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Using PHP and MYSQL create a form that: user Full name user Comment (as many lines...

Using PHP and MYSQL create a form that:

  1. user Full name
  2. user Comment (as many lines as they want)
  3. allows the user to attach one image
  4. uploads the Full name, comment, and image to a database
  5. Displays the user's full name, comment and user image on the website by selecting them from the database.
  6. Include the server connection file as well any necessary codes to connect and maintain the database
engineering   Computer-Science   
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Homework Answers

Answer #1

solution :- following code will connect database and store data into database and display the data

index.php

------------------code starts from here-----------------------------

<?php

$connect = mysqli_connect("localhost", "root", "", "testing");

if(isset($_POST["insert"]))

{

$file = addslashes(file_get_contents($_FILES["image"]["tmp_name"]));

$query = "INSERT INTO tbl_images(fullname,comments,name) VALUES ('$fullname','$comments','$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="text" name="fullname" id="fullname"/>

<br />

<input type="text" name="comments" id="comments"/>

<br />

<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);

while($row = mysqli_fetch_array($result))

$fullname = $row['fullname'];

$comments = $row['comments'];

{

echo '

<tr>

<td>

<h2>$fullname<h2/>

</td>

</tr>

<tr>

<td>

<h2>$comments<h2/>

</td>

</tr>

<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>

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