Default image won't show when database is still empty - php

Question

Basically, the whole thought here is changing profile pictures and it will be uploaded in the database.
I want to show a default icon first when the database is still empty but my codes just displays a broken image thumbnail. What could have gone wrong?
<form method="post" enctype="multipart/form-data">
<div class="row form-group">
<div class="col col-md-3">
<label for="text-input" class="form-control-label">Profile Picture</label>
</div>
<div class="col-9 col-md-9">
<div class="input-group">
<?php
$query = "SELECT image FROM member WHERE member_id = '$ID'";
$result = mysqli_query($con, $query);
if(mysqli_num_rows($result)>0){
while($row = mysqli_fetch_array($result))
{
echo '<img src="data:image/jpeg;base64,'.base64_encode($row['image']).'">';
}
} else{
echo "<img src='images/icon.jpg' style='height:200px'draggable='false'>";
}
?>
<div class="col-12">
<input type="file" name="image" id="image" style ='margin-left:-10px;margin-top:5px'>
</div>
<div class="col-12">
<input type="submit" name="insert" id="insert" value="Change Profile Picture" class="btn btn-primary" style ='margin-left:-10px;margin-top:5px'>
</div>
<?php
if(isset($_POST["insert"])){
$file = addslashes(file_get_contents($_FILES["image"]["tmp_name"]));
$query = "UPDATE member SET image = '$file' WHERE member_id='$ID'";
if(mysqli_query($con, $query)){
echo ("<script LANGUAGE='JavaScript'>
window.alert('Profile Picture has been successfully changed!');
window.location.href='edit.php';
</script>");
}
}
?>
</div>
</div>
</div>
</form>

Answer 1

if your database is empty upload query will not work. if your image row is empty then your code should be like if image row is not empty display image else display the default image.
if(mysqli_num_rows($result)>0){
while($row = mysqli_fetch_array($result)) {
if (isset($row['image']) && !empty($row['image'])) {
echo '<img src="data:image/jpeg;base64,'.base64_encode($row['image']).'">';
} else {
echo "<img src='images/icon.jpg' style='height:200px'draggable='false'>";
}
} // endwhile
} else{
echo "no result found";
}

Answer 2

While writing HTML or CSS code inside PHP you should put a \ before all double quotations and you need change all quotations to double quotation.like shown below:
echo "<img src=\"images/icon.jpg\" style=\"height:200px;\" draggable=\"false\">";
I didn't check your PHP codes, If you've coded them properly the problem is obviously what I mentioned.