1

我在下面有一个通过 php 上传单个文件输入的工作示例，现在我想通过遍历每个输入 type="file" 来上传多个文件。

我已经读过在 IE10 下不支持 input 标签上的 multiple 属性，所以我认为最好的方法是有几个 input type=files 以便循环它们

任何帮助表示赞赏

<input type="file" name="FilesUpload1" class="filesUpload" />
<input type="file" name="FilesUpload2" class="filesUpload" />
<input type="file" name="FilesUpload3" class="filesUpload" />

<?php
//Сheck that we have a file
if((!empty($_FILES["FilesUpload1"])) && ($_FILES['FilesUpload1']['error'] == 0)) {

    //Check if the file is JPEG image and it's size is less than 350Kb
    $filename = basename($_FILES['FilesUpload1']['name']);
    $ext = substr($filename, strrpos($filename, '.') + 1);

    //check file extension
    if ((($ext == "gif") 
    || ($ext == "jpeg")
    || ($ext == "jpg")
    || ($ext == "png")
    || ($ext == "doc")
    || ($ext == "docx")
    || ($ext == "rtf")
    || ($ext == "txt")          
    || ($ext == "pdf"))

    //check file mime
    && (($_FILES["FilesUpload1"]["type"] == "image/gif")
    || ($_FILES["FilesUpload1"]["type"] == "image/jpeg")
    || ($_FILES["FilesUpload1"]["type"] == "image/jpg")
    || ($_FILES["FilesUpload1"]["type"] == "image/pjpeg")
    || ($_FILES["FilesUpload1"]["type"] == "image/x-png")
    || ($_FILES["FilesUpload1"]["type"] == "image/png")
    || ($_FILES["FilesUpload1"]["type"] == "application/msword")
    || ($_FILES["FilesUpload1"]["type"] == "application/vnd.openxmlformats-officedocument.wordprocessingml.document")
    || ($_FILES["FilesUpload1"]["type"] == "application/rtf")
    || ($_FILES["FilesUpload1"]["type"] == "text/plain")
    || ($_FILES["FilesUpload1"]["type"] == "application/pdf"))


    //check file size is less than 1048576 bytes [1 MB]
    && (($_FILES["FilesUpload1"]["size"] < 1048576))) {

        //Determine the path to which we want to save this file
        $newname = dirname(__FILE__).'/../entries/'.$_POST["CompanyName"].'-'.$filename;    

        //Check if the file with the same name is already exists on the server
        if (!file_exists($newname)) {

            //Attempt to move the uploaded file to it's new place
            if ((move_uploaded_file($_FILES['FilesUpload1']['tmp_name'],$newname))) {
                echo "It's done! Your file has been saved.";

            } else {
                echo "Error: A problem occurred during file upload!";
            }

        } else {
            echo "Error: File ".$_FILES["FilesUpload1"]["name"]." already exists";
        }

    } 

    else {
        echo "Error: Only .gif, .jpeg, .jpg, .png, .doc, .docx, .rtf, .txt, .pdf files under 1 MB are accepted for upload.";
    }

} else {
    echo "Error: No file uploaded";
}

?>
4

2 回答 2

3

试试这个方法。

HTML：

<input name="upload[]" type="file" multiple="multiple" />

PHP代码：

$total = count($_FILES['upload']['name']);

// Loop through each file
for($i=0; $i<$total; $i++) {

  //Get the temp file path
  $tmpFilePath = $_FILES['upload']['tmp_name'][$i];

  //Make sure we have a filepath
  if ($tmpFilePath != ""){
    //Setup our new file path
    $newFilePath = "./uploadFiles/" . $_FILES['upload']['name'][$i];

    //Upload the file into the temp dir
    if(move_uploaded_file($tmpFilePath, $newFilePath)) {

      //Handle other code here

    }
  }
}
于 2017-04-25T13:17:23.567 回答
2

你试过这个逻辑吗？

HTML

<form method="POST" action="desired_url" enctype="multipart/form-data">

        <input type="file" name="FilesUpload[]" class="filesUpload" />
        <input type="file" name="FilesUpload[]" class="filesUpload" />
        <input type="file" name="FilesUpload[]" class="filesUpload" />

        <button type="submit" name="upload_file" class="btn btn-primary">Upload</button>
</form>

PHP

if( isset($_POST["upload_file"] ) ){

    $get_files = $_FILES["FilesUpload"];

    foreach( $get_files['name'] as $key => $name ){

        $files = $_FILES['FilesUpload'];

        if( $files['error'][$key] == 0 ){

            $ext          = substr($name, strrpos($name, '.') + 1);
            $filename     = rand(11111,99999).'.'.$ext;
            $destination = "uploads/".$filename;

            if( move_uploaded_file( $files['tmp_name'][$key], $destination ) ){
                echo "Your file has been uploaded.<br>";
            }
        }

    }   

}
于 2017-04-25T13:04:50.417 回答