SELECT image from a database and display on webpage (PHP)
Question
I'm SELECT ing an images column from my database to display in an HTML <table> . I am unsure if the $sql = "SELECT Images" . "FROM koop"; $sql = "SELECT Images" . "FROM koop"; code I have below will work to load up images.
Example page how it should look like
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "woodstocks";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Type, Prijs, Adres, Woonopp, Grondopp, Slaapkamers, Badkamers, Bebouwing, Bouwjaar, Images, Text "
. "FROM koop";
$htmlTabel = "Er is niks gevonden terug te vinden!";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
$htmlTabel = "<table class='table'>";
$htmlTabel .= "<tr class='thead-dark'>";
$htmlTabel .= "<th>Type</th>" .
"<th>Prijs</th>" .
"<th>Adres</th>" .
"<th>Woonopp</th>" .
"<th>Grondopp</th>" .
"<th>Slaapkamers</th>" .
"<th>Badkamers</th>" .
"<th>Bebouwing</th>" .
"<th>Bouwjaar</th>" .
"<th>Images</th>" .
"<th>Text</th>";
$htmlTabel .= "</tr>";
while ($row = $result->fetch_assoc()) {
$htmlTabel .= '<tr>';
$htmlTabel .= '<td>' . $row['Type'] . '</td>' .
'<td>' . $row['Prijs'] . '</td>' .
'<td>' . $row['Adres'] . '</td>' .
'<td>' . $row['Woonopp'] . '</td>' .
'<td>' . $row['Grondopp'] . '</td>' .
'<td>' . $row['Slaapkamers'] . '</td>' .
'<td>' . $row['Badkamers'] . '</td>' .
'<td>' . $row['Bebouwing'] . '</td>' .
'<td>' . $row['Bouwjaar'] . '</td>' .
'<td>' . $row['Images'] . '</td>' .
'<td>' . $row['Text'] . '</td>';
$htmlTabel .= '</tr>';
}
$htmlTabel .= "</table>";
}
$conn->close();
?>
<!DOCTYPE html>
<html>
<head>
<!-- Required meta tags -->
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<!--BOOTSTRAP-->
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.2.1/css/bootstrap.min.css" integrity="sha384-GJzZqFGwb1QTTN6wy59ffF1BuGJpLSa9DkKMp0DgiMDm4iYMj70gZWKYbI706tWS" crossorigin="anonymous">
<!-- COSTUM STYLE.CSS -->
<link href="Classes/Model/Koop.css" rel="stylesheet" type="text/css"/>
<title>Woodstocks Immo</title>
<link rel="shortcut icon" type="image/png" href="img/Woodstocks.png">
</head>
<body class="d-flex flex-column">
<nav class="navbar navbar-expand-md navbar-dark sticky-top" style="background-color: rgba(0,0,0,0.3);">
<a href="http://www.facebook.com" style="margin-left: 15px;"><img width="50" src="img/facebook.png" alt=""/></a>
<a href="http://www.instagram.com" style="margin-left: 15px;"><img width="40" src="img/instagram.png" alt=""/></a>
<a href="http://www.twitter.com" style="margin-left: 15px;"><img width="50" src="img/twitter.png" alt=""/></a>
<a style="color: mintcream; margin-left: 15px;" class="navbar-brand" href="#">Woodstocks</a>
<button class="navbar-toggler navbar-toggler-right" type="button" data-toggle="collapse" data-target="#navb" aria-expanded="true">
<span class="navbar-toggler-icon"></span>
</button>
<div id="navb" class="navbar-collapse collapse hide">
<ul class="navbar-nav ml-auto">
<li class="nav-item">
<a style="color: mintcream; padding-left: 45px; padding-right: 45px;" class="nav-link" href="Home">Home</a>
</li>
<li class="nav-item">
<a style="color: mintcream; padding-left: 45px; padding-right: 45px;" class="nav-link" href="Huur">Huur</a>
</li>
<li class="nav-item">
<a style="color: mintcream; padding-left: 45px; padding-right: 45px;" class="nav-link" href="Koop">Koop</a>
</li>
<li class="nav-item">
<a style="color: mintcream; padding-left: 45px; padding-right: 45px;" class="nav-link" href="Verkoop-Verhuur">Verkopen/Verhuren</a>
</li>
<li class="nav-item">
<a style="color: mintcream; padding-left: 45px; padding-right: 45px;" class="nav-link" href="Team">Team</a>
</li>
<li class="nav-item">
<a style="color: mintcream; padding-left: 45px; padding-right: 45px;" class="nav-link" href="Contact">Contact</a>
</li>
</ul>
<ul class="nav navbar-nav ml-auto">
<li class="nav-item">
<a style="color: mintcream; padding-left: 45px; padding-right: 45px;" class="nav-link" href="Login">Inloggen/Registreren</a>
</li>
</ul>
</div>
</nav>
<div class="background-item">
<img src="img/pexels-photo-453201.jpeg" style="height: auto; max-height: 400px; object-fit: cover; width: 100%;" alt=""/>
<div class="col-md-12" style="width: 100%;">
</div>
</div>
<div class="container" style="margin-top: 25px;">
<?php echo $htmlTabel ?>
</div>
</body>
</html>
2 answers
solution1
0 2019-03-14 21:28:02
I am unsure if the $sql = "SELECT Images" . "FROM koop"; code I have below will work to load up images.
You'll have to go ahead and try it to see if it will work, but it should work if you have your database details correct, as well as a correct query.
If you have the images inside a table called Koop , and the images urls (probably) are stored under a column called Images under the same table, then the query SELECT Images FROM koop; should work.
solution2
0 2019-03-15 20:05:22
So I have made the next example of and upload images and show image but I come close to it PICTURE 1 UPLOAD Picture 2 HTML
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "woodstocks";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
if (isset($_POST['Upload'])) {
$target = "img/" . basename($_FILES['images']['name']);
$image = $_FILES['images']['name'];
$sql = "INSERT INTO huur (images) VALUES ('$images')";
mysqli($conn, $sql);
if (move_uploaded_file($_FILES['images']['tmp_name'], $target)) {
$msg = "Image uploaded successfully";
} else {
$msg = "There was a problem uploading image";
}
}
?>
<!DOCTYPE html>
<!--
To change this license header, choose License Headers in Project Properties.
To change this template file, choose Tools | Templates
and open the template in the editor.
-->
<html>
<head>
<meta charset="UTF-8">
<title></title>
</head>
<body>
<form enctype="multipart/form-data" action="upload" method="POST">
<p>Upload van een bestand</p>
<input type="file" name="images"/>
<br>
<input type="submit" value="Upload"/>
</form>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "woodstocks";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT images" . "FROM images";
$result = msqli($conn, $sql);
while ($row = mysqli_fetch_all($result)) {
echo "<img src='img/".$row['images']."'>";
}
?>
</body>
</html>
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