上传文件并将其名称显示为帖子

Question

我正在尝试上传文件并将其名称替换为标题名称，但我无法通过控制器中的 echo 获取文件名，即使不是作为分析器中的 POST。

我还需要重命名它，但在此之前我需要知道发布值，即文件名。

我在这里发布我的代码。

我的看法

<?php echo form_open_multipart('emailtemplate/do_upload');?>

<tr class='odd gradeX'>
    <td class='center'>
        <input type="file" name="userfile" size="20" />
    </td>
    <td class='center'>
        <input placeholder='Title For Your File' name='title' class='form-control'>
    <td class='center'>
        <input placeholder='Comment On File' name='comment' class='form-control'>
    </td>
</tr>

<input type="submit" value="upload" />

</form>

我的控制器

function do_upload()
{
    $this->load->helper('form');
    $this->load->helper('html');
    $config['upload_path'] = './uploads/';
    $config['allowed_types'] = 'gif|jpg|png';
    $config['max_size'] = '1000';
    $config['encrypt_name']     = true;
    //$file=$this->input->post('userfile');
    $this->load->library('upload', $config);
    $this->upload->data();
    $file_name = $this->upload->do_upload('userfile');
    echo $file_name;


    if ( ! $this->upload->do_upload())
    {
        $error = array('error' => $this->upload->display_errors());

        $this->load->view('add/addfile', $error);
    }
    else
    {
        $data = array('upload_data' => $this->upload->data());

        $this->load->view('add/addfile', $data);
    }
    $this->output->enable_profiler(true);
}

Answer 1

使用 POST 值重命名上传的文件：

//get extension 
  $ext = end(explode(".", $_FILES["userfile"]['name']));
  $config['file_name'] = $this->input->post("title").'.'.$ext;

删除$config['encrypt_name'] = true;。否则你的文件名将被加密。

确保您的title 不存在于您上传的文件夹中。或者在标题后面附加一些随机数/字符。

Answer 2

我已经创建了一个上传文件的函数：-

它需要参数，即 field_name

    public function fileUpload($field) {
    //get file extension
    $exts = @split("[/\\.]", $_Files[$field][name]);
    $n = count($exts) - 1;
    $ext = $exts[$n];
    //create image name  or replace it with your desired name
    $imageName = date("Ymdhis") . time() . rand();
    $config['file_name'] = $imageName; //set the desired file name
    $config['overwrite'] = true;        
    $config['allowed_types'] = str_replace(',', "|", $ext);
    $config['upload_path'] = ABSOLUTEPATH . 'product/original/';
    $config['optional'] = true;
    $config['max_size'] = '1000';
    $config['max_width'] = '00';
    $this->load->library('upload', $config);
    $this->upload->initialize($config);
    if (!$this->upload->do_upload($field, true)) {
        $error = array('error' => $this->upload->display_errors());
        $r = $error;
    } else {
        $data = array('upload_data' => $this->upload->data());
        $r = "sucess";
    }        
    return $r;
}

Answer 3

下面我写的代码对你有帮助

$config['upload_path'] = './uploads/path';
        $config['allowed_types'] = 'png|jpg|jpeg|gif';
        $config['max_size']    = '1000000';
        $config['max_width']  = '1000000';
        $config['max_height']  = '1000000000';

        $this->load->library('upload', $config);
        $files = $_FILES;

        if(isset($files['file']['name']))
        {   
            $filename = $files['file']['name'];
            $ext = end(explode(".", $filename));
            $_FILES['file']['name']=  date('d').date('m').date('y').'_'.date(time()).'.'.$ext; //here i am changing file name with current date as your wish to change your logic   

        //  print_r($_FILES);exit; check the changed name

           if ($this->upload->do_upload('file')) 
             {
              $upload = array('upload_data' => $this->upload->data());
              $imagename = $upload['upload_data']['file_name']; //uploaded your image name
             }
             else
             {
                    $error = array('error' => $this->upload->display_errors());
                    print_r($error); //error 
             }
        }

Answer 4

注意：嘿 user3345331 ，请参阅该行

$this->upload->do_upload() 

用这个替换

$this->upload->do_upload('userfile');

此用户文件是您的 HTML 代码中具有文件类型的输入框的 ID。