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Home/ Questions/Q 7440307
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Editorial Team
  • 0
Editorial Team
Asked: May 29, 20262026-05-29T10:49:53+00:00 2026-05-29T10:49:53+00:00

I’m exploring the file upload and text parsing capabilities of PHP, but my first

  • 0

I’m exploring the file upload and text parsing capabilities of PHP, but my first step is buggy, and I can’t figure it out. The goal of the code is to display a form to upload a text, and then display the value of the corresponding $_FILES array. However, it doesn’t work – it runs without error but doesn’t display print_r($_FILES['upload']). What am I missing?

This is my Index_txtParsing file:

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
    <head>
        <title>test</title>
    </head>
    <body>
        <form enctype="multipart/form-data" class="Txt_upload" method='POST'  action='index_txtParsing.php'>
            <label for="file">Enter upload here:</label>
            <input type='file' name='upload'/>
            <input type='submit' name='submit' value="upload here"/>
        </form>
    </body>
</html>

<?php
    if(isset($_POST['upload'])){
    $upload=$_POST['upload'];
    print_r($_FILES['upload']);
    }else{
    $upload="unknown";
    }
?>

EDIT: After incorporating the recommendations below, this code works:

<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
    <head>
        <title>test</title>
    </head>
    <body>
        <form enctype="multipart/form-data" class="Txt_upload" method='POST'  action='index_txtParsing.php'>
            <label for="file">Enter upload here:</label>
            <input type='file' name='upload'/>
            <input type='submit' name='submit' value="upload here"/>
        </form>
    </body>
</html>

<?php
    if(isset($_FILES['upload'])){
    echo "Value of FILES['upload'] ";
    print_r($_FILES['upload']);
    echo "<br/>";
    }else{
    }
?>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-29T10:49:54+00:00Added an answer on May 29, 2026 at 10:49 am

    To get the uploaded file use $_FILES instead of $_POST:

    <?php
        if(isset($_FILES['upload'])){
        $upload=$_FILES['upload'];
        print_r($_FILES['upload']);
        }else{
        $upload="unknown";
        }
    ?>
    

    Also please read the manual on File Uploads.

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