加入两个表codeigniter mysql php时出现“消息：未定义变量：查询”错误

Question

我有两个名为 'addexpense' 和 'addcategory' 的表。 我已成功加入每个表，但在我的视图页面上没有查看数据，并且屏幕上传递了一条错误消息。请帮忙。

这是我的模特

public function getExpenses(){
 $this->db->select("addexpense.exp_date, addexpense.exp_amount, addexpense.exp_note, addexpense.exp_created, addcategory.category_name");
 $this->db->from('addexpense');
 $this->db->join('addcategory', 'addcategory.category_id = addexpense.category_id');
 $query = $this->db->get();
 return $query->result();
}

这是我的控制器

public function join(){
  $query = $this->base_model->getExpenses();
  //$data['result'] = null;
    if($query)
    {
        $data['query'] =  $query;
    }
  $this->load->view('listexpense', $data);
}

这是我的视图代码

<tr>
    <td><strong>Date</strong></td>
    <td><strong>Amount</strong></td>
    <td><strong>Note</strong></td>
    <td><strong>Created</strong></td>
    <td><strong>Category Name</strong></td>
</tr> 
<?php foreach($query as $expenses){?>
<tr>
    <td><?=$expenses->exp_date;?></td>
    <td><?=$expenses->exp_amount;?></td>
    <td><?=$expenses->exp_note;?></td>
    <td><?=$expenses->exp_created;?></td>
    <td><?=$expenses->category_name;?></td>
</tr>     
<?php }?>

Answer 1

这是代码，请检查并更新 型号

public function getExpenses(){
   $this->db->select("addexpense.exp_date, addexpense.exp_amount, addexpense.exp_note, addexpense.exp_created, addcategory.category_name");
   $this->db->from('addexpense');
   $this->db->join('addcategory', 'addcategory.category_id = addexpense.category_id');
   $query = $this->db->get();
   $query_rows = $query->num_rows();
   if($query_rows > 0){
          return $query->result();
   } else{
          return 0;
   }
}

对于控制器

public function join(){
    $this->load->model('base_model');
    $query = $this->base_model->getExpenses();
    //print_r($query);die();
    $data['query'] = '';
    if($query != 0){
            $data['query'] = $query;
    }

    $this->load->view('listexpense', $data);
}

供查看

<tr>
    <td><strong>Date</strong></td>
    <td><strong>Amount</strong></td>
    <td><strong>Note</strong></td>
    <td><strong>Created</strong></td>
    <td><strong>Category Name</strong></td>
</tr> 
<?php
    if($query != ''){ 
    foreach($query as $expenses){?>
        <tr>
        <td><?=$expenses->exp_date;?></td>
        <td><?=$expenses->exp_amount;?></td>
        <td><?=$expenses->exp_note;?></td>
        <td><?=$expenses->exp_created;?></td>
        <td><?=$expenses->category_name;?></td>
    </tr>     
<?php } }?>

删除注释“//print_r($query);die();”并检查模型是否正在发送数据，请更新..

Answer 2

模型将 result() 更改为 result_array()

public function getExpenses(){
    $this->db->select("addexpense.exp_date, addexpense.exp_amount, addexpense.exp_note, addexpense.exp_created, addcategory.category_name");
    $this->db->from('addexpense');
    $this->db->join('addcategory', 'addcategory.category_id = addexpense.category_id');
    $query = $this->db->get();
    return $query->result_array();
}

控制器尝试下面的代码

public function join()
{
    $results = $this->base_model->getExpenses();

    $data['expenses'] = array();

    foreach ($results as $result) {
        $data['expenses'][] = array(
            'exp_date' => $result['exp_date'],
            'exp_amount' => $result['exp_amount'],
            'exp_note' => $result['exp_note'],
            'exp_created' => $result['exp_created'],
            'category_name' => $result['category_name']
        );
    }

    $this->load->view('listexpense', $data);

}

查看确保是正确的视图

<table>
<thead>
</thead>
<tbody>
<?php foreach($expenses as $expense){?>
<tr>
    <td><?php echo $expense['exp_date'];?></td>
    <td><?php echo $expense['exp_amount'];?></td>
    <td><?php echo $expense['exp_note'];?></td>
    <td><?php echo $expense['exp_created'];?></td>
    <td><?php echo $expense['category_name'];?></td>
</tr>     
<?php }?>
</tbody>
</table>

Answer 3

这样设置控制器功能

   public function join(){
      $data['query'] = $this->base_model->getExpenses();
      $this->load->view('listexpense', $data);
    }

Answer 4

您似乎还没有在controller 中初始化model。

在您的controller 中，您需要先加载model，然后才能使用它的properties。请检查下面的控制器方法。

public function join() {
    $this->load->model('base_model');
    $query = $this->base_model->getExpenses();
    $data = array();
    if ($query) {
        $data['query'] = $query;
    }
    $this->load->view('listexpense', $data);
}

让我知道它仍然不起作用。

Answer 5

请您可以通过在视图开始时打印来检查您的数据是否在视图文件中

<?php echo '<pre>';
 print_r($query);
 ?>