您没有选择要上传的文件。 PHP 代码点火器

Question

我正在尝试实现发布功能并希望从 php 视图中选择消息和图像。我的消息功能运行良好。但是在图片上传时我收到一个错误You did not select a file to upload。这是我的控制器功能

function post_func()
{
    session_start();
    echo $post_message=$_POST['post'];
    echo $share_with=$_POST['share_with'];
    echo $image=$_POST['image'];
    if($image==null){
        echo "<br/>no image<br/>";  
    }
    else{
        ////////////////////////////////////////////////////////////////////////////////////////

            $config['upload_path'] = './application/css';
            $config['allowed_types'] = 'gif|jpg|png';
            $config['max_size'] = '100';
            $config['max_width']  = '1024';
            $config['max_height']  = '768';


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


            $this->upload->initialize($config);



            if ( ! $this->upload->do_upload())
            {
                $error = array('error' => $this->upload->display_errors());

                echo "<br/>";
                echo $this->upload->display_errors();
                echo "<br/> image error<br/>";
            }
            else
            {

                echo "<br/> reached <br/>";
                session_start();
                $this->membership_model->insert_images($this->upload->data(),$email);
                $data = array('upload_data' => $this->upload->data());

                echo "<br/ problem<br/>";
            }




        ////////////////////////////////////////////////////////////////////////////////////////
    }
    $public;
    if($share_with=="public"){
        echo "1";
        $public=true;
    }else{
        echo "0";
        $public=false;
    }echo "-----------------------------<br/>";
    echo $user=$this->session->userdata('user_identification');
    $data = array 
            (
                'userid'=> $user,
                'public' => $public,
                'message' => $post_message,
                'picname' => "None"
            );
    $this->load->model('membership_model');
    $this->membership_model->add_message($data);
    echo "</br>";
    echo $user=$this->session->userdata('user_identification');
}

这是我的看法。

<?php echo form_open('search/post_func');?>

<!--<form id="loginForm" action="../search/post_func" method="post" enctype="multipart/form-data" >-->
<div id="your_post">
<div id="post_image">
<img   id ="post_img" src="<?php  echo $this->config->item('base_url'); ?><?php echo '/application/css/'. $img ?>"/>
</div>
<textarea name="post" rows="5" cols="30" placeholder="Share an update..." id="post_text" rows="2" value=""></textarea>

//other view items
            <?php
    echo form_close();
    ?> 

请帮帮我

Answer 1

变化：

$this->upload->do_upload()

到这里：

$this->upload->do_upload('my_file_input_name')

它会工作的！ :)

Answer 2

变化：

<?php echo form_open_multipart('search/post_func');?>

Answer 3

您的表单类型应为“多部分”
将您的表单标签更改为：

 <?php echo form_open_multipart('search/post_func');?>

Answer 4

enctype='multipart/form-data'

你应该把之前的属性放在表单标签中

$this->upload->do_upload('file_name')

将输入标记文件名传递到控制器中的前一行

<input type="file"   name ="file_name" >

Answer 5

示例代码是

    $config['upload_path'] = FCPATH.'upload/';
    $config['allowed_types'] = 'gif|jpg|png';
    $config['max_size'] = '100';
    $config['max_width']  = '1024';
    $config['max_height']  = '768';

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

    if ( ! $this->upload-> do_upload('userfile'))
    {

                   $errors = $this->upload->display_errors();
                   $this->session->set_flashdata('error', $errors);
                       redirect('ur_link');
    }
    else
    { 
                            $data = array('upload_data' => $this->upload->data());
                            $fullpath= $data['upload_data']['full_path'];
                            $file_name = $data['upload_data']['file_name'];
                    }
    }

Answer 6

这样做：

if(isset($_FILES['profile_pic']['name']) && !empty($_FILES['profile_pic']['name'])){

    $config['upload_path']= './assets/img/';
    $config['allowed_types'] = 'gif|jpg|png';
    $config['max_size'] = 10000;
    $config['max_width'] = 10000;
    $config['max_height'] = 10000;

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

    if ( ! $this->upload->do_upload('profile_pic'))
    {
        $error = array('error' => $this->upload->display_errors());

        print_r($error);
        exit;
    }
    else
    {
        $data = array('upload_data' => $this->upload->data());

        if(!empty($data)){

            $img1 =  $data['upload_data']['file_name'];                           

        }
    }
}

$insert_data['profile_pic'] = $img1;

Answer 7

我发现发生此问题的另一个原因。如果您在加载库后忘记初始化：

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

$this->upload->initialize($config); //Make this line must be here.

希望对你有所帮助。

Answer 8

所以请注意，我正在执行 mysql 上传导入，并且收到“您没有选择要上传的文件”消息。原来我的导入 sql 文件有额外的列导致失败并返回此错误消息。