Codeigniter form_helper使数据库行成为选择菜单中的值 [英] Codeigniter form_helper getting database rows to be values in select menu

问题描述

我正在写一个表单，其中有一个选择菜单，我想从数据库中提取值，所以我认为这将是这样的东西：

我的检视

 <？php 
 echo form_open（'admin / save_content'）; 
 echo form_fieldset（）; 
 echo form_dropdown（'categories'，$ select_options）; 
 echo form_submit（'category_submit'，'Submit'）; 
 echo form_fieldset_close（）; 
 echo form_close（）; 
？> 
  

我的控制器

 code> function add_content（）{
 $ data = array（）; 
 $ this-> is_logged_in（）; 
 $ this-> load-> model（'category_model'）; 
 $ data ['select_options'] = $ this-> category_model-> get_all_online（）; 
 $ this-> load-> view（'admin / content / add_content'，$ data）; 
} 
  

我的模型

  public function get_all_online（）{
 $ this-> db-> select（'*'）; 
 $ this-> db-> from（'category'）; 
 $ this-> db-> where（'category_online'，1）; 
 $ query = $ this-> db-> get（）; 
 
 return $ query-> result（）; 
 
} 
  

select_options 在下拉菜单中我遇到此错误

遇到PHP错误

严重性：4096

消息：stdClass类的对象
无法转换为字符串

文件名：helpers / form_helper.php

行号：331

解决方案

您需要向下拉列表传递一个数组，其中数组键的值为POSTed，值将为显示的文本。 p>

要实现这一点，请改变你的控制器：

  function add_content ）{
 $ data = array（）; 
 $ this-> is_logged_in（）; 
 $ this-> load-> model（'category_model'）; 
 $ data ['select_options'] = $ this-> category_model-> get_all_online_select（）; 
 $ this-> load-> view（'admin / content / add_content'，$ data）; 
} 
  

并将此函数添加到您的模型

  public function get_all_online_select（）{
 $ this-> db-> select（'id，name'）; //将此更改为要使用的两个主要值
 $ this-> db-> from（'category'）; 
 $ this-> db-> where（'category_online'，1）; 
 $ query = $ this-> db-> get（）; 
 foreach（$ query-> result_array（）as $ row）{
 $ data [$ row ['id']] = $ row ['name']; 
} 
 return $ data; 
} 
  

应该这样做

I am writing a form, which has a select menu in it, I want the values to pulled from the database, so I thought it would be something along these lines:

My view

<?php
   echo form_open('admin/save_content');
   echo form_fieldset();
   echo form_dropdown('categories', $select_options);
   echo form_submit('category_submit', 'Submit');
   echo form_fieldset_close();
   echo form_close();
?>

My controller

function add_content() {
    $data = array();
    $this->is_logged_in();
    $this->load->model('category_model');
    $data['select_options'] = $this->category_model->get_all_online();
    $this->load->view('admin/content/add_content', $data);
}

my model

public function get_all_online() {
    $this->db->select('*');
    $this->db->from('category');
    $this->db->where('category_online', 1);
    $query = $this->db->get();

    return $query->result();

}

now when I place the $selected_options in the form dropdown I get this error,

A PHP Error was encountered

Severity: 4096

Message: Object of class stdClass could not be converted to string

Filename: helpers/form_helper.php

Line Number: 331

解决方案

You need to pass an array to your dropdown, where the array key will be the value that is POSTed and the value will the text that is displayed.

To achieve this, change your controller like so:

function add_content() {
        $data = array();
        $this->is_logged_in();
        $this->load->model('category_model');
        $data['select_options'] = $this->category_model->get_all_online_select();
        $this->load->view('admin/content/add_content', $data);
}

and add this function to your model

public function get_all_online_select() {
        $this->db->select('id, name'); //change this to the two main values you want to use
        $this->db->from('category');
        $this->db->where('category_online', 1);
        $query = $this->db->get();
        foreach($query->result_array() as $row){
            $data[$row['id']]=$row['name'];
        }
        return $data;
}

That should do the trick

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