Parse error

[Plagel]

Hi all. Trying to learn from a book I received..pretty sure I'm missing a character somewhere. The parse error says it's on the last line of the doc. Please help me find it! Thanks in advance.

 

<?php

$link = mysql_connect("localhost", "root", "pass")

or die(mysql_error());

mysql_select_db("moviesite")

or die(mysql_error());

$query = "SELECT movie_name, movie_director, movie_leadactor " .

"FROM movie";

 

$result = mysql_query($query, $link)

or die(mysql_error());

$num_movies = mysql_num_rows($result);

 

$movie_header=<<<EOD

<h2><center>Movie Review Database</center></h2>

<table width="70%" border="1" cellpadding="2"

cellspacing="2" align="center">

<tr>

<th>Movie Title</th>

<th>Year of Release</th>

<th>Movie Director</th>

<th>Movie Lead Actor</th>

<th>Movie Type</th>

</tr>

 

EOD;

function get_director() {

global $movie_director;

global $director;

 

$query_d = "SELECT people_fullname " .

  "FROM people " .

  "WHERE people_id='$movie_director'";

$results_d = mysql_query($query_d)

or die(mysql_error());

$row_d = mysql_fetch_array($results_d);

extract($row_d);

$director = $people_fullname;

}

 

function get_leadactor() {

global $movie_leadactor;

global $leadactor;

 

$query_a = "SELECT people_fullname " .

  "FROM people " .

  "WHERE people_id='$movie_leadactor'";

$results_a = mysql_query($query_a)

or die(mysql_error());

$row_a = mysql_fetch_array($results_a);

extract($row_a);

$leadactor = $people_fullname;

}

 

while ($row = mysql_fetch_array($result)) {

$movie_name = $row['movie_name'];

$movie_director = $row['movie_director'];

$movie_leadactor = $row['movie_leadactor'];

//get director's name from people table

get_director();

//get lead actor's name from people table

get_leadactor();

$movie_details .=<<<EOD

<tr>

<td>$movie_name</td>

<td>$director</td>

<td>$leadactor</td>

</tr>

EOD;

}

$movie_details .=<<<EOD

<tr>

<td>Total :$num_movies Movies</td>

</tr>

EOD;

 

$movie_footer ="</table>";

 

$movie =<<<MOVIE

$movie_header

$movie_details

$movie_footer

MOVIE;

 

echo "There are $num_movies movies in our database";

echo $movie;

?>

[Plagel]

I commented a lot of stuff..the part that seems to be generating the error is within the while part.

[oni-kun]

Please add your code within

 .. 

tags! But add this to after your <?php opening tag:

ini_set ("display_errors", "1"); 
error_reporting(E_ALL);

 

This should reveal the error rather than just halt execution, show us the error.

[Plagel]

<?php
ini_set ("display_errors", "1"); 
error_reporting(E_ALL);
$link = mysql_connect("localhost", "root", "pass")
or die(mysql_error());
mysql_select_db("moviesite")
or die(mysql_error());
$query = "SELECT movie_name, movie_director, movie_leadactor " .
	 "FROM movie";

$result = mysql_query($query, $link)
or die(mysql_error());
$num_movies = mysql_num_rows($result);

$movie_header =<<<EOD
<h2><center>Movie Review Database</center></h2>
<table width="70%" border="1" cellpadding="2" cellspacing="2" align="center">
	<tr>
		<th>Movie Title</th>
		<th>Year of Release</th>
		<th>Movie Director</th>
		<th>Movie Lead Actor</th>
		<th>Movie Type</th>
	</tr>

EOD;

function get_director() {
global $movie_director;
global $director;
$query_d = "SELECT people_fullname " .
		   "FROM people " .
		   "WHERE people_id='$movie_director'";
$results_d = mysql_query($query_d)
	or die(mysql