我尝试了所有可能的方法但没有成功。只是尝试使用代码点火器上传文件,但没有处理我得到的错误
<pre>Array
(
[error] => <p>You did not select a file to upload.</p>
)
我已经尝试在我的本地主机上的正常核心php工作正常但不使用代码点火器。它根本就没有选择文件。如果我查询var_dump($_FILES['fileToUpload']);
,结果将是array(0)
。
表格代码
<form id="contact_form" enctype="multipart/form-data" method="post" action="<?php echo base_url();?>Main/do_upload">
<input type="file" class="form-control" name="fileToUpload" id="fileToUpload">
</form>
控制器代码
$config = array(
'upload_path' => "./uploads/",
'allowed_types' => "gif|jpg|png|jpeg|pdf",
'overwrite' => TRUE,
'max_size' => "2048000", // Can be set to particular file size , here it is 2 MB(2048 Kb)
'max_height' => "768",
'max_width' => "1024"
);
$this->load->library('upload', $config);
if($this->upload->do_upload())
{
$data = array('upload_data' => $this->upload->data());
echo "<pre>";
var_dump($data);
// $this->load->view('upload_success',$data);
}else{
$error = array('error' => $this->upload->display_errors());
echo "<pre>";
print_r($error);
}
配置
$autoload['libraries'] = array("session", "email", "database");
$autoload['helper'] = array("url", "file", "form");
有什么我不知道的吗?请指导我被困在这里。
你错过了do_upload()
中的输入文件名:
使用 :
if(!$this->upload->do_upload('image_file'))
{
//$this->upload->display_errors()
}
else
{
//$this->upload->data()
}
代替:
if($this->upload->do_upload())
你错过了qazxsw poi中的参数请检查下面的代码。
$this->upload->do_upload
在public function do_upload(){
$config = array(
'upload_path' => "assets/uploads/",
'allowed_types' => "gif|jpg|png|jpeg|pdf",
'overwrite' => TRUE,
'max_size' => "2048000", // Can be set to particular file size , here it is 2 MB(2048 Kb)
'max_height' => "768",
'max_width' => "1024"
);
$this->load->library('upload', $config);
$this->upload->initialize($config);
if($this->upload->do_upload('fileToUpload'))
{
$data = array('upload_data' => $this->upload->data());
echo "<pre>";
var_dump($data);
// $this->load->view('upload_success',$data);
}else{
$error = array('error' => $this->upload->display_errors());
echo "<pre>";
print_r($error);
}
}
中传递您的文件上传名称