我有以下codeigniter代码,这取决于导演的数量。意味着如果我的任何董事都是2,那么下面的代码将是2次。所以我用了唯一的id。我想一键上传所有图像( 1个1个图像的1个导演)。他们的名字应该以 director1 _
开头<div class="row">
<h5 style="padding: 0 15px;">Details for Director 1</h5>
<div class="form-group col-md-4">
<label>Full Name</label>
<input type="hidden" name="director[0][cid]" value="9">
<input type="text" class="form-control" name="director[0][name]" required="required" placeholder="Enter Your Full Name" autocomplete="off">
</div>
<div class="form-group col-md-4">
<label>Father's Name</label>
<input type="text" class="form-control" name="director[0][fname]" required="" placeholder="Enter Father's Name">
</div>
<div class="form-group col-md-4">
<label>Address</label>
<input type="text" class="form-control" name="director[0][address]" placeholder="Enter Address">
</div>
</div>
<div class="row">
<div class="form-group col-md-4">
<label>Color Photograph</label>
<input type="file" class="form-control" name="image[0][photo]">
</div>
<div class="form-group col-md-4">
<label>PAN Card</label>
<input type="file" class="form-control" name="image[0][pan]">
</div>
<div class="form-group col-md-4">
<label>Address Proof</label>
<input type="file" class="form-control" name="image[0][address]">
</div>
</div>
在我的控制器中,我使用了以下代码
function upload_files($path, $title, $files)
{
$config = array(
'upload_path' => $path,
'allowed_types' => 'jpg|gif|png',
'overwrite' => 1,
);
$this->load->library('upload', $config);
$images = array();
foreach ($files['image'] as $key => $image) {
$_FILES['images[]']['name']= $files['name'][$key];
$_FILES['images[]']['type']= $files['type'][$key];
$_FILES['images[]']['tmp_name']= $files['tmp_name'][$key];
$_FILES['images[]']['error']= $files['error'][$key];
$_FILES['images[]']['size']= $files['size'][$key];
$fileName = $title .'_'. $image;
$images[] = $fileName;
$config['file_name'] = $fileName;
$this->upload->initialize($config);
if ($this->upload->do_upload('images[]')) {
$this->upload->data();
} else {
return false;
}
}
return $images;
}
但是,它不起作用。任何帮助将不胜感激