Question

我想分别上传2张图片。我有一个条件，我需要上传文件和一些时间bohth和一些时间只有图像我需要使用单独的按钮如何区分图像和文件。

我的代码是

<form action="http://localhost/cod_login/club/test2" enctype="multipart/form-data" method="post" accept-charset="utf-8">
  <input type="file" name="userfile" size="20">
  <input type="file" name="userfile" size="20">
  <input type="submit" name="submit" value="upload">
</form>

这是控制器

function ddoo_upload(){
    $config['upload_path'] = './uploads/';
    $config['allowed_types'] = 'gif|jpg|png';
    $config['max_size'] = '100';
    $config['max_width']  = '1024';
    $config['max_height']  = '768';

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

    if ( ! $this->upload->do_upload()) {
        $error = array('error' => $this->upload->display_errors());
        $this->load->view('upload_form', $error);
    } else {
        $data = array('upload_data' => $this->upload->data());
        $this->load->view('upload_success', $data);
    }
}

Answer 1

<form action="http://localhost/cod_login/club/test2" enctype="multipart/form-data" method="post" accept-charset="utf-8">


<input type="file" name="userfile" size="20" multiple="">
<input type="submit" name="submit" value="upload">
</form>

Answer 2

如果你想要2个不同的文件按钮，你需要给它们不同的名字。

<form action="" enctype="multipart/form-data" method="post" accept-charset="utf-8">
<input type="file" name="userfile1" size="20">
<input type="file" name="userfile2" size="20">
<input type="submit" name="submit" value="upload">

比你必须修改你的函数ddoo_upload，如下所示

function ddoo_upload($filename){
$config['upload_path'] = './uploads/';
$config['allowed_types'] = 'gif|jpg|png';
$config['max_size'] = '100';
$config['max_width']  = '1024';
$config['max_height']  = '768';

$this->load->library('upload', $config);
if ( ! $this->upload->do_upload($filename)) {
    $error = array('error' => $this->upload->display_errors());
return false;
// $this->load->view('upload_form', $error);
} else {
$data = array('upload_data' => $this->upload->data());
return true;
//$this->load->view('upload_success', $data);
}

}

注意：我们将$ filename作为变量传递，而不是使用它来上传不同的文件。

现在在表单操作重定向的控制器中，您需要编写下面的代码。

if ($this->input->post('submit')){
if (isset($_FILES['userfile1']) && $_FILES['userfile1']['name'] != ''){
    $file1 = $this->ddoo_upload('userfile1');
}

if (isset($_FILES['userfile2']) && $_FILES['userfile2']['name'] != ''){
    $file2 = $this->ddoo_upload('userfile2');
}

}

Answer 3

这是我已申请在codeigniter中上传两张图片的控制器代码

公共职能指数（） {

if($this->input->post('Submit')){

//-----------Image File Section Start Here -----------//

 $config['upload_path'] = './uploads/';   // Directory 
$config['allowed_types'] = 'jpg|jpeg|bmp|png';  //type of images allowed
$config['max_size'] = '30720';   //Max Size
$config['encrypt_name'] = TRUE;   // For unique image name at a time

$this->load->library('upload', $config);  //File Uploading library
$this->upload->do_upload('userfile');  // input name which have to upload 
$video_upload=$this->upload->data();   //variable which store the path

//--------------End of Image File  Section------------------------//



//---------Thumbnail Image Upload Section Start Here -----------//

$config2['upload_path'] = './thumb/';   // Directory 
$config2['allowed_types'] = 'jpg|jpeg|bmp|png'; //type of images allowed
$config2['max_size'] = '30720';   //Max Size
$config2['encrypt_name'] = TRUE;   // For unique image name at a time


$this->upload->initialize($config2); //we can not use upload library again and again it will not initialize again and again so thats why i have used initialize 
$this->upload->do_upload('txt_thumb');  // File Name
$thumbnail_upload=$this->upload->data(); // store the name of the file 

//--------End of Thumbnail Upload Section-----------//



    $date=date("d-m-Y");  // Store current date in variable 

   // Here the database query to insert

    $data = array(
    'parent_id'=> $this->input->post('txt_parent'),
    'cat_id' => $this->input->post('txt_category'),
    'title'=> $this->input->post('txt_title'),
    'status' => $this->input->post('txt_status'),
    'featured' => $thumbnail_upload['file_name'],
    'image' => $video_upload['file_name'],
    'time'=>$date
    );

       $sql_ins= $this->Insimage->insertimage($data);
       if($sql_ins)
       {
           $data['Success'] = "Image has been succesfully inserted!!";
       }

}

此代码肯定可以上传2张图片请享用！！！！： - ）

Answer 4

1-在您的控制器中创建image_uploader

function image_uploader($filename){
  $config['upload_path'] = './assets/uploads/setting/';
  $config['allowed_types'] = 'gif|jpg|png';
  $config['max_size'] = '100';
  $config['max_width']  = '2000';
  $config['max_height']  = '2000';
  $this->load->library('upload', $config);
  if ( ! $this->upload->do_upload($filename)) {
  $error = array('error' => $this->upload->display_errors());
  } else {
  $data_foto = $this->upload->data();
  return $data_foto['file_name'];
}

此函数返回图像文件的名称，如果上传成功

2-在其他函数中调用image_uploader（视图中的表单动作）

public function slideroneupload() {
  $this->site_security->_make_sure_is_admin();
  $data = $this->input->post();
  $sliderimage = $this->image_uploader('image');
  $thumbimage = $this->image_uploader('thumb');

  // save all data to database 
  .
  .
  .

}

如何在codeigniter中上传2个单独的图像

4 个答案: