Displaying images in php from database

[Matthias Utrata]

Hello. I got the path of images stored in my database with a blob. I wanna access them in php and display them aswell. However, instead of displaying the picture, it displays the select command.

$picture = 'SELECT * FROM fragenquelle.bilder where BildID = 1'; echo $picture;

I just tried these 2 little lines as a test, if the display of the image even works. It does not sadly.

DROP TABLE IF EXISTS `bilder`; CREATE TABLE IF NOT EXISTS `bilder` (    `BildID`      int     NOT NULL AUTO_INCREMENT PRIMARY KEY,    `FK_FragenID` int(10) NOT NULL,    `Bild`        BLOB    NOT NULL ); INSERT INTO `bilder` (BildID, FK_FragenID, Bild) VALUES (1, 1, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Muscle.jpg'),       (2, 1, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Rennwagen.jpg'),       (3, 1, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/SUV.jpg'),       (4, 1, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Polizei.jpg'),       (5, 2, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Fischerboot.jpg'),       (6, 2, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Frachtschiff.jpg'),       (7, 2, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Schlachtschiff.jpg'),       (8, 2, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Yacht.jpg'),       (9, 3, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Bomber.jpg'),       (10, 3, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Düsenjet.jpg'),       (11, 3, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Passagierflugzeug.jpg'),       (12, 3, 'C:/xampp/htdocs/UTR/quiz/Bilder-Quiz/Propellerflugzeug.jpg');

[Rob Spoor]

Matthias Utrata wrote:Hello. I got the path of images stored in my database with a blob. I wanna access them in php and display them aswell. However, instead of displaying the picture, it displays the select command.

Sure, because that's what's your code is doing - it creates a string variable with the SQL command, and then it echoes that string variable. There is no code to actually execute the command. I suggest you read about PDO, then try to change your code. To help you, here's a few steps that you need to take:

* Execute the command.
* From the result, get the path. You still shouldn't echo the path because that too is a string.
* Retrieve the file from disk and display its contents.


Edit: here's a PDO guide that's apparently a bit easier to use: https://phpdelusions.net/pdo

[Matthias Utrata]

$stmt = $pdo->prepare('SELECT * FROM fragenquelle.bilder where FK_FragenID = :Frage');    $stmt->bindParam(':Frage',$row['BildID']);    $stmt->execute();    echo "<br />";    while ($row = $stmt->fetch()) {        echo $row['Bild'];        echo "<br />";      }
I tried that now. It does not echo anything though.

[Rob Spoor]

Matthias Utrata wrote:$stmt->bindParam(':Frage',$row['BildID']);

What's the value of $row['BildID'] before you execute the query? Because if you're not seeing anything, maybe your query simply yields zero results.

[Matthias Utrata]

I can post a link of my database, if that helps

https://pastebin.com/qAxv74nV

[Rob Spoor]

I don't think that's necessary.

Can you add the following before the code you provided last time, and show us what the outcome is?
echo "$row['BildID']: " . $row['BildID'] . "<br />";

[Matthias Utrata]

Parse error: syntax error, unexpected '' (T_ENCAPSED_AND_WHITESPACE), expecting '-' or identifier (T_STRING) or variable (T_VARIABLE) or number (T_NUM_STRING) in C:\xampp\htdocs\UTR\quiz\fragen.php on line 65



That is my outcome

[Rob Spoor]

Sorry, I forgot that PHP tries to expand variables in double-quoted strings. Try the following instead:
echo "\$row['BildID']: " . $row['BildID'] . "<br />";

But I think I already know the issue. By default, bindParam uses a string type. Try explicitly using an int type:
$stmt->bindParam(':Frage', $row['BildID'], PDO::PARAM_INT);

[Matthias Utrata]

Notice: Trying to access array offset on value of type bool in C:\xampp\htdocs\UTR\quiz\fragen.php on line 77
$row['BildID']:

That is the error message for the first piece of code.

The second piece of code displayed nothing.

[Rob Spoor]

So $row is not an array, and it does not have an index called BildId. So when you pass it to PDO, it gets turned into something that does not exist in the database. Try hard-coding the value first (BildId 1 should be valid based on your inserts):

$stmt->bindParam(':Frage', 1, PDO::PARAM_INT);

Get this working first. Then focus on getting the BildId from somewhere else. Hint: it's not from $row.

[Matthias Utrata]

I found the issue. It is something completely different, but I found it. The error was in my database