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Php 未在codeigniter中上载文件_Php_Codeigniter - Fatal编程技术网

Php 未在codeigniter中上载文件

Php 未在codeigniter中上载文件,php,codeigniter,Php,Codeigniter,我试图通过codeigniter控制器上传2个文件。当我选择文件并点击提交时,它总是返回错误。但是当我做一个var_dump$文件时;它显示文件正在传递,但没有被codeigniter控制器捕获 有人能告诉我我做错了什么吗?下面是我的代码 $config['upload_path'] = './docs/'; $config['allowed_types'] = 'jpg|doc|docx';

我试图通过codeigniter控制器上传2个文件。当我选择文件并点击提交时,它总是返回错误。但是当我做一个var_dump$文件时;它显示文件正在传递,但没有被codeigniter控制器捕获

有人能告诉我我做错了什么吗?下面是我的代码

            $config['upload_path']          = './docs/';
            $config['allowed_types']        = 'jpg|doc|docx';
            $config['max_size']             = 10000;
            $config['max_width']            = 3000;
            $config['max_height']           = 3000;

            $this->load->library('upload', $config);
if ( !$this->upload->do_upload('userfile1') || !$this->upload->do_upload('userfile2'))
{

echo "error";

} else {

$f1= $this->upload->data('userfile1');
$f2= $this->upload->data('userfile2');

echo $f1['file_name'];
echo $f2['file_name'];

}

对于CI中的多个文件上载,请按照以下方式操作-

$config['upload_path']   = 'uploads/photos/'; 
$config['allowed_types'] = 'jpg|jpeg|png|gif'; 
$this->load->library('upload', $config);
for ($i=0; $i < count($_FILES['photos']['name']); $i++) { 
    $_FILES['photos[]']['name']     = $_FILES['photos']['name'][$i];
    $_FILES['photos[]']['type']     = $_FILES['photos']['type'][$i];
    $_FILES['photos[]']['tmp_name'] = $_FILES['photos']['tmp_name'][$i];
    $_FILES['photos[]']['error']    = $_FILES['photos']['error'][$i];
    $_FILES['photos[]']['size']     = $_FILES['photos']['size'][$i]; 
    if ($this->upload->do_upload('photos[]')) {
        $photos_files = array('upload_data' => $this->upload->data()); 
        $photos_arr[] = $photos_files['upload_data']['file_name'];
    }else{
        $error[] = $this->upload->display_errors();
    }
}

你好谢谢你的回复。。。我有两个单独的输入字段,用于两种不同类型的文档。那么,如何在不使用循环的情况下获取文件呢?您可以将两个输入的名称属性值定义为like-name=photo[]