如果文件上传，则发送文件名，否则在 PHP 中发送空文件

Question

我不知道我正在尽我所能，否则将来会出现问题。

我必须在我的页面上显示 7 种文件类型，并且将来会增加。 因此，我获取了数组并存储了标签名称，并根据标签计数显示了文件类型。所以我在数组中有 7 个标签，我得到了 7 种文件类型。

下面是我正在使用的代码。

<div class="input_fields_wrap">
      <?php
      $labelname = array('Label 1','Label 2','Label 3','Label 4','Label 5','Label 6','Label 7');
      $labelCount=count($labelname);
        ?>
   <div class="row mt-3">
      <?php 
      for ($i=0; $i < $labelCount; $i++) { 
       ?>
       <div class="col-lg-4 mb-4 ">
         <div class="documentUploadWrap">
            <label><?php echo $labelname[$i];?></label>

         <div class="upload_doc">
          <input type="hidden" name="docUploadLabel[]" value="<?php echo $labelname[$i];?>">
          <input type="file" name="docUpload[]"  class="fileupload">
          <input type="hidden" name="docUpload[]" value="">

          <div class="uploadInfo">
             <div class="upload_icon"></div>
             <p>Drop your image here, or <span>browse</span></p> 
             <span>Supports: JPEG, PNG, PPT, PPTX, PDF <br />Max Size: 5MB</span>
          </div>
       </div>
         </div>
      </div> 
    <?php } ?>

function imagerename($image,$foldername,$pdo){
$image1  = $image['name'];
$filename  = basename($image1);
$onlyfile = pathinfo($filename, PATHINFO_FILENAME);
$extension = pathinfo($filename, PATHINFO_EXTENSION);

$file      = mt_rand();// random number 
$newname   = $onlyfile.'_'.$file.'.'.$extension;
 
$location='/images/'.$foldername.'/'.$newname;

if($extension=='png' || $extension=='jpg' || $extension=='jpeg') { 
  compressImage($image['tmp_name'],$location,60); 
} 
else{ 
  move_uploaded_file($image['tmp_name'], $location); 
}


return  $newname;

}

// Compress image
function compressImage($source, $destination, $quality) {

  $info = getimagesize($source);

  if ($info['mime'] == 'image/jpeg') 
    $image = imagecreatefromjpeg($source);

  elseif ($info['mime'] == 'application/pdf') 
    $image = imagecreatefrompng($source);

  elseif ($info['mime'] == 'image/png') 
    $image = imagecreatefrompng($source);

  imagejpeg($image, $destination, $quality);

}

现在，我上传了前两种文件类型的文件，其余 5 个为空。然后我点击提交按钮，我收到以下错误。

1)

2) 请注意，如果我上传 1 或 2 个文件，第二个名称会继续重复，直到 7 次迭代。

Answer 1

<?php

$filetypes_array = [
  "jpg" => [
    "label" => "Label 1", 
    "name" => "Name 1"
  ],
  "png" => [
    "label" => "Label 2", 
    "name" => "Name 2"
  ],
  "gif" => [
    "label" => "Label 3", 
    "name" => "Name 3"
  ],
  "pdf" => [
    "label" => "Label 4", 
    "name" => "Name 4"
  ]
];

// Build an array of your uploads with your preferred method.

$uploaded_files = [];

// Example
$uploaded_files[] = $_POST["inputname"];
$uploaded_files[] = $_POST["inputname2"];
// etc...

// Or inside a loop
foreach($somearray as $key => $value) {
  $uploaded_files[] = $value;
}

// Then run a loop ONLY as many time as you have uploads.
foreach($uploaded_files as $key => $value) {
  switch ($value) {
    case "pdf":
      // do your thing here
      $filetypes_array[$value].label; // Then do something with it
      $filetypes_array[$value].name; // Then do something with it
      break;
    case "jpg": 
      // do your thing here
      // You can access your prepeared array
      $filetypes_array[$value].label; 
      break;
    case "png":
      // another case
      break;
    case "gif":
      // another case
      break;
  }
}

没有看到你的其余代码这只是伪代码。基本上，您希望循环只运行所需的次数，这样您就不会尝试访问不存在的值。