Question

首先我正在使用cakePHP 2.3框架，我创建了一个表格，我也有一些关系。好的是，如果字段不是如下所示的数组形式，我能够保存数据。

<input type="text" name="data['Academicrecord']['school'][]" />
<input type="text" name="data['Academicrecord']['award'][]" />
<input type="text" name="data['Academicrecord']['from'][]" />
<input type="text" name="data['Academicrecord']['to'][]" />

但是当我提交上面的表单数据时，我收到此错误。

错误：SQLSTATE [42S22]：未找到列：1054未知列＆＃39;数组＆＃39;在＆＃39;字段列表＆＃39;

查看内置调试函数中cakePHP生成的信息。

debug($this->request->data);

'Academicrecord' => array(
    'school' => array(
        (int) 0 => 'Kansanga Primary School',
        (int) 1 => 'Tropical High School'
    ),
    'award' => array(
        (int) 0 => 'P.L.E',
        (int) 1 => 'U.C.E'
    ),
    'from' => array(
        (int) 0 => '1997',
        (int) 1 => '2003'
    ),
    'to' => array(
        (int) 0 => '2002',
        (int) 1 => '2006'
    )
)

在我解决这个问题的路上，我想我可以使用for循环来迭代提交的项目，然后将它们（数据）发送到模型。不幸的是，我的for循环不能迭代直到第二行。

          echo  "The counts: ".count( $this->request->data['Academicrecord']['school']).'<br/>';

       for($i=0;$i<count( $this->request->data['Academicrecord']['school']);$i++){

         $this->request->data['Academicrecord']['staff_id'] = $this->Staff->id;
         echo "{$i} :".$this->request->data['Academicrecord']['school'] =   $this->request->data['Academicrecord']['school'][$i];
         echo ' '.$this->request->data['Academicrecord']['award']  =   $this->request->data['Academicrecord']['award'][$i];
         echo ' '.$this->request->data['Academicrecord']['from']   =   $this->request->data['Academicrecord']['from'][$i];
         echo ' '.$this->request->data['Academicrecord']['to']     =   $this->request->data['Academicrecord']['to'][$i];
         echo "<br/>";

           //$this->Staff->Academicrecord->save($this->request->data);
       }

提交后的示例结果。

行数：2

第0行：Kansanga小学PLE 1990 2004

Answer 1

在循环中，您需要创建如下数据：

echo  "The counts: ".count( $this->request->data['Academicrecord']['school']).'<br/>';

       for($i=0;$i<count( $this->request->data['Academicrecord']['school']);$i++){

         $this->request->data['Academicrecord']['staff_id'] = $this->Staff->id;
         echo "{$i} :".$this->request->data['Academicrecord']['school'] =   $this->request->data['Academicrecord']['school'][$i];
         echo ' '.$this->request->data['Academicrecord']['award']  =   $this->request->data['Academicrecord']['award'][$i];
         echo ' '.$this->request->data['Academicrecord']['from']   =   $this->request->data['Academicrecord']['from'][$i];
         echo ' '.$this->request->data['Academicrecord']['to']     =   $this->request->data['Academicrecord']['to'][$i];
         //echo "<br/>";
         **$this->Staff->Academicrecord->create();**

         $this->Staff->Academicrecord->save($this->request->data);
       }

我不确定你为什么使用$ this-＆gt; Staff-＆gt; Academicrecord而不是$ this-＆gt; Academicrecord。

Answer 2

我终于从这里得到了答案。 http://topdsoft.com/stories/view/186

如何在cakePHP 2.3中保存这种数据？

2 个答案: