Question

In: Computer Science

Create a new SQL table in your phpMyAdmin account that collects the following fields (columns) - (10...

Create a new SQL table in your phpMyAdmin account that collects the following fields (columns) -

1) The name of a medicine, sanitizer, or tip/trick for protecting yourself against transmission or fight the disease - STRING

2) The amount of money the item costs - NUMBER

3) The UPC code of the item - NUMBER

4) Manufacturing country of the item - STRING

Connect to the SQL database with your HTML code, and make sure to show the results of ALL columns show! -

Create a form where the user can insert their items into the 4 columns mentioned above -

Build the HTML site with at least 5 pictures and a few paragraphs explaining your creativity (and how important it is to be safe this flu season!) - make sure to use proper APA guidelines for your in-text citations and a bibliography section! -

Solutions

Expert Solution

SQL Code :

CREATE TABLE table1 (
Medicine_name varchar(255),

Amount int,

    FirstName varchar(255),
    Address varchar(255),
    City varchar(255)
);

Html code : //I have written basic html code for creating the from. Hope you can insert pictures and write relevant text.

<!DOCTYPE html>

<html>

<head>

</head>

<body>

Cure : <input type="text" name="Medicine"><br><br>

Cost : <input type="number " name="Cost"><br><br>

UPC : <input type="number " name="UPC"><br><br>

Country : <input type="text" name="Country"><br><br>

</form>

</body>

</html>

Php code:

<?php

$cure = filter_input(INPUT_POST, 'cure');

$cost = filter_input(INPUT_POST, 'cost');

if (!empty($cure)){

if (!empty($cost)){

$host = "localhost";

$dbusername = "root";

$dbpassword = "";

$dbname = "my_db";

// Create connection

$conn = new mysqli ($host, $dbcure, $dbcost, $dbname);

if (mysqli_connect_error()){

die('Connect Error ('. mysqli_connect_errno() .') '

. mysqli_connect_error());

}

else{

$sql = "INSERT INTO account (cure, cost)

values ('$cure','$cost')";

if ($conn->query($sql)){

echo "New record is inserted sucessfully";

}

else{

echo "Error: ". $sql ."

". $conn->error;

}

$conn->close();

}

else{

echo "Values should not be empty";

die();

}

else{

echo "Values should not be empty";

die();

}

venereology answered 1 month ago

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