Getting complete PATH of uploaded file - PHP

Question

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 Code:

<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 Script:

<?php
   .....
......
   $handle = fopen($_FILES["UploadFileName"]["name"], "r"); # fopen(test.csv) [function.fopen]: failed to open stream: No such file or directory
?>

old question, still, I think such comment is missing here: `$_FILES["UploadFileName"]["name"]` does not exist. `$_FILES["UploadFileName"]["tmp_name"]` does and `move_uploaded_file` is meant to move from tmp name to whatever location you wish. — Jiří, Oct 19 '22 at 13:41
however, sometimes you NEED the complete path of the uploaded file (client side), for example, some file formats use directory hierarchy - this will be available as of 8.1 — Jiří, Oct 19 '22 at 13:43

score 37 · Accepted Answer · answered Mar 27 '11 at 17:03

37

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');

answered Mar 27 '11 at 17:03

netcoder

66,435
19
125
142

score 4 · Answer 2 · answered Nov 08 '12 at 10:33

4

$target='uploads/'.basename($_FILES['UploadFileName']['name']);

if(move_uploaded_file($_FILES['UploadFileName']['tmp_name'],$target)) {

     //Insert into your db

     $fp = fopen($target, "r");
}

answered Nov 08 '12 at 10:33

ShivarajRH

902
12
24

score 4 · Answer 3 · answered Jan 16 '14 at 11:19

I wrote like this:

$filePath = realpath($_FILES["file"]["tmp_name"]);

This gave me the full path to the uploaded file in PHP. If you find 0 bytes problem in file download, just modify this content-lenght line like this

header("Content-Length: ".filesize($filePath));

Where $filePath should be absolute path to file not just file handle.

score 2 · Answer 4 · answered Jan 09 '12 at 12:15

2

Use the following code,

$handle = fopen($_FILES["UploadFileName"]["tmp_name"], 'r');

answered Jan 09 '12 at 12:15

Vijin Paulraj

4,469
5
39
54

score 1 · Answer 5 · answered Mar 12 '18 at 19:47

1

I use like this...

<?php
$NameOriginal = $_FILES["UploadFileName"]['name'];
$Typo_Image = $_FILES["UploadFileName"]['type'];
$name_Temp = $_FILES["UploadFileName"]['tmp_name']; 
?>

answered Mar 12 '18 at 19:47

Jetro Bernardo

49
4

Getting complete PATH of uploaded file - PHP

5 Answers5

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