Question

我尝试回拨用户上传的图像路径（到文件夹目录＆＃39; gambar＆＃39;），方法也将图像数据发送到数据库表。
在下面的代码中，如果图像文件名已经存在，它将以计数作为最后一个字符保存（例如： my-path.jpg ，my-path.png， MY-path1.jpg ）。
但是使用相同的代码，它不会像上面那样生成具有相同路径和扩展名的图像计数器。
我使用Codeigniter 3.1.2和PHP 5.3

问题：如何确保此代码也插入与第2点相同的路径（如果 file-name.extension 已经存在，则添加计数器是唯一的）

控制器＆gt; multiple_upload.php

<?php
class multiple_upload extends CI_Controller {

    function __construct() {
        parent::__construct();
        $this->load->helper(array('form', 'url'));
    }

    function index() {
        //load file upload form
        $this->load->view('multiple_upload_view');
    }

    function upload() {

        // set upload preferences
        $config['upload_path'] = './gambar/';
        $config['allowed_types'] = 'png|jpg|gif|jpeg|JPG|JPEG|GIF|PNG';
        $config['max_size']    = '150';

        //initialize upload class
        $this->load->library('upload', $config);

        $upload_error = array();

        for($i=0; $i<count($_FILES['usr_files']['name']); $i++) {

            $_FILES['userfile']['name']= $_FILES['usr_files']['name'][$i];
            $_FILES['userfile']['type']= $_FILES['usr_files']['type'][$i];
            $_FILES['userfile']['tmp_name']= $_FILES['usr_files']['tmp_name'][$i];
            $_FILES['userfile']['error']= $_FILES['usr_files']['error'][$i];
            $_FILES['userfile']['size']= $_FILES['usr_files']['size'][$i];

            if (!$this->upload->do_upload()) {
                // fail
                $upload_error = array('error' => $this->upload->display_errors());
                $this->load->view('multiple_upload_view', $upload_error);
                break;
            } else {

              $filename = $_FILES['usr_files']['name'][$i];
              $haha = $_FILES['usr_files']['tmp_name'][$i];

              //-- CODE SEND TO DATABASE START

              $kepalabana = array(

                  'img_path' => $filename,
                  'img_name' => 'lol',
                  'login_session_id' => 'wtf',
                  'user_id' => 'wth'

              );

              $this->db->insert('img_tbl', $kepalabana); // my table name is img_tbl

              //-- CODE SEND TO DATABASE END

            }
        }

        // success
        if ($upload_error == NULL) {
            $data['success_msg'] = '<div class="alert alert-success text-center">Done, sila upload gambar lain..</div>';
            $this->load->view('multiple_upload_view', $data);
        }


    }
}

观看＆gt; multiple_upload_view.php

<div class="container">
    <div class="row">
        <div class="col-xs-8 col-xs-offset-2 well">
            <?php echo form_open_multipart('multiple_upload/upload');?>
                <legend>Select Files to Upload:</legend>
                <div class="form-group">
                    <input name="usr_files[]" type="file" multiple="" />
                    <span class="text-danger"><?php if (isset($error)) { echo $error; } ?></span>
                </div>
                <div class="form-group">
                    <input type="submit" value="Upload" class="btn btn-primary btn-block"/>
                </div>
            <?php echo form_close(); ?>
            <?php if (isset($success_msg)) { echo $success_msg; } ?>
        </div>
    </div>
</div>

我的目标：我不打算回拨图片网址（例如： mysite.com/gambar/image-path.jpg 或 mysite.com/gambar/image- path1.jpg ），由用户在我的WYSIWYG上传，而不通过提供从该表生成的json数据来刷新页面。

Answer 1

我知道如何做到这一点...... 因为您为用户上传的所有文件名保存到数据库，然后你可以检查文件名是否已经存在然后如果存在，你可以得到最后一个计数器然后用1添加它并用当前文件名连接它

前：

for($i=0; $i<count($_FILES['usr_files']['name']); $i++) {

    /*
    //you should use id to get max id 
    $strsql = "select max(id) as id from img_tbl where img_path like '".$_FILES['usr_files']['name'][$i]."%'";
    $checkcounter = $this->db->query($strsql); 

    $newcounter = '';
    if (count($checkcounter) > 0) {             
        $lastcounter = $this->db->query("select img_path from img_tbl where id = ".$checkcounter[0]['id']);
        $newcounter = getcounter from lastcounter //=> you can use subtring and length to get the last counter from filename
        $newcounter = $newcounter + 1
    }

    */

    $newfilename = $_FILES['usr_files']['name'][$i] . '-' . $newcounter;  //=> this is new filename contain the counter         

    $_FILES['userfile']['name']= $_FILES['usr_files']['name'][$i];
    $_FILES['userfile']['type']= $_FILES['usr_files']['type'][$i];
    $_FILES['userfile']['tmp_name']= $_FILES['usr_files']['tmp_name'][$i];
    $_FILES['userfile']['error']= $_FILES['usr_files']['error'][$i];
    $_FILES['userfile']['size']= $_FILES['usr_files']['size'][$i];

    if (!$this->upload->do_upload()) {
        // fail
        $upload_error = array('error' => $this->upload->display_errors());
        $this->load->view('multiple_upload_view', $upload_error);
        break;
    } else {

        $filename = $_FILES['usr_files']['name'][$i];
        $haha = $_FILES['usr_files']['tmp_name'][$i];

        //-- CODE SEND TO DATABASE START

        $kepalabana = array(
            'img_path' => $filename,
            'img_name' => 'lol',
            'login_session_id' => 'wtf',
            'user_id' => 'wth'
        );

        $this->db->insert('img_tbl', $kepalabana); // my table name is img_tbl

        //-- CODE SEND TO DATABASE END

    }
}

如果数据库中已存在该值，则向图像文件名添加计数（Codeigniter PHP）

1 个答案: