获取上载文件的完整PATH-PHP [英] Getting complete PATH of uploaded file - PHP
问题描述
我有一个表单(HTML，PHP)，允许最终用户上传文件，以使用上传文件(特别是.csv)中的记录来更新数据库(MySQL).但是，在phpscript中，我只能获取文件名，而不能获取特定文件的完整路径.由于这个原因，fopen()失败.谁能让我知道如何找到完整的路径?
I have a form(HTML, PHP) that lets the end user upload a file to update the database(MySQL) with the records in the uploaded file(specifically .csv). However, in the phpscript, I can only get the filename and not the complete path of the file specificed. fopen() fails due to this reason. Can anyone please let me know how I can work on finding the complete path?
HTML代码:
<html>
<head>
</head>
<body>
<form method="POST" action="upload.php" enctype="multipart/form-data">
<p>File to upload : <input type ="file" name = "UploadFileName"></p><br />
<input type = "submit" name = "Submit" value = "Press THIS to upload">
</form>
</body>
</html>
PHP脚本:
<?php
.....
......
$handle = fopen($_FILES["UploadFileName"]["name"], "r"); # fopen(test.csv) [function.fopen]: failed to open stream: No such file or directory
?>
推荐答案
name是指客户端上的文件名.要在服务器端上获取文件名(包括完整路径)，您需要使用tmp_name:
name refers to the filename on the client-side. To get the filename (including the full path) on the server-side, you need to use tmp_name:
$handle = fopen($_FILES["UploadFileName"]["tmp_name"], 'r');
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