Display image from database as background of a div element in php - php

Question

Currently I'm learning about PHP and database, I wrote the script below to display the image as background for my div element, but the output is actually nothing! the error is from the line:
echo "<div class=\"post\" style='background-image: url('\"<?php echo $data[Image];\"?>')'";?>
The quotes mess up! Can someone tell me how to correct this? I tried to change double quotes to single quotes, but still doesn't work at all.
This is my full script:
<div class="dashboardA">
<?php
$con = mysqli_connect("localhost", "Dave", "password");
if (!$con){
die ("Could not connect to database: " . mysqli_connect_error());
}
mysqli_select_db($con, "my_blog");
$sql = mysqli_query($con, "select * from article");
while ($data=mysqli_fetch_array($sql)){
echo "<div class=\"post\" style='background-image: url('\"<?php echo $data[Image];\"?>')'";?>
<?php echo "<p>" . $data["Title"] . "</p>";
echo "<p>" . $data["Category"] . "</p>";
echo "<p>" . $data["Published"] . "</p>";
echo "</div>";
}
?>
</div>

Answer 1

You are getting a little mixed up. You already have PHP opening tags, you don't need them again. Just concatenate your variable:
echo "<div class='post' style='background-image: url(\"$data[Image]\")'>";
Note: You also need to close your opening <div> tag.

Answer 2

There is a mistake in the line
echo "<div class=\"post\" style='background-image: url('\"<?php echo $data[Image];\"?>')'";?>
You are already in php, so the opening tags are wrong there.
Try:
echo "<div class=\"post\" style='background-image: url(\"" . $data[Image] . "\")'";?>

Answer 3

Try this
while ($data=mysqli_fetch_array($sql)){
echo "<div class=\"post\" style='background-image: url('\"<?php echo $data[Image];\"?>')'";?>
<?php echo "<p>" . $data["Title"] . "</p>";
echo "<p>" . $data["Category"] . "</p>";
echo "<p>" . $data["Published"] . "</p>";
echo "</div>";
}
Replace this
with this
while ($data=mysqli_fetch_array($sql)){
$img = $data["Image"];?>
<div class="post" style="background-image: url('<?php echo $img;?>')" >
<?php echo "<p>" . $data["Title"] . "</p>";
echo "<p>" . $data["Category"] . "</p>";
echo "<p>" . $data["Published"] . "</p>";
echo "</div>";
}

Answer 4

You actually did it right in the following echo commands. To combine the value of your variable and a string you use the "." that basically substitutes for "+" that is used in other languages for this purpose. Opening a new php tag is therefore not needed here.
The correct code is
echo '<div class="post" style="background-image: url(\'' . $data[Image] . '\')">';

Answer 5

Try this, outside php tags.
<div class="post" style="background-image: url('<?php echo "$data[Image]"; ?>');">