Question

我正在创建一个表单，用户最多可以上传4张图片并更改上传的图片。照片会上传到文件夹上传/ 。

如果文件夹中已存在任何图片，表单将显示现有图片，以便用户可以检查是否更改图片。

HTML＆amp; PHP代码如下：

     <form name="upload_form" action="upload.php">
      <div id="upload_block0" class="upload_block" style="margin: 1rem;">
         <?php 
         if( file_exists("upload/pic0.jpg")){
           echo "<img src='upload/pic0.jpg'>";
         }
         ?>
         <input type="file" name="picture[]" id="picture" />
         <button onclick="removeBlock()">Remove</button>
      </div>

      <div id="upload_block1" class="upload_block" style="margin: 1rem;">
         <?php 
         if( file_exists("upload/pic1.jpg")){
           echo "<img src='upload/pic1.jpg'>";
         }
         ?>
        <input type="file" name="picture[]" id="picture" />
        <button onclick="removeBlock()">Remove</button>
      </div>

      <div id="upload_block2" class="upload_block" style="margin: 1rem;">
        <?php 
         if( file_exists("upload/pic2.jpg")){
           echo "<img src='upload/pic2.jpg'>";
         }
         ?>
        <input type="file" name="picture[]" id="picture" />
        <button onclick="removeBlock()">Remove</button>
      </div>

      <div id="upload_block3" class="upload_block" style="margin: 1rem;">
        <?php 
         if( file_exists("upload/pic3.jpg")){
           echo "<img src='upload/pic3.jpg'>";
         }
         ?>
        <input type="file" name="picture[]" id="picture" />
        <button onclick="removeBlock()">Remove</button>
      </div>

      <input type="submit" />

      <button onclick="addBlock()">Upload More</button>

    </form>

<?php  //upload.php
for ($i = 0; $i < count($_FILES["picture"]["name"]); $i++) {
  move_uploaded_file($_FILES["picture"]["tmp_name"][$i], "upload/pic".$i
}
?>

如果用户不想要旧图片，可以点击<input type="file">上传新图片;

如果他们仍然想要旧的，我怎样才能让<input type="file">获取现有图片的tmp_name，以便 upload.php 可以上传图片？

非常感谢！

Answer 1

在没有上传文件时忽略move_uploaded_file()

for ($i = 0; $i < count($_FILES["picture"]["name"]); $i++) {
  if ($_FILES["picture"]['size'] > 0 && $_FILES['picture']['error'] == 0) {
      // upload file when there are uploaded file
      move_uploaded_file($_FILES['picture']['tmp_name'][$i], 'upload/pic' . $i);
  }
}

如何在表单中获取上载文件的tmp_name？

1 个答案: