我正在做这件事来获取数据库的值,我一直在使用这个,但现在我收到了这个错误。这个正在研究我制作的每个模块,但是我的模块没有工作,为什么呢?
错误是,
localStorage.setItem(keyName, keyValue);
模型是
Severity: Notice
Message: Trying to get property of non-object
Filename: views/add_view.php
Line Number: 18
Backtrace:
工作正常。
控制器,
public function idgenerator(){
$sql = "SELECT * FROM sequence_generator where DESCRIPTION = 'ID' AND IN_USE = 'YES'";
$query = $this->db->query($sql);
return $query->result();
}
我也检查并且工作得很好
查看模板
public function add_view(){
//redirecting and passing data for combobox
$data['content_view'] = 'Employees/add_view';
$data['content1'] = $this->Employees_Model->nationality('nationality');
$data['content'] = $this->Employees_Model->idgenerator('idgenerator');
$data['content4'] = $this->Employees_Model->provinces('provinces');
$this->templates->admin_template($data);
}
获取此视图,其中包含错误。
<?php $this->load->view($content_view, $content = NULL, $content1 = NULL, $content2 = NULL, $content3 = NULL, $content4 = NULL, $content5 = NULL, $pagination = NULL); ?>
这里是我有几乎相同代码但没有任何错误的部分。
<div class="box-body">
<div class="form-group">
<label for="ID_NUM" class="col-sm-2 control-label col-sm-offset-2">ID Number:</label>
<div class="col-sm-5">
<input type="text" class="form-control" id="ID_NUM" name="ID_NUM" value="<?php echo $content->SEQUENCE_CODE.'-'.$content->NEXT_A.'-'.$content1->NEXT_B; ?>" disabled>
</div>
</div>
</div>
答案 0 :(得分:0)
<?php $this->load->view($content_view, $content = NULL, $content1 = NULL, $content2 = NULL, $content3 = NULL, $content4 = NULL, $content5 = NULL, $pagination = NULL); ?>
也许我落后于PHP时代,但使用那些$variable = NULL
通常用于声明一个函数,而不是使用一个函数。我有一种感觉,你传递了一堆布尔语句,而不是你的变量。
如果您尝试这样做怎么办:
<?php $this->load->view($content_view, $content, $content1, $content2, $content3 , $content4, $content5, $pagination); ?>
答案 1 :(得分:0)
Codeigniter的view方法最多需要三个参数:
view($view, $vars = array(), $return = FALSE)
您需要将数组或对象数组传递给视图:
$data = array('content1' => $content1, 'content2' => $content2); //etc
$this->load->view($content_view, $data);