Ch. 6 commit.php

MySequel · January 2nd, 2012, 08:06 PM

Ok so this is the code from the book on Ch. 6 for the first commit.php,

Quote:

<?php
$db = mysql_connect('localhost', 'bp6am', 'bp6ampass') or
die ('Unable to connect. Check your connection parameters.');
mysql_select_db('moviesite', $db) or die(mysql_error($db));
?>
<html>
<head>
<title>Commit</title>
</head>
<body>
<?php
switch ($_GET['action']) {
case 'add':
switch ($_GET['type']) {
case 'movie':
$query = 'INSERT INTO
movie
(movie_name, movie_year, movie_type, movie_leadactor,
movie_director)
VALUES
("' . $_POST['movie_name'] . '",
' . $_POST['movie_year'] . ',
' . $_POST['movie_type'] . ',
' . $_POST['movie_leadactor'] . ',
' . $_POST['movie_director'] . ')';

break;
}
break;
}

if (isset($query)) {
$result = mysql_query($query, $db) or die(mysql_error($db));
}
?>
<p>Done!</p>
</body>
</html>

I've been getting an error and I think there's something wrong with the VALUES part of the query, can somebody tell me whether there is something wrong with this code and how to correct it?

MySequel · January 15th, 2012, 02:35 AM

Can anybody help? Let me add some more details about what my question is and be more specific.

Code:

VALUES
("' . $_POST['movie_name'] . '",
' . $_POST['movie_year'] . ',
' . $_POST['movie_type'] . ',
' . $_POST['movie_leadactor'] . ',
' . $_POST['movie_director'] . ')';

I suspect that there's something wrong with the particular code above. My question is why is the first $_POST value encapsulated with both double-quotes and single quotes but the rest aren't? Also why does each $_POST have a period before and after it, isn't that used to concatenate something in PHP?

UseLess · January 15th, 2012, 10:00 AM

Greetings,

Try the following, I've altered it slightly so that the query is wrapped in " and any text value being stored in the db is wrapped in ' I am unable to test it as I haven't got the db setup etc and the book is at work.

PHP Code:


			
<?php
    $db = mysql_connect('localhost', 'bp6am', 'bp6ampass') or
    die ('Unable to connect. Check your connection parameters.');
    mysql_select_db('moviesite', $db) or die(mysql_error($db));
?>
<html>
<head>
<title>Commit</title>
</head>
<body>
<?php
switch ($_GET['action'])
{
    case 'add':
        switch ($_GET['type'])
        {
            case 'movie':
                $query    =    "INSERT INTO movie (movie_name, movie_year, movie_type, movie_leadactor, movie_director)
                            VALUES ('" . $_POST['movie_name'] . "',
                                    '" . $_POST['movie_year'] . "',
                                    '" . $_POST['movie_type'] . "',
                                    '" . $_POST['movie_leadactor'] . "',
                                    '" . $_POST['movie_director'] . "')";
            break;
        }
    break;
}

if (isset($query))
{
    $result = mysql_query($query, $db) or die(mysql_error($db));
}
?>
<p>Done!</p>
</body>
</html>

And yes the period . is used to add some text to the $query string thus once complete you should get something like the following in $query;

Code:

INSERT INTO movie (movie_name, movie_year, movie_type, movie_leadactor, movie_director)
VALUES ('Dreamscape', '1988', 'Horror', 'Dennis Quaid', 'Joseph Ruben')

MySequel · January 15th, 2012, 04:48 PM

Quote:

Originally Posted by UseLess

Greetings,

Try the following, I've altered it slightly so that the query is wrapped in " and any text value being stored in the db is wrapped in ' I am unable to test it as I haven't got the db setup etc and the book is at work.

