【发布时间】:2012-01-18 04:23:02
【问题描述】:
我无法从 $_FILES 检索我的 php 代码中的多个文件。 这是输入表单:
<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,一条错误消息指出数组 _FILES['uploaded'][$i] 中的 $i 索引无效。
这是为什么呢?我需要获取这 7 个文件名并能够将它们保存在服务器上。我该怎么做:
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。
(P.S. 我正在使用我在Retrieving file names out of a multi-file upload control with javascript 看到的启用多个文件上传的示例中的 input type="file" multiple="multiple" name="uploaded" id="id_upload" 代码)
【问题讨论】:
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看到这个了吗?可能会有所帮助(检查 cmets):verens.com/2009/12/28/multiple-file-uploads-using-html5
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是的,太可怕了!我不知道为什么很难找到有关它如何工作的信息,以及为什么它的工作方式如此不直观。我很难找到那个信息。很高兴你现在可以正常工作了。