PHP display image from mysql

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I’m developing a CMS where the user can insert articles and if he wants to upload an image to that same article, the text uploads work properly but the images don’t, the only thing that appears is the name of the image in the place where the image is supposed to appear, I followed the same logic I did for uploading text (articles).

PHP: 

    <?php
session_start();
include_once('../includes/connection.php');
if(isset($_SESSION['logged_in'])) {
    // display add page
        if (isset($_POST['title'], $_POST['content'])) {
        $title = $_POST['title'];
        $content = $_POST['content'];
        $image = $_FILES['image']['name'];

            if (empty($title) || empty($content)) {
                $error = 'Todos os campos têm de ser preenchidos!';
            }

            else {
                $query = $pdo->prepare('INSERT INTO articles(article_title, article_content, article_img, article_timestamp) VAlUES (?, ?, ?, ?)');

                $query->bindValue(1, $title);
                $query->bindValue(2, $content);
                $query->bindValue(3, $image);
                $query->bindValue(4, time());

                $query->execute();

                header('Location:index.php');
            }
        }

}
        else {
        header('Location:index.php');
    }
    ?>

    <html>
    <head>
        <title>AdminPT</title>
        <meta charset="UTF-8">
        <link rel ="stylesheet" href="../assets/style.css"/>
    </head>

        <body>
            <div class="container">
                CMS
                <br>

                <h4 id ="add">Adicionar Artigo</h4>

                <?php
                if (isset($error)) { ?>
                    <small style="color:#aa0000"><?php echo $error; ?></small>
                <?php } ?>

                <form action="add.php" method="post" autocomplete="off" enctype="multipart/form-data">
                    <input id="title" type="text" name="title" placeholder="Título"/><br><br>
                    <textarea id="content" rows="15" placeholder="Conteúdo" name="content"></textarea>
                    <input type="file" name="image"/><br><br>
                    <input id="addBtn" type="submit" value="Adicionar Artigo"/>
                </form>
                <a id="back" href="../admin">&larr; Voltar</a>
            </div>
        </body>
    </html>

Display image (HTML display):

<div id="news">
            <?php foreach ($articles as $article) { ?>
            <div><h2><?php echo utf8_encode($article['article_title']); ?></h2><div id="newsImg"><?php echo $article['article_img']; ?></div><br><span id="date">Publicado 
                    <?php
                        setlocale(LC_ALL, 'pt_BR.utf8', 'Portuguese_Brazil');
                        //setlocale(LC_ALL, NULL);
                        date_default_timezone_set('Europe/Lisbon');

                        $uppercaseMonth = ucfirst(gmstrftime('%B'));
                        echo utf8_encode(strftime( '%a, '.$uppercaseMonth.' %d de %Y'/* - %H:%M'*/, $article['article_timestamp']));
                    ?></span><p><?php echo utf8_encode($article['article_content']); ?><br><br><span id="print"><a onclick="javascript:window.print();" href="#">Imprimir</a></span><span id="link"><a href="#">Enviar link</a></p></div>
                    <?php } ?>

        </div>
php mysql image display

2 answers

3


Instead of

$image = $_FILES['image']['name'];

You need to read the file with

$image = file_get_contents( $_FILES['image']['tmp_name'] );

to be able to store.

To show, you would need a separate PHP that did the reverse, and put this PHP as the image source, something like that:

<img src="mostraimagem.php?id=286318936">

However, remember that I am simplifying the example, first of all you need to learn how to use the correct BIND for Blobs, and have the database prepared for this amount of data.

the screenshot.php would be simple:

<?php

//conecte na base de dados antes, e guarde em $con

$query = "SELECT * FROM articles WHERE id = ?";
$stmt = $conexao->prepare( $query );
$stmt->bindParam(1, $_GET['id']);
$stmt->execute();

if( $stmt->rowCount() ){
   $resultado = $stmt->fetch(PDO::FETCH_ASSOC);
   header(“Content-type: image/jpg”); //acerte pro tipo de imagem
   print $resultado['article_img'];
   die();
}else{
   //pode servir uma imagem padrão, ou um 404
}

?>

The ideal solution would be to reconsider the whole system and just move the file to a folder, and save only this name in the database, thus avoiding an overload in the base and processing of PHP.

To learn how to do this, a great start is this:

PHP - Manage file upload

0

To view images with PHP from the image code there is the GD library, from which you can use the method imagecreatefromstring(string $image).

So to upload these images you will need to create a PHP file to upload them from the database as you saved them:

<?php
header('Content-Type: image/jpeg');

$articleID = settype($_GET['id'], 'integer');
$pdo = new PDO("...");
$query = $pdo->query("SELECT article_img FROM articles WHERE article_id = $articleID");

while($row = $query->fetch(PDO::FETCH_ASSOC)){
    $image = imagecreatefromstring($row['article_img']);
    imagepng($image);
}

?>

Then you would call this file according to the other user’s example:

<img src="mostraimagem.php?id={article_id}">

Meanwhile, This is not a very recommended practice, it would be more interesting for you to upload the file yourself, according to what is shown in php.net.

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