Php codeigniter在标题上显示图像
我使用了会话数据。但是我想在DB上做sessiondata=一些事情。比如说Php codeigniter在标题上显示图像,php,codeigniter,session,login,codeigniter-3,Php,Codeigniter,Session,Login,Codeigniter 3,我使用了会话数据。但是我想在DB上做sessiondata=一些事情。比如说 <a data-toggle="dropdown" class="dropdown-toggle" href="#"> <img alt="" src="<?php if(($this->session->userdata('people_id')) == 'cosId') { echo base_url
<a data-toggle="dropdown" class="dropdown-toggle" href="#">
<img alt="" src="<?php
if(($this->session->userdata('people_id')) == 'cosId') { echo base_url().'upload/customer/'.$this->session->userdata('people_img');} else{ echo base_url().'upload/user/'.$this->session->userdata('people_img');} ?>">
<span class="username"><?php echo $this->session->userdata('people_name'); ?> <?php echo $this->session->userdata('people_surname'); ?></span>
<b class="caret"></b>
</a>
我可以为客户这样编辑吗
<?php $cus_data = $this->db->get_where('customer');
foreach ($cus_data->result_array() as $row) {
if($this->session->set_userdata('people_id',$row['cusId']))
{
$src = base_url('upload/customer/'.$this->session->userdata('people_img'));
}
else
{
$src = base_url('upload/user/'.$this->session->userdata('people_img'));
}}
?>
这样做: 第一种方式: 在您的登录方法中,还为image
url设置一个会话
对于客户:
$cust_src = base_url('upload/customer/'.$row['cusImg']);
$this->session->set_userdata('image_src',$cust_src);
对于用户:
$user_src = base_url('upload/customer/'.$row['userImg']);
$this->session->set_userdata('image_src',$user_src);
你的主播img应该是这样的:
<a data-toggle="dropdown" class="dropdown-toggle" href="#">
<img alt="" src="<?=$this->session->userdata('image_src');;?>">
<span class="username"><?php echo $this->session->userdata('people_name'); ?> <?php echo $this->session->userdata('people_surname'); ?></span>
<b class="caret"></b>
</a>
第二种方法是这样做:
<a data-toggle="dropdown" class="dropdown-toggle" href="#">
<?php if($this->session->userdata('people_id') == 'cosId')
{
$src = base_url('upload/customer/'.$this->session->userdata('people_img'));
}
else
{
$src = base_url('upload/user/'.$this->session->userdata('people_img'));
}
?>
<img alt="" src="<?php echo $src;?>" >
<span class="username"><?php echo $this->session->userdata('people_name'); ?> <?php echo $this->session->userdata('people_surname'); ?>
</span>
<b class="caret"></b>
</a>
您遇到了什么错误?我没有遇到任何错误。客户登录时,标题上没有图像。您希望实现什么?条件看起来正常,那么什么还不起作用?我想显示当客户登录时,客户从客户目录中看到自己的图片,或者当用户登录时,用户可以从用户目录中看到自己的图片。不,我看不到客户图片。cosId是数据库上customer表中的一个字段。你认为我能和这个相等吗?($this->session->userdata('people_id')=='cosId')如果会话值与您的db customer_id匹配,它肯定会向您显示结果,请确保两者都是相同的图像路径是正确的。我在上面添加了userdata的控制器部分。有点困惑,如果$cus\u data
和$user\u data
都有数据,您的会话会是什么?我怎么做?。
<a data-toggle="dropdown" class="dropdown-toggle" href="#">
<img alt="" src="<?=$this->session->userdata('image_src');;?>">
<span class="username"><?php echo $this->session->userdata('people_name'); ?> <?php echo $this->session->userdata('people_surname'); ?></span>
<b class="caret"></b>
</a>
<a data-toggle="dropdown" class="dropdown-toggle" href="#">
<?php if($this->session->userdata('people_id') == 'cosId')
{
$src = base_url('upload/customer/'.$this->session->userdata('people_img'));
}
else
{
$src = base_url('upload/user/'.$this->session->userdata('people_img'));
}
?>
<img alt="" src="<?php echo $src;?>" >
<span class="username"><?php echo $this->session->userdata('people_name'); ?> <?php echo $this->session->userdata('people_surname'); ?>
</span>
<b class="caret"></b>
</a>