How to get the file path in html <input type="file"> in PHP?

Question

Can somebody pls tell me how to get the filepath using html <input type="file"> in PHP?

Here are my codes:

index.php

<form action="csv_to_database.php" method="get" >
 <input type="file" name="csv_file" />
 <input type="submit" name="upload" value="Upload" />
</form>

and in csv_to_database.php

<?php

 if (isset($_GET['csv_file'])) {

 $row = 1;

  if (($handle = fopen($_GET['csv_file'], "r")) !== FALSE) {
   while (($data = fgetcsv($handle, 1000, ",")) !== FALSE) {
    $num = count($data);
    echo "<p> $num fields in line $row: <br /></p>\n";
    $row++;
    for ($c=0; $c < $num; $c++) {
        echo $data[$c] . "<br />\n";
    }
   }
   fclose($handle);
  }

 } 
?>

My problem is, it only works when the csv data is in the same directory as my php files. I think I need to get the file path but I don't know how to do it.

php.net/manual/en/features.file-upload.php

Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams

2011-04-28 05:51:39 +00:00
Commented Apr 28, 2011 at 5:51 — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Apr 28, 2011 at 5:51

Nanne · Accepted Answer · 2011-04-28 05:53:27Z

10

You shouldn't just use the $_GET you've got now. Your file is based in $_FILES["csv_file"]["tmp_name"].

Best you review this tutorial, that basically says you need to do something like this:

<?php
if ($_FILES["csv_file"]["error"] > 0)
  {
  echo "Error: " . $_FILES["csv_file"]["error"] . "<br />";
  }
else
  {
  echo "Upload: " . $_FILES["csv_file"]["name"] . "<br />";
  echo "Type: " . $_FILES["csv_file"]["type"] . "<br />";
  echo "Size: " . ($_FILES["csv_file"]["size"] / 1024) . " Kb<br />";
  echo "Stored in: " . $_FILES["csv_file"]["tmp_name"];
  }
?>

And you can go from there. Use move_uploaded_file if you want to move the file from the temp location, also explained in the tutorial :)

answered Apr 28, 2011 at 5:53

Nanne

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Comments

Kokos · Accepted Answer · 2011-04-28 05:53:49Z

3

I think you would gain a lot from taking a look at the following link: POST method uploads.

First of all, you should change your form method to post, and add enctype="multipart/form-data".

Then you can get the temporary file path from $_FILES['csv_file']['tmp_name'].

answered Apr 28, 2011 at 5:53

Kokos

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Comments

Mahmoud Gamal · Accepted Answer · 2012-02-05 07:04:44Z

-1

In your call to fopen, use $_GET['csv_file']['tmp_name'] - this points to the file on the server immediately after upload.

edited Feb 5, 2012 at 7:04

Mahmoud Gamal

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answered Apr 28, 2011 at 5:54

David Fells

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