Interview Question in Sql Server Views
Interview Question :: How do i make this php code go into the products id html box defined at the bottom
<html>
<head>
<title>Sign Discussion Board</title>
<link href="index1.css" rel="stylesheet" type="text/css"/>
</head>
<body>
<img src="guestbook.png" align="middle" alt="advert"" width="200" height="50"/>
<?php
$self = $_SERVER['PHP_SELF'];
$name = $_POST['name'];
$email = $_POST['email'];
$comments = $_POST['comments'];
$submit = $_POST['submit'];
#the html form
$form = "<form action=\"$self\" method=\"post\">";
$form.= "Name: <input type=\"text\" name=\"name\" ";
$form.= "size=\"50\" value=\"$name\"> <br>";
$form.= "Email: <input type=\"text\" name=\"email\" ";
$form.= "size=\"50\" value=\"$email\"> <br>";
$form.= "Comments:<br>";
$form.= "<textarea name=\"comments\" cols=\"45\" ";
$form.= "rows=\"4\">$comments</textarea> <br>";
$form.= "<input type=\"submit\" name=\"submit\" ";
$form.= "value=\"Sign\"> </form>";
#on first opening display the form
if( !$submit)
{ $msg = $form; }
else
#redisplay a message and the form if incomplete
if( !$name or !$email or !$comments)
{
$msg = "<b>Please complete all fields</b><br><br>";
$msg.= $form;
}
else
#add the form data to the guestbook database table
{
#connect to MySQL
$conn = mysql_connect("localhost", "ewood","woodie")
or die("Count not connect to database");
#select the database
$rs = mysql_select_db("ewood",$conn)
or die ("Could not select database");
#create the SQL query
if($name and $comments)
{
$sql = "insert into guestbook (name,email, comments)
values (\"$name\",\"$email\",\"$comments\")";
$rs = mysql_query($sql,$conn)
or die ("Could not execute SQL query");
}
#confirm the entry and display a link to view guestbook
if($rs)
{
$msg = "<h3>Thank you - your entry has been saved.</h3>";
$msg.= "<h3><a href = \"guestbook-view.php\">";
$msg.= "View My Guestbook</a></h3>";
}
}
echo($msg);
?>
<div id="products"align = center>
</div>
</body>
</html>
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