I am unfortunately getting this error.Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\xampp\htdocs\work\index.php on line 41.Any idea what it means please? Still on the learning curve of php sorry.... here is my code with databse testing table name tbl_image.
<?php $connect = mysqli_connect("localhost", "root", "", "testing"); if(isset($_POST["insert"])) {
$file = addslashes(file_get_contents($_FILES["image"]["tmp_name"]));
$query = "INSERT INTO tbl_images(name) VALUES ('$file')";
if(mysqli_query($connect, $query))
{
echo '<script>alert("Image Inserted into Database")</script>';
}}?>
<!DOCTYPE html><html>
<head>
<title>Webslesson Tutorial | Insert and Display Images From Mysql Database in PHP</title>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/js/bootstrap.min.js"></script>
</head>
<body>
<br /><br />
<div class="container" style="width:500px;">
<h3 align="center">Insert and Display Images From Mysql Database in PHP</h3>
<br />
<form method="post" enctype="multipart/form-data">
<input type="file" name="image" id="image" />
<br />
<input type="submit" name="insert" id="insert" value="Insert" class="btn btn-info" />
</form>
<br />
<br />
<table class="table table-bordered">
<tr>
<th>Image</th>
</tr>
<?php
$query = "SELECT * FROM tbl_images ORDER BY id DESC";
$result = mysqli_query($connect, $query);
$start = 1;
while($row = mysqli_fetch_array($result))
{
echo '
<tr>
<td>
<img src="data:image/jpeg;base64,'.base64_encode($row['name'] ).'" height="200" width="200" class="img-thumnail" />
</td>
</tr>
';
}
?>
</table>
</div>
</body></html><script>$(document).ready(function(){
$('#insert').click(function(){
var image_name = $('#image').val();
if(image_name == '')
{
alert("Please Select Image");
return false;
}
else
{
var extension = $('#image').val().split('.').pop().toLowerCase();
if(jQuery.inArray(extension, ['gif','png','jpg','jpeg']) == -1)
{
alert('Invalid Image File');
$('#image').val('');
return false;
}
}
});});</script>
