Question

在我的表单中，我有3个用于文件上传的输入字段：

<input type=file name="cover_image">
<input type=file name="image1">
<input type=file name="image2">

如何检查cover_image是否为空 - 没有文件要上传？

Answer 1

您可以使用size数组上的$_FILES字段进行检查，如下所示：

if ($_FILES['cover_image']['size'] == 0 && $_FILES['cover_image']['error'] == 0)
{
    // cover_image is empty (and not an error)
}

（我也在这里检查error，因为如果出现问题，它可能是0。我不会使用name进行此检查，因为可以覆盖此项< / em>的

Answer 2

方法1

if($_FILES['cover_image']['name'] == "") {
// No file was selected for upload, your (re)action goes here
}

方法2

if($_FILES['cover_image']['size'] == 0) {
// No file was selected for upload, your (re)action goes here
}

Answer 3

您可以通过执行以下操作来检查是否有值，以及图像是否有效：

if(empty($_FILES['cover_image']['tmp_name']) || !is_uploaded_file($_FILES['cover_image']['tmp_name']))
{
   // Handle no image here...
}

Answer 4

if (empty($_FILES['cover_image']['name']))

Answer 5

简单：

if($_FILES['cover_image']['error'] > 0)
    // cover_image is empty

Answer 6

在表格发布后检查以下

$_FILES["cover_image"]["size"]==0

Answer 7

 if( ($_POST) && (!empty($_POST['cover_image'])) )    //verifies  if post exists and cover_image is not empty
    {
    //execute whatever code you want
    }

Answer 8

<input type="file" class="custom-file-input" id="imagefile" name="imagefile[]"  multiple lang="en">
<input type="hidden" name="hidden_imagefile[]" value="<?=$row[2]; ?>" class="form-control border-input" >

    if($_FILES['imagefile']['name'] == '')
        {
          $img = $_POST['hidden_imagefile'];
        }
        else{
          $img = '';
          $uploadFolder = 'uploads/gallery/';
          foreach ($_FILES['imagefile']['tmp_name'] as $key => $image) {
            $imageTmpName = time() .$_FILES['imagefile']['tmp_name'][$key];
            $imageName = time() .$_FILES['imagefile']['name'][$key];
            $img .= $imageName.',';
            $result = move_uploaded_file($imageTmpName, $uploadFolder.$img);
          }
          
        }

Answer 9

这会起作用

if ($_FILES['cover_image']['size'] == 0 && $_FILES['cover_image']['error'] == 0)

// 检查文件是否被选中而不是错误

{

//文件没有被选中，不是错误

}

Answer 10

$_FILES 是一个关联的 POST 方法数组，如果您想检查有关 $_FILES 的任何内容，您必须考虑索引...我尝试了很多建议的选项，唯一对我有用的方法是我在验证方法中包含了一个索引。

$_FILES['Your_File']['name'][0];

再见了：

 if(empty($_FILES['Your_File']['name'][0])){
    print('this thing is empty');
 }else{
    print('Something, something, something');
 }

最好的旧实验和大量阅读。

Answer 11

if($_FILES['img_name']['name']!=""){
   echo "File Present";
}else{
  echo "Empty file";
}

Answer 12

if ($_FILES['cover_image']['size'] == 0 && $_FILES['cover_image']['error'] == 0)
{ 
      // Code comes here
}

这件事对我有用........

Answer 13

if(!empty($_FILES)) { // code if not uploaded } else { // code if uploaded }

Answer 14

    if (!$_FILES['image']['size'][0] == 0){ //}

检查特定输入文件是否为空

14 个答案: