Database query - how to put the result in the HTML Code? - php

Question

I have a database where the path of pictures are stored. Now I want to get the path of the picture and the description but I dont know how to surround the data with HTML-Code. A "(Html-Code) " indicates Html-Code but in my example it is a bit more complicated. Maybe it is wrong the way I want to do it but I dont know another way.
Here the Php-Code
<?php include ("data_con.php");
$conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Dateipfad FROM last_upload WHERE ID = 1";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
$row = $result->fetch_assoc(); {
}
} else {
echo "0 results";
}
$conn->close();
echo "Connected successfully";?>
Here the Html-Code
<div class="col-xs-5 col-sm-5 col-md-5 col-lg-5 no-gutter">
<div class="rahmen"> <img src="Pic/OP.jpg" class="img-fluid" alt="Responsive image">
<p>Episode 1</p></div>
First of all I just want to have the path of the file (Dateipfad) ecchoed in
<img src="Pic/Op.jp">
therefore I need to make it like that
<img="echo .row[Dateipfad]."
But all the " to indicate the html code is really hard.
The code have to be in
$row = $result->fetch_assoc(); {
}
but with all the HTML-Code it is really hard and doesnt work for me. I think i do it completly wrong. Can anyone help me?

Answer 1

Method 1
<?php $conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Dateipfad FROM last_upload WHERE ID = 1";
if ( $result = $conn->query($sql) ) {
while ($row = mysqli_fetch_row($result)){
echo '<img src ="$row['value']">'$row['value']'</img>';
}
} else {
echo "0 results";
}
$conn->close();
echo "Connected successfully";
Method Two
<?php $conn = new mysqli($servername, $username, $password);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT Dateipfad FROM last_upload WHERE ID = 1";
if ( $result = $conn->query($sql) ) {
while ($row = mysqli_fetch_row($result)){ ?>
// write html here
<ul>
<li>Just a test</li>
<li>Another test</li>
</ul>
<img src ="<?php echo $row['link']; ?>"><?php echo $row['value']; ?></img>
<?php }
} else {
echo "0 results";
}
$conn->close();
echo "Connected successfully";

Answer 2

Actually what you are looking for is something akin to
<img src="<?= $row[Dateipfad] ?>" />
Also, are you sure that you don't want to iterate on all the results? Will your query always return a single result or none at all?

Answer 3

<img src="<?php echo row[Dateipfad]; ?>" class="img-fluid" alt="Responsive image">
And don't use short handed echo statement <?= ?> for my sake.

Answer 4

update
if($result->num_rows > 0)
{
while ($row = mysqli_fetch_row($result)){?>
<div class="col-xs-5 col-sm-5 col-md-5 col-lg-5 no-gutter">
<div class="rahmen"> <img src="<?php echo $row['Dateipfad']; ?>" class="img-fluid" alt="Responsive image">
<p>Episode 1</p>
</div>
</div>
<?php}
}