如何上传照片到我的托管服务器文件夹目录 [英] How to upload photo to my hosting server folder directory

问题描述

你好，我试图从我制作的网站上传照片到网站的托管服务器的文件夹目录（/ public_html / upload /）使用php。但该文件夹总是显示为空。我不知道我做错了什么。任何人都可以帮忙吗？

这是html文件：

 << ; div id =forms> 
< form action =add_data.phpmethod =postform enctype =multipart / form-data> 
 
< div id =info1> 
< p> < H2>事故报告表格< / h2> < / p为H. 
< / div> 
< div id =info2>< p>请填写以下表格以举报意外< / p>< / div> 
 
 
 
 
< li> < label for =name>名称：< / label> 
 
< input" type =textname =nameid =name/>< / li> 
 
< li>< label事故发生地点：< / label> 
 
< inputtype =textname =locationid =location/> < /锂> 
 
< li> < label for =road>道路名称​​：< / label> 
 
< input" type =textname =roadid =road/>< / li> 
 
 
< li> ;< label for =image>图片：< / label> 
 
  
 $ b< / ol> 
< div id =submit>< button type =submit>提交< / button>< / div> ;< br> 
 
 
< / form> 
 
 
< / div> 
  

这是 add_data.php 文件：

<？php 
 $ target =/ public_html / upload; 
 $ target = $ target。basename（$ _FILES ['photo'] ['name' ]）; 
 $ pic =（$ _ FILES ['photo'] ['name']）; 
？> 
  

解决方案

我已经更新了你的代码。格式化得更好，试试一个在线格式化工具
http://jsbeautifier.org/ 。
此外，表格的开标签应该像这样

 < form action =add_data.phpmethod =postenctype =multipart / form-data> 
  

PHP - 评论中提供的解释

 <？php 
 
 //上传的文件将存储在临时位置，需要
 // be移动到目标目录并给出原始文件名
 // ---------------------------------- ------------------------------- 
 
 //准备目标路径名
 // 
 $ target =/ public_html / upload; 
 $ targetName = $ _FILES ['photo'] ['name']; 
 
 //获取临时文件名
 // 
 $ tmp = $ _FILES ['photo'] ['tmp_name']; 
 $ b $ //使用php的move_uploaded_file（）函数
 //将临时文件复制到最终目的地
 move_uploaded_file（$ tmp，$ targetDir。$ targetName）; 
  

Hello I was trying to upload a photo from a website I made, to the website's hosting server's folder directory (/public_html/upload/) using php. but the folder always shows up empty. I dont know what am I doing wrong. can anyone help?

this is the html file:

 <div id = "forms">
  <form action="add_data.php" method="post" form enctype="multipart/form-data">

   <div id = "info1">
   <p> <h2> Accident Report Form </h2> </p>
   </div>
    <div id = "info2"><p> Please fill in the form below to report an accident </p></div>

      <ol>

       <li> <label for = "name" >  Name: </label> 

          <input "type = "text" name = "name" id = "name"/> </li>

         <li> <label for = "location" > Location of the accident: </label>

          <input "type = "text" name = "location" id = "location" /> </li>

         <li> <label for = "road" > Road Name: </label>

          <input "type = "text" name = "road" id = "road"/> </li>


                 <li> <label for = "image" > Image: </label>

          <input type="file" name="photo"><br> </li>


      </ol>
     <div id = "submit"><button type="submit" >Submit</button> </div><br>    


  </form>


 </div>

And, this is the add_data.php file:

 <?php
$target = "/public_html/upload"; 
 $target = $target . basename( $_FILES['photo']['name']);
 $pic=($_FILES['photo']['name']);
   ?> 

解决方案

I have updated your code to work.

HTML - is ok but could be formatted better, try a online formatter tool http://jsbeautifier.org/.
Also, form opening tag should look like this

<form action="add_data.php" method="post" enctype="multipart/form-data">

PHP - Explanation provided in the comments

<?php

// Uploaded file will be stored in a temporary location and needs to
// be moved to a destination directory and given the original filename
//-----------------------------------------------------------------

//  prepare target pathname
//
$target = "/public_html/upload";
$targetName = $_FILES['photo']['name'];

// get temp file name
//
$tmp = $_FILES['photo']['tmp_name'];

// use php's move_uploaded_file() function
// to copy temp file to final destination
move_uploaded_file($tmp, $targetDir . $targetName);

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