Question

我正在尝试创建一种可以上载文件的表单，但是看来我无法将此输入设为不需要的输入。每当我尝试验证是否没有上传的文件时，代码都会不断进入条件之中。

 if (isset($_FILES['fileToUpload']) && !empty($_FILES['fileToUpload']['name'])){
    // Set files location
    $target_dir = "media/";
    $total = count($_FILES['fileToUpload']['name']);
    $uploadOk = 1;
    // Check if there is files to be uploaded.
      if($total != 0) {
           //$uploadOk = 1;
           for($i=0; $i<$total; $i++) {

           $target_file = $target_dir . basename($_FILES["fileToUpload"]["name"][$i]);
           $imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));
           $namee =$_FILES["fileToUpload"]["name"][$i];
           // Check if file already exists
           if (file_exists($target_file)) {
               header('location: new_report.php?upload=fileexist');
               $uploadOk = 0;
           }
           // Check file size
           if ($_FILES["fileToUpload"]["size"][$i] > 500000) {
               header('location: new_report.php?upload=toobig');
           }
           // Allow certain file formats
           if($imageFileType != "png" && $imageFileType != "jpg"  ) {
               var_dump($_FILES['fileToUpload']['name']);
              // exit;
             header('location: new_report.php?upload=filenotallowed');
               $uploadOk = 0;
           }
           // Check if $uploadOk is set to 0 by an error
           if ($uploadOk == 0) {
               echo "Sorry, your file was not uploaded.";
           // if everything is ok, try to upload file
           } else {
               if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"][$i], $target_file)) {
               Do something..
               } else {
                   header('location: new_report.php?upload=error');
               }
           }
           }

      }
    }

即使我不上传文件，我也会继续进行代码的某些工作。

html代码是：

<form action="submit_report.php" method="POST" enctype="multipart/form-data">
    <input class="btn btn-primary" name="fileToUpload[]" type="file" multiple="multiple" id="fileToUpload">
</form>

是用于多选文件

Answer 1

设置文件isset的条件并尝试。

此代码已通过测试

<?php

    if (isset($_FILES['fileToUpload']))
    {
        if($_FILES['fileToUpload']['tmp_name'][0] == "") {
            die("No files to upload");
        }

        foreach($_FILES['fileToUpload']['tmp_name'] as $key => $tmp_name)
        {
            $file_name = $key . $_FILES['fileToUpload']['name'][$key];
            $file_size = $_FILES['fileToUpload']['size'][$key];
            $file_tmp = $_FILES['fileToUpload']['tmp_name'][$key];
            $file_type = $_FILES['fileToUpload']['type'][$key];
            move_uploaded_file($file_tmp, getcwd() . "/" . time() . $file_name);
        }

        echo "Success";
    }
    else
    {
        echo "<form enctype='multipart/form-data' action='' method='POST'>";
        echo "File:<input name='fileToUpload[]' multiple='multiple' type='file'/><input type='submit' value='Upload'/>";
        echo "</form>";
    }
?>

确保您的HTML表单应包含enctype="multipart/form-data"

Answer 2

由于输入名称为fileToUpload[]，所以将始终发布数组，因此您需要检查if count($_FILES['fileToUpload']) { ...而不是使用isset($_FILES['fileToUpload'])。

不需要使用PHP上传文件-PHP FORM不起作用

2 个答案: