simple file upload script

Question

I have written a simple file upload script but it gives me the error of undefined index file1.

<html>
    <body>
        <form method="post">
            <label for="file">Filename:</label>
            <input type="file" name="file1" id="file1" /> 
            <br />
            <input type="submit" name="submit" value="Submit" />
        </form>
    </body>
</html>
<?php
if(isset($_POST['submit'])) {
    if ($_FILES["file1"]["error"] > 0) {
        echo "Error: " . $_FILES["file1"]["error"] . "<br />";
    } else {
        echo "Upload: " . $_FILES["file1"]["name"] . "<br />";
        echo "Type: " . $_FILES["file1"]["type"] . "<br />";
        echo "Size: " . ($_FILES["file1"]["size"] / 1024) . " Kb<br />";
        echo "Stored in: " . $_FILES["file1"]["tmp_name"];
    }
}
?>

What is the problem in code?

try to add `enctype="multipart/form-data"` to form tag – k102 Feb 09 '12 at 08:18 — k102, Feb 09 '12 at 08:18

score 13 · Accepted Answer · answered Feb 09 '12 at 08:17

13

You lack enctype="multipart/form-data" in your <form> element.

answered Feb 09 '12 at 08:17

Wesley van Opdorp

14,888
4
41
59

T.Todua · Answer 2 · 2013-04-23T10:01:20.347

Another solution for simple php file upload script is here :
(make a yourfile.php and insert the below code. then put that yourfile.php on your website)

<?php
$pass = "YOUR_PASSWORD";
session_start();
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html><head><meta http-equiv="Content-Type" content="text/html; charset=windows-1256" /></head><body>
<?php
if (!empty($_GET['action']) &&  $_GET['action'] == "logout") {session_destroy();unset ($_SESSION['pass']);}

$path_name = pathinfo($_SERVER['PHP_SELF']);
$this_script = $path_name['basename'];
if (empty($_SESSION['pass'])) {$_SESSION['pass']='';}
if (empty($_POST['pass'])) {$_POST['pass']='';}
if ( $_SESSION['pass']!== $pass) 
{
    if ($_POST['pass'] == $pass) {$_SESSION['pass'] = $pass; }
    else 
    {
        echo '<form action="'.$_SERVER['PHP_SELF'].'" method="post"><input name="pass" type="password"><input type="submit"></form>';
        exit;
    }
}
?>


<form enctype="multipart/form-data" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
Please choose a file: <input name="file" type="file" /><br />
<input type="submit" value="Upload" /></form>


<?php 

if (!empty($_FILES["file"]))
{
    if ($_FILES["file"]["error"] > 0)
       {echo "Error: " . $_FILES["file"]["error"] . "<br>";}
    else
       {echo "Stored file:".$_FILES["file"]["name"]."<br/>Size:".($_FILES["file"]["size"]/1024)." kB<br/>";
       move_uploaded_file($_FILES["file"]["tmp_name"],$_FILES["file"]["name"]);
       }
}

    // open this directory 
    $myDirectory = opendir(".");
    // get each entry
    while($entryName = readdir($myDirectory)) {$dirArray[] = $entryName;} closedir($myDirectory);
    $indexCount = count($dirArray);
        echo "$indexCount files<br/>";
    sort($dirArray);

    echo "<TABLE border=1 cellpadding=5 cellspacing=0 class=whitelinks><TR><TH>Filename</TH><th>Filetype</th><th>Filesize</th></TR>\n";

        for($index=0; $index < $indexCount; $index++) 
        {
            if (substr("$dirArray[$index]", 0, 1) != ".")
            {
            echo "<TR>
            <td><a href=\"$dirArray[$index]\">$dirArray[$index]</a></td>
            <td>".filetype($dirArray[$index])."</td>
            <td>".filesize($dirArray[$index])."</td>
                </TR>";
            }
        }
    echo "</TABLE>";
    ?>

score 2 · Answer 3 · answered Feb 09 '12 at 08:21

2

Make the following changes and try.

<form method="post" action="" enctype="multipart/form-data" >

answered Feb 09 '12 at 08:21

Ibrahim Azhar Armar

25,288
35
131
207

score 1 · Answer 4 · answered Nov 07 '14 at 14:08

Html

 <!DOCTYPE html>
<html>
<body>

<form action="upload.php" method="post" enctype="multipart/form-data">
    Select image to upload:
    <input type="file" name="fileToUpload" id="fileToUpload">
    <input type="submit" value="Upload Image" name="submit">
</form>

</body>
</html>

Php

<?php
$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submit"])) {
    $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
    if($check !== false) {
        echo "File is an image - " . $check["mime"] . ".";
        $uploadOk = 1;
    } else {
        echo "File is not an image.";
        $uploadOk = 0;
    }
}
?>

Could you explain what the difference between your answer and e.g. @Solutioners is, please? What does your answer include that the others haven't already mentioned? — kkuilla, Nov 07 '14 at 14:30
using `getimagesize()` to check if the uploaded file is really an image — ismnoiet, Jul 28 '15 at 07:47

score 0 · Answer 5 · answered Sep 24 '15 at 19:56

<html>
<body>
<form action="" method="post" ectype="multipart/form-data">
<label for="file">Filename:</label>
<input type="file" name="file" id="file"></br>
<input type="submit" name="submit" value="Submit">
</form>
<?php
if(isset($_POST['submit']))
{
$allowedExts = array("gif", "jpeg", "jpg", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 20000000)
&& in_array($extension, $allowedExts))
  {
  if ($_FILES["file"]["error"] > 0)
    {
    echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
    }
  else
    {
    echo "Upload: " . $_FILES["file"]["name"] . "<br>";
    echo "Type: " . $_FILES["file"]["type"] . "<br>";
    echo "Size: " . ($_FILES["file"]["size"] / 1024) . " kB<br>";
    echo "Temp file: " . $_FILES["file"]["tmp_name"] . "<br>";
   if (file_exists("upload/" . $_FILES["file"]["name"]))
      {
      echo $_FILES["file"]["name"] . " already exists. ";
      }
    else
      {
      move_uploaded_file($_FILES["file"]["tmp_name"],
      "upload/" . $_FILES["file"]["name"]);
      echo "Stored in: " . "upload/" . $_FILES["file"]["name"];
      }
    }
  }
else
  {
  echo "Invalid file";
  }enter code here
}
?>
</body>
</html>

Generally, answers are much more helpful if they include an explanation of what the code is intended to do, and why that solves the problem without introducing others. — Jose Ricardo Bustos M., Sep 24 '15 at 20:12

score -1 · Answer 6 · answered Feb 09 '12 at 11:51

Primary issue is your form does not have option to send file content over http . To send binary data along with the text data from input elements you need to add an extra attribute in form tag .

<form method="post" enctype="multipart/form-data">

Then in php code

try this line

<?php

print_r($_FILES);

?>

above code will display all information regarding file uploading from your form .

simple file upload script

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