我在仪表板上的学生晋升功能有问题,我很头晕,但还没有找到解决通知的解决方案:
错误号:1048列'class_id'不能为null INSERT INTO
enroll
(enroll_code
,student_id
,class_id
,year
,date_added
)值('a18b241','254',NULL,'2019 -2020',1532682443)文件名:C:/xampp/htdocs/sisnef/application/controllers/Admin.php
行号:860
<script type="text/javascript">
$(document).ready(function() {
if($.isFunction($.fn.selectBoxIt))
{
$("select.selectboxit").each(function(i, el)
{
var $this = $(el),
opts = {
showFirstOption: attrDefault($this, 'first-option', true),
'native': attrDefault($this, 'native', false),
defaultText: attrDefault($this, 'text', ''),
};
$this.addClass('visible');
$this.selectBoxIt(opts);
});
}
});
</script>
<?php
$students = $this->db->get_where('enroll' , array(
'class_id' => $class_id_from , 'year' => $running_year
))->result_array();
foreach($students as $row):
$query = $this->db->get_where('enroll' , array(
'student_id' => $row['student_id'],
'year' => $promotion_year
));
?>
<tr>
<td align="center">
<?php echo $this->db->get_where('student' , array('student_id' => $row['student_id']))->row()->name;?>
</td>
<td align="center">
<?php if($row['section_id'] != '' && $row['section_id'] != 0)
echo $this->db->get_where('section' , array('section_id' => $row['section_id']))->row()->name;
?>
</td>
<td align="center"><?php echo $row['roll'];?></td>
<td>
<?php if($query->num_rows() < 1):?>
<select class="form-control selectboxit" name="promotion_status_<?php echo $row['student_id'];?>" id="promotion_status">
<option value="<?php echo $class_id_to;?>">
<?php echo get_phrase('Promotion-to') ." - ". $this->crud_model->get_class_name($class_id_to);?>
</option>
<option value="<?php echo $class_id_from;?>">
<?php echo get_phrase('Promotion-to') ." - ". $this->crud_model->get_class_name($class_id_from);?>
</select>
<?php endif;?>
<?php if($query->num_rows() > 0):?>
<center>
<button class="btn btn-success">
<i class="entypo-check"></i> <?php echo get_phrase('Already');?>
</button>
</center>
<?php endif;?>
</td>
</tr>
<?php endforeach;?>
</tbody>
</table>
</div>
</div>
</div>
<br>
<div class="row">
<center>
<button type="submit" class="btn btn-info">
<i class="fa fa-check"></i> <?php echo get_phrase('Promotion-Selected');?>
</button>
</center>
</div>
<!-- begin snippet: js hide: false console: true babel: false -->
我真的希望为这种情况找到一个好的解决方案。谢谢
答案 0 :(得分:0)
您需要为class_id
定义一个值。
$class_id_from
在哪里定义?确保已设置值。
答案 1 :(得分:0)
我可以在您的屏幕截图中看到
https://drive.google.com/file/d/1mFNH8TsdCjhUaFxGcgTw1dnHYyDl6ote/view
您必须
`class_id` varchar(20) NULL,
class_id可以为null,只需在表结构中将其声明为NULL