Question

我无法从$ _FILES中检索我的php代码中的多个文件。这是输入表格：

<form enctype="multipart/form-data" action="file-upload.php" method="POST">
  Upload the several files:<input type="file" multiple="multiple" name="uploaded" id="id_upload" />
  <input type="submit" value="Upload" />
</form>

这是来自file-upload.php的php代码：

 // first let's find out how many files were uploaded..
 $numUploadedfiles = count($_FILES['uploaded']);
 $num_FILES = count($_FILES); 
        // BOTH COUNTS ARE 5.  I SELECT 7 FILE NAMES FOR UPLOADING THOUGH.


 echo "<br>" . "The number of uploaded files is == " . $numUploadedfiles;
 echo "<br>" . "Here is the name of _FILES['uploaded']: " . $_FILES['uploaded'];
     // THE NAME REPORTED IS 'array' AND THE COUNT IS 5..


 echo "<br>" . "The count size of _FILES is == " . $num_FILES;
 echo "<br>" . "Here is the name of _FILES => " . $_FILES;
       // HERE ALSO, THE NAME REPORTED IS 'array' AND THE COUNT IS 5.


 echo "<br>file temp_name " . $i . " is: " . $_FILES['uploaded']['tmp_name'];
 echo "<br>file name " . $i . " is: " . $_FILES['uploaded']['name'];
        // THE NAME REPORTED HERE IS THE FILENAME OF LAST OF THE 7 FILES I UPLOADED (not sure why.)


 echo "<br>" . "Here are the filenames: ";
 for($i = 0; $i < $numUploadedfiles; $i++)
 {
    echo "<br>filename " . $i . " is: " . $_FILES['uploaded'][$i];
 }
 exit();

当我运行它时会发生什么，当'for'循环开始时，一条错误消息说$ i索引到数组_FILES ['uploaded'] [$ i]无效。

为什么？我需要获取这7个文件名，并能够将它们保存在服务器上。我怎么能：

1）获得文件数量的准确“计数”？当我上传7个文件时，上面的代码计数为5

2）如何在'for'循环中正确索引_FILES数组？ PHP告诉我$ i值0,1,2,3 ...... 无效。

（PS我使用输入类型=“文件”multiple =“multiple”name =“uploaded”id =“id_upload”代码来自我看到的用于在Retrieving file names out of a multi-file upload control with javascript启用多个文件上传的示例

Answer 1

你的名字应该是数组：

<input type="file" name="uploaded[]" id="id_upload" />

Answer 2

此代码未实现您的目标。你应该使用几个输入标签。

如果您想上传多个文件但文件数量是可变的，您可以使用java脚本获取文件数量

无法从PHP中的$ _FILES获取多个上传的文件

2 个答案: