Undefined index in $_FILES

Question

I am getting the error that my index is undefined in $_FILES. I am trying to copy a picture with php on my server.

Error:

PHP Notice:  Undefined index: bilddatei in C:\inetpub\wwwroot\aufgabe8.php on line 85
PHP Notice:  Undefined variable: namefile in C:\inetpub\wwwroot\aufgabe8.php on line 86

I've already tried with isset($_FILES["bilddatei"]) but this returns false too.

Code:

<?php
session_start();

if (!isset($_SESSION['Datenbank']) and
    !isset($_SESSION['Kennung']) and !isset($_SESSION['Passwort']) )
{
    echo "<p>Bitte starten Sie zuerst mit der Eingabe-Startseite. Sie werden automatisch nach 5 Sekunden weitergeleitet.</p>";
    // automatisches Zurueckspringen nach 5 Sekunden nach 'index.html': 
    echo '<meta http-equiv="Refresh" content="5;url=login.php">';
} else {
          echo "<form action=\"aufgabe8.php\" method=\"post\" enctype=\"multipart/form-data\">
                    <p>Datei von PC hochladen
                        <input type=\"file\" name=\"bilddatei\" size\"60\" accept=\"image/*\"/>
                    </p>

                    <p><input type=\"submit\" value=\"Bild anzeigen\" name=\"bildBtn\"/></p>
                </form>";


    if(isset($_POST["bildBtn"])) {
            echo "<p>BILD IF</p>";

            $nameFile = $_FILES['bilddatei']['name'];
            echo "<p>$namefile</p>";

            //copy($fileSource,$destination); 

    }

Do you have any ideas? Thanks! :-)

just because you echo the form, doesn't mean it's submitted. the code is confusing as to weather you think this or the form is displayed on submission. — ArtisticPhoenix
– ArtisticPhoenix, Commented Jun 28, 2017 at 12:57

Bhunesh Satpada · Accepted Answer · 2017-06-28 12:51:13Z

2

Variable name is different. Both php variable is case-sensitive.check $nameFile <=> $namefile.

https://www.w3schools.com/php/php_variables.asp

$namefile = $_FILES['bilddatei']['name'];
echo "<p>$namefile</p>";

answered Jun 28, 2017 at 12:51

Bhunesh Satpada

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1 Comment

Bhunesh Satpada Over a year ago

you have to take care about variable definition.

ArtisticPhoenix · Accepted Answer · 2017-06-28 13:44:06Z

1

I think these things need to be done separately instead...

Render form

if (!isset($_SESSION['Datenbank']) and
    !isset($_SESSION['Kennung']) and !isset($_SESSION['Passwort']) )
{
    echo "<p>Bitte starten Sie zuerst mit der Eingabe-Startseite. Sie werden automatisch nach 5 Sekunden weitergeleitet.</p>";
    // automatisches Zurueckspringen nach 5 Sekunden nach 'index.html': 
    echo '<meta http-equiv="Refresh" content="5;url=login.php">';
} else {
    echo "<form action=\"aufgabe8.php\" method=\"post\" enctype=\"multipart/form-data\">
            <p>Datei von PC hochladen
                <input type=\"file\" name=\"bilddatei\" size\"60\" accept=\"image/*\"/>
            </p>

            <p><input type=\"submit\" value=\"Bild anzeigen\" name=\"bildBtn\"/></p>
        </form>";
}  ///missing } close I am assuming it goes here

Second php file ( say aufgabe8.php to match the form action )

if(isset($_POST["bildBtn"])) {
    echo "<p>BILD IF</p>";

    $nameFile = $_FILES['bilddatei']['name'];
    echo "<p>$nameFile</p>";  //spelling  $nameFile is not $namefile

    //copy($fileSource,$destination); 

}

Just because you echo out the form is very different then when it's submitted. The code is confusing on whether or not you understand this. In any case they are 2 distinct actions.

Besides the spelling mistakes as mentioned in the other answer.

Now you can do them in the same page like this

<?php
session_start();

if (!isset($_SESSION['Datenbank']) and
    !isset($_SESSION['Kennung']) and !isset($_SESSION['Passwort']) )
{
    echo "<p>Bitte starten Sie zuerst mit der Eingabe-Startseite. Sie werden automatisch nach 5 Sekunden weitergeleitet.</p>";
    // automatisches Zurueckspringen nach 5 Sekunden nach 'index.html': 
    echo '<meta http-equiv="Refresh" content="5;url=login.php">';
} else {

    if(isset($_POST["bildBtn"])) {
        ///submitted so process it
        echo "<p>BILD IF</p>";

        $nameFile = $_FILES['bilddatei']['name'];
        echo "<p>$nameFile</p>";  //spelling  $nameFile is not $namefile

        //copy($fileSource,$destination); 
    }else{
        ///not submitted so rendor the form
        echo "<form action=\"aufgabe8.php\" method=\"post\" enctype=\"multipart/form-data\">
            <p>Datei von PC hochladen
                <input type=\"file\" name=\"bilddatei\" size\"60\" accept=\"image/*\"/>
            </p>

            <p><input type=\"submit\" value=\"Bild anzeigen\" name=\"bildBtn\"/></p>
        </form>";

    }
}

You can even render the form, and process it in the same block ( without the else ), but just rendering it doesn't mean someone submitted it. In this case I would do away with the echo and just us plain old html to output the form.

edited Jun 28, 2017 at 13:44

answered Jun 28, 2017 at 13:00

ArtisticPhoenix

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3 Comments

Sandro_V Over a year ago

So you mean I have to do it in two separate files? Because normally I do everything in one file and check if a button is clicked or a variable is set with isset.

ArtisticPhoenix Over a year ago

Technically, you can do it in the same file, but just echoing the form, is not the same as submitting it. Echoing it out, is a sever operation, and submitting it is something the client does.

ArtisticPhoenix Over a year ago

In other words, you can output the form, but until a user submits it you can't process the data in it, right. With that missing } its hard to tell if you were making that mistake or not. For ease of coding and readabillity, it's often better to handle them separately, then you can avoid having a condition to either render or process.

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