Php 在Codeigniter视图中上载和检索图像路径
我对codeigniter是新手。我试着上传一个产品的图片。现在我正在尝试上传每个产品只有一个图像。我已经在我的控制器中完成了这项工作:Php 在Codeigniter视图中上载和检索图像路径,php,codeigniter,Php,Codeigniter,我对codeigniter是新手。我试着上传一个产品的图片。现在我正在尝试上传每个产品只有一个图像。我已经在我的控制器中完成了这项工作: <?php defined('BASEPATH') OR exit('No direct script access allowed'); class Users extends CI_Controller { function __construct() { parent::__construct();
<?php defined('BASEPATH') OR exit('No direct script access allowed');
class Users extends CI_Controller
{
function __construct() {
parent::__construct();
$this->load->model('user');
}
function add(){
if($this->input->post('userSubmit')){
//Check whether user upload picture
if(!empty($_FILES['picture']['name'])){
$config['upload_path'] = 'uploads/images/';
$config['allowed_types'] = 'jpg|jpeg|png|gif';
$config['file_name'] = $_FILES['picture']['name'];
//Load upload library and initialize configuration
$this->load->library('upload',$config);
$this->upload->initialize($config);
if($this->upload->do_upload('picture')){
$uploadData = $this->upload->data();
$picture = $uploadData['file_name'];
}else{
$picture = '';
}
}else{
$picture = '';
}
//Prepare array of user data
$userData = array(
'name' => $this->input->post('name'),
'email' => $this->input->post('email'),
'picture' => $picture
);
//Pass user data to model
$insertUserData = $this->user->insert($userData);
//Storing insertion status message.
if($insertUserData){
$this->session->set_flashdata('success_msg', 'User data have been added successfully.');
}else{
$this->session->set_flashdata('error_msg', 'Some problems occured, please try again.');
}
}
//Form for adding user data
$data['data']=$this->user->getdata();
$this->load->view('show',$data);
}
public function show()
{
#code
$data['data']=$this->user->getdata();
$this->load->view('show',$data);
}
现在的问题是看。我正在尝试动态访问存储的图像,如下所示:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>show</title>
</head>
<body>
<table border='1' cellpadding='4'>
<tr>
<td><strong>User_Id</strong></td>
<td><strong>Name</strong></td>
<td><strong>Image</strong></td>
<td><strong>Option</strong></td>
</tr>
<?php foreach ($data as $p): ?>
<tr>
<td><?php echo $p['id']; ?></td>
<td><?php echo $p['name']; ?></td>
<td><img src="../uploads/images/<?php echo $p['picture']; ?>" /> </td>
<td>
<a href="#">View</a> |
</td>
</tr>
<?php endforeach; ?>
</table>
<img src="../uploads/images/image-0-02-03-15420e726f6e9c97408cbb86b222586f7efe3b002e588ae6bdcfc3bc1ea1561e-V.jpg" />
or like this
<img src="../../uploadsimages/image-0-02-03-15420e726f6e9c97408cbb86b222586f7efe3b002e588ae6bdcfc3bc1ea1561e-V.jpg" />
显示
用户Id
名称
图像
选项
" />
|
使用此图像源,图像不会出现。只看到imge crack缩略图。我还尝试手动提供图像的根文件夹,如下所示:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<title>show</title>
</head>
<body>
<table border='1' cellpadding='4'>
<tr>
<td><strong>User_Id</strong></td>
<td><strong>Name</strong></td>
<td><strong>Image</strong></td>
<td><strong>Option</strong></td>
</tr>
<?php foreach ($data as $p): ?>
<tr>
<td><?php echo $p['id']; ?></td>
<td><?php echo $p['name']; ?></td>
<td><img src="../uploads/images/<?php echo $p['picture']; ?>" /> </td>
<td>
<a href="#">View</a> |
</td>
</tr>
<?php endforeach; ?>
</table>
<img src="../uploads/images/image-0-02-03-15420e726f6e9c97408cbb86b222586f7efe3b002e588ae6bdcfc3bc1ea1561e-V.jpg" />
or like this
<img src="../../uploadsimages/image-0-02-03-15420e726f6e9c97408cbb86b222586f7efe3b002e588ae6bdcfc3bc1ea1561e-V.jpg" />
还是像这样
但这没用。有人能帮我吗?欢迎提供任何建议和建议。谢谢。试试这个
<img src="<?php echo base_url('uploads/images/'.$p['picture']); ?>"/>
insted of this line
<img src="../uploads/images/<?php echo $p['picture']; ?>" />
“/>
代替这条线
" />
还可以正确设置文件夹位置试试这个
<img src="<?php echo base_url('uploads/images/'.$p['picture']); ?>"/>
insted of this line
<img src="../uploads/images/<?php echo $p['picture']; ?>" />
“/>
代替这条线
" />
还可以正确设置文件夹位置。这是一个愚蠢的错误。我让虚拟主机指向根文件夹的index.html。因此,只要指向根文件夹,问题就解决了。这是一个愚蠢的错误。我让虚拟主机指向根文件夹的index.html。因此,只要指向根文件夹,问题就解决了