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I am trying to upload the name of the selected file to the database. For this I made a HTML page . The code of HTML page is

<!DOCTYPE html>
<html lang="en">
<head>
    <title>Photo Input</title>
</head>

<body>
    <form action="input-photo.php" method="POST" enctype="multipart/form-data" >

        <h3>Insert a photo</h3>
        <input type="file" name="image" id="1"  >
        <br>
        <br>
        <input type="submit" name="upload" value="upload image">
    </form>
    
</body>
</html>

The action of this form in HTML is in input-photo.php file . The code of this php file is

<?php
    $server = "localhost";
    $username ="root";
    $password = "";
    $conn = mysqli_connect($server,$username,$password,"form");
    if(isset($_POST['upload']))
    {
        $name = $_FILES['image']['name'];
        echo $name;
    }
?>

But when i run this segment of code it shows two warnings:-
Warning: Undefined array key "image"
Warning: Trying to access array offset on value of type null

The warning is shown for the .php file where $_FILES is used. It says Undefined array key "image"
I dont know how to resolve this.

Improve this question
6
  • When do you see this error? After submitting the form? When opening the page directly? Commented Oct 21, 2021 at 18:52
  • When i click the upload image then the error is shown. Commented Oct 21, 2021 at 18:56
  • Do any of the answers here help? stackoverflow.com/questions/19027992/… Commented Oct 21, 2021 at 18:58
  • 1
    It looks like it should work. You have selected a file? Try var_dump($_FILES); Commented Oct 21, 2021 at 19:09
  • 1
    Make sure to check your file_uploads=On in php.ini file Commented Oct 21, 2021 at 19:26
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