我有兩個名爲'addexpense'和'addcategory'的表。 我已成功加入每個表格,但在我的查看頁面上數據未查看,並且錯誤消息在屏幕上傳遞。請幫忙。「消息:未定義的變量:查詢」加入兩個表的錯誤codeigniter mysql php
這是我的模型
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();
}
這是我的控制器
public function join(){
$query = $this->base_model->getExpenses();
//$data['result'] = null;
if($query)
{
$data['query'] = $query;
}
$this->load->view('listexpense', $data);
}
這是我的看法代碼
<tr>
<td><strong>Date</strong></td>
<td><strong>Amount</strong></td>
<td><strong>Note</strong></td>
<td><strong>Created</strong></td>
<td><strong>Category Name</strong></td>
</tr>
<?php foreach($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 }?>
'$ query = $ this-> base_model-> getExpenses(); print_r($ query);退出;'檢查值是否來自模態或不? – JYoThI
是的,我alredy檢查和通過我的控制器頁測試\t \t功能測試() \t \t { \t \t \t $查詢= $這個 - > base_model-> getExpenses(); \t \t \t的foreach($查詢作爲$查詢) \t \t \t { \t \t \t \t回聲$查詢 - > EXP_DATE; \t \t \t \t回聲$查詢 - > exp_amount; \t \t \t \t回聲$查詢 - > exp_note; \t \t \t \t回聲$查詢 - > exp_created; \t \t \t \t回聲$查詢 - > CATEGORY_NAME; \t \t \t} \t \t} – Yahiya
的結果是什麼。你獲得了價值嗎? – JYoThI