22. Consider the following entities and their relationships
Movie (movie_no,
movie_name, release_year)
Actor (actor_no,
name)
Relationship between movie and actor is many
– many with attribute rate in Rs. Create a RDB in 3 NF for the above and solve
following Using above database, write PHP scripts for the
following: (Hint: Create HTML form having two
radio buttons)
a) Accept actor name and display the names of
the movies in which he has acted.
b) Insert new movie information
html file :
<html>
<body>
<form
action="slip_22.php" method="get">
<h3>Enter
Actor Name : <input type=text name=nm> </h3>
<input
type=radio name=a value=1>Display Movie Name<br><br>
<h3>Enter
movie no :<input type=text name=m_no> </h3>
<h3>Enter
movie name :<input type=text name=m_nm></h3>
<h3>Enter
release year :<input type=text name=r_yr> </h3>
<h3>Enter
actor no :<input type=text name=a_no> </h3>
<h3>Enter
actor name :<input type=text name=a_nm> </h3>
<input
type=radio name=a value=2>Insert New movie info<br><br>
<input
type=submit value=OK>
</form>
<div
id="place"></div>
</body>
</html>
php file :
<?php
$r
= $_GET['a'];
$con
= mysql_connect("localhost","root","");
$d
=
mysql_select_db("bca_programs",$con);
if($r == 1)
{
$actor_name =
$_GET['nm'];
$q =
mysql_query("select m_name from movie,actor,movie_actor where
movie.m_no=movie_actor.m_no and actor.a_no=movie_actor.a_no and
a_name='$actor_name'");
echo "<br>Movie
Name </br>";
while($row=mysql_fetch_array($q))
{
echo
$row[0]."<br>";
}
}
else if($r == 2)
{
$m_no =
$_GET['m_no'];
$m_name =
$_GET['m_nm'];
$r_yr =
$_GET['r_yr'];
$a_no =
$_GET['a_no'];
$a_name =
$_GET['a_nm'];
$q =
mysql_query("insert into movie
values($m_no,'$m_name',$r_yr)");
$q1 =
mysql_query("insert into actor values($a_no,'$a_name')");
echo "Value
Inserted";
}
mysql_close();
?>
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