data);
$this->load->view('templates/footer');
}
public function view_plan()
{
$sql="SELECT * FROM your table";
$query = $this->db->query($sql);
return $query->result_array();
}
<select class="form-control" name="plan_type" required>
<option value="">--- Select ---</option>
<?php
foreach ($plan_data as $plan){
if($plan['plan_id']==$select_plan){
?>
<option value="<?php echo $plan['plan_id'];?>" selected><?php echo $plan['plan_name'];?></option>
<?php
}else{
?>
<option value="<?php echo $plan['plan_id'];?>"><?php echo $plan['plan_name'];?></option>
<?php
}
}
?>
</select>
在型号中:
public function abc()
{
$data['plan_data']=$this->plan_model->view_plan();
$this->load->view('templates/header',$header);
$this->load->view('xyz',$data);
$this->load->view('templates/footer');
}
public function view_plan()
{
$sql="SELECT * FROM your table";
$query = $this->db->query($sql);
return $query->result_array();
}
<select class="form-control" name="plan_type" required>
<option value="">--- Select ---</option>
<?php
foreach ($plan_data as $plan){
if($plan['plan_id']==$select_plan){
?>
<option value="<?php echo $plan['plan_id'];?>" selected><?php echo $plan['plan_name'];?></option>
<?php
}else{
?>
<option value="<?php echo $plan['plan_id'];?>"><?php echo $plan['plan_name'];?></option>
<?php
}
}
?>
</select>
视图中:
public function abc()
{
$data['plan_data']=$this->plan_model->view_plan();
$this->load->view('templates/header',$header);
$this->load->view('xyz',$data);
$this->load->view('templates/footer');
}
public function view_plan()
{
$sql="SELECT * FROM your table";
$query = $this->db->query($sql);
return $query->result_array();
}
<select class="form-control" name="plan_type" required>
<option value="">--- Select ---</option>
<?php
foreach ($plan_data as $plan){
if($plan['plan_id']==$select_plan){
?>
<option value="<?php echo $plan['plan_id'];?>" selected><?php echo $plan['plan_name'];?></option>
<?php
}else{
?>
<option value="<?php echo $plan['plan_id'];?>"><?php echo $plan['plan_name'];?></option>
<?php
}
}
?>
</select>
---挑选---
这可能对你有帮助。你可以这样试试
在控制器中:
public function abc()
{
$data['plan_data']=$this->plan_model->view_plan();
$this->load->view('templates/header',$header);
$this->load->view('xyz',$data);
$this->load->view('templates/footer');
}
public function view_plan()
{
$sql="SELECT * FROM your table";
$query = $this->db->query($sql);
return $query->result_array();
}
<select class="form-control" name="plan_type" required>
<option value="">--- Select ---</option>
<?php
foreach ($plan_data as $plan){
if($plan['plan_id']==$select_plan){
?>
<option value="<?php echo $plan['plan_id'];?>" selected><?php echo $plan['plan_name'];?></option>
<?php
}else{
?>
<option value="<?php echo $plan['plan_id'];?>"><?php echo $plan['plan_name'];?></option>
<?php
}
}
?>
</select>
在型号中:
public function abc()
{
$data['plan_data']=$this->plan_model->view_plan();
$this->load->view('templates/header',$header);
$this->load->view('xyz',$data);
$this->load->view('templates/footer');
}
public function view_plan()
{
$sql="SELECT * FROM your table";
$query = $this->db->query($sql);
return $query->result_array();
}
<select class="form-control" name="plan_type" required>
<option value="">--- Select ---</option>
<?php
foreach ($plan_data as $plan){
if($plan['plan_id']==$select_plan){
?>
<option value="<?php echo $plan['plan_id'];?>" selected><?php echo $plan['plan_name'];?></option>
<?php
}else{
?>
<option value="<?php echo $plan['plan_id'];?>"><?php echo $plan['plan_name'];?></option>
<?php
}
}
?>
</select>
视图中:
public function abc()
{
$data['plan_data']=$this->plan_model->view_plan();
$this->load->view('templates/header',$header);
$this->load->view('xyz',$data);
$this->load->view('templates/footer');
}
public function view_plan()
{
$sql="SELECT * FROM your table";
$query = $this->db->query($sql);
return $query->result_array();
}
<select class="form-control" name="plan_type" required>
<option value="">--- Select ---</option>
<?php
foreach ($plan_data as $plan){
if($plan['plan_id']==$select_plan){
?>
<option value="<?php echo $plan['plan_id'];?>" selected><?php echo $plan['plan_name'];?></option>
<?php
}else{
?>
<option value="<?php echo $plan['plan_id'];?>"><?php echo $plan['plan_name'];?></option>
<?php
}
}
?>
</select>
---挑选---
这可能会对您有所帮助。我了解您提供的服务,但我是否可以将此代码与我已经创建的其他项目一起放在我的菜单中?如果可以,我应该将您的视图放在我的视图代码中的何处,因为此选择选项应该靠近其他菜单项目?如果我理解正确,我应该将您的视图代码放在视图代码中的某个位置,我之前已经发布了视图代码。但是在哪里?我了解您提供的内容,但是我可以将此代码与我已经创建的其他项目一起放在我的菜单中吗?如果可以,我应该将您的视图放在我的视图代码中的哪里,因为此选择选项应该靠近其他菜单项目?如果我理解正确,我应该将您的视图代码放在视图代码中的某个位置,我之前已经发布了视图代码。但是在哪里呢?