PHP Code:


			
<?php

    $db = mysql_connect('localhost', 'bp6am', 'bp6ampass') or

    die ('Unable to connect. Check your connection parameters.');

    mysql_select_db('moviesite', $db) or die(mysql_error($db));

?>

<html>

<head>

<title>Commit</title>

</head>

<body>

<?php

switch ($_GET['action'])

{

    case 'add':

        switch ($_GET['type'])

        {

            case 'movie':

                $query    =    "INSERT INTO movie (movie_name, movie_year, movie_type, movie_leadactor, movie_director)

                            VALUES ('" . $_POST['movie_name'] . "',

                                    '" . $_POST['movie_year'] . "',

                                    '" . $_POST['movie_type'] . "',

                                    '" . $_POST['movie_leadactor'] . "',

                                    '" . $_POST['movie_director'] . "')";

            break;

        }

    break;

}



if (isset($query))

{

    $result = mysql_query($query, $db) or die(mysql_error($db));

}

?>

<p>Done!</p>

</body>

</html>

And yes the period . is used to add some text to the $query string thus once complete you should get something like the following in $query;

Code:

INSERT INTO movie (movie_name, movie_year, movie_type, movie_leadactor, movie_director)
VALUES ('Dreamscape', '1988', 'Horror', 'Dennis Quaid', 'Joseph Ruben')

Thanks, it worked!

Stamatis · February 5th, 2012, 04:06 PM

Hello all,

I have a problem also in ch 6 but in movie.php. I'm on page 157 i writed the code but i get each record of the Movie type,Year,Lead Actor,Director
two times.For example if a record for movie type is Action i get in list Action Action. Why ? Here is the code:

Code:

<?php
$db = mysql_connect('localhost','','') or 
	die('Unable to connect.Check your connection parameters');
	mysql_select_db('moviesite' ,$db) or die(mysql_error($db));
	
?>
<html>
<head>
<title>Add movie</title>
</head>
<body>
<form method="post" action="commit.php?action=add&type=movie">
  <table width="408" height="142" border="1">
    <tr>
      <td width="147">Movie Name</td>
      <td width="182">
      <input type="text" name="textfield" id="moviename"></td>
    </tr>
    <tr>
      <td>Movie Type</td>
      <td>
        <select name="movie_type">
	  <?php
	  $query = 'select movietype_id, movietype_label 
	  from movietype
	  ORDER BY 
	  movietype_label';
	  $result=mysql_query($query, $db) or die(mysql_error($db));

while ($row = mysql_fetch_assoc($result)){ 
foreach ($row as $value) {
	
	   
      echo '<option value="' . $row['movietype_id'] . '">';
		     echo $row['movietype_label'] . '</option>';
	}
}
?>     
        </select>
     </td>
    </tr>
    <tr>
      <td>Movie Year</td>
      <td>
        <select name="movie_year" >
        <?php
		for($yr=date("Y");$yr>=1970;$yr--){
			echo '<option value="' . $yr . '">' . $yr . '</option>';
		}
		?>
        
      </select>      </td>
    </tr>
    <tr>
      <td>Lead Actor</td>
      <td>
        <select name="movie_leadactor" >
        <?php
		$query= 'select people_id,people_fullname
		from people
		where people_isactor=1
		order by people_fullname';
		  $result=mysql_query($query, $db) or die(mysql_error($db));
		  while($row = mysql_fetch_assoc($result)) {
			  foreach ($row as $value){
				  echo '<option value="' . $row['people_id'] . '">';
				  echo $row['people_fullname'] . '</option>';
			  }
		  }
		  ?>
      </select></td>
    </tr>
    <tr>
      <td>Director</td>
      <td>
        <select name="movie_director" >
        <?php
		$query = 'SELECT
        people_id, people_fullname
    FROM
        people
    WHERE
        people_isdirector = 1
    ORDER BY
        people_fullname';
$result = mysql_query($query, $db) or die(mysql_error($db));

// populate the select options with the results
while ($row = mysql_fetch_assoc($result)) {
    foreach ($row as $value) {
    	  echo '<option value="' . $row['people_id'] . '">';
	  	  echo $row['people_fullname'] . '</option>';
	}
}
    ?>
	  </select></td>
	 
    </tr>
	
  </table> <input type="submit" name="submit" value="Add"/>
</form>
</body>
</html>

UseLess · February 5th, 2012, 04:54 PM

Greetings,

That's because you have a foreach inside the while loop, and you're not even using the foreach so remove it so it's like this:

Code:

while ($row = mysql_fetch_assoc($result))
{ 
	echo '<option value="' . $row['movietype_id'] . '">' . $row['movietype_label'] . '</option>';
}

Stamatis · February 5th, 2012, 04:57 PM

its ok now.
perhaps it was a wrong of the book while the code was the same of the book

thank you