Question

我在搜索脚本时遇到问题。问题是我想创建过滤器。过滤器是下拉菜单，我从此下拉菜单中输入数据到数组。我认为这是问题但是如何以不同的方式进入它。

或者如何将模型中的数据作为数组传递而不是obj。（我认为这是对象）

我的搜索：

          <?php echo form_open("otk/triedit", 'class="form-inline"'); ?>
          <th scope="col"></th>
          <th scope="col">ČÍSLO ZÁKAZKY</th>
          <th scope="col"><?php echo form_dropdown(['class' => 'form-control', 'name' => 'p'], $pozicia); ?>POZÍCIA</th>
          <th scope="col"><?php echo form_dropdown(['class' => 'form-control', 'name' => 's'], $stav); ?>STAV</th>
          <th scope="col">PORADOVÉ ČÍSLO</th>
          <th scope="col"><?php echo form_dropdown(['class' => 'form-control', 'name' => 'st'], $technologia); ?>TECHNOLÓGIA</th>
          <th scope="col"><?php echo form_dropdown(['class' => 'form-control', 'name' => 'd'], $datum); ?>DÁTUM</th>
          <th scope="col">DOKUMENT</th>
          <th scope="col"><?php echo form_dropdown(['class' => 'form-control', 'name' => 'z'], $zariadenie); ?>ZARIADENIE</th>
          <th scope="col">OPERÁTOR</th>
          <th scope="col"><center></center></th>
          <?php echo form_submit(['type' => 'submit', 'class' => 'btn btn-warning', 'value' => 'Hľadať']); ?>
          <?php echo form_close(); ?>

我的控制器：

function triedit()
    {
        $this->load->model('base_model');

    $udaje = array(
        'hladat' => $this->input->post('p'),
        'hladat' => $this->input->post('s'),
        'hladat' => $this->input->post('st'),
        'hladat' => $this->input->post('d'), 
        'hladat' => $this->input->post('z')
    );

    $keyword    =   $udaje;
    $data['resulte']    =   $this->base_model->triedit($keyword);
    $this->load->view('base_triedit_view',$data);
}

我的模特：

function triedit($keyword)
    {
        $this->db->select('*');
        $this->db->from('otk');
        $this->db->like('ckod_otk',$keyword);
        $this->db->join('technologia', 'technologia.id_technologia = otk.technologia_otk', 'left');
        $this->db->join('zariadenie', 'zariadenie.id_zariadenie = otk.zariadenie_otk', 'left');
        $this->db->join('zamestnanci', 'zamestnanci.id_zamestnanci = otk.operator_otk', 'left');
        $this->db->join('stav', 'stav.id_stav = otk.stav_otk', 'left');
        $query  =   $this->db->get();
        return $query->num_rows();
    }

我的观点预览：

<?php if(is_array($resulte)): ?>
        <?php 
            foreach ($resulte as $row) {?>
        <tr>
          <td></td>
          <td><?php echo $row->czakazky_otk; ?></td>
          <td><?php echo $row->pozicia_otk; ?></td>
          .....
          <?php } ?>
     <?php endif; ?>

Answer 1

据我了解你的问题

form_dropdown的数据选项格式始终位于array，如果您使用form_dropdown

，则无法替代

如果您希望您的模型应该将数据作为数组返回

这样做：

function triedit($keyword)
{
    $this->db->select('*');
    $this->db->from('otk');
    $this->db->like('ckod_otk',$keyword);
    $this->db->join('technologia', 'technologia.id_technologia = otk.technologia_otk', 'left');
    $this->db->join('zariadenie', 'zariadenie.id_zariadenie = otk.zariadenie_otk', 'left');
    $this->db->join('zamestnanci', 'zamestnanci.id_zamestnanci = otk.operator_otk', 'left');
    $this->db->join('stav', 'stav.id_stav = otk.stav_otk', 'left');
    $query = $this->db->get();
    if ($query->num_rows() > 0 )
    {
       return $query->result_array();
    }
}

视图中的更改：

<?php if(is_array($resulte)): ?>

<?php 
    foreach ($resulte as $row) {?>
    <tr>
      <td></td>
      <td><?php echo $row['czakazky_otk']; ?></td>
      <td><?php echo $row['pozicia_otk']; ?></td>
      ..............
<?php } ?>
<?php endif; ?>

更多信息：https://www.codeigniter.com/user_guide/database/results.html#result-arrays

数组到字符串转换（在codeigniter中搜索）

1 个答案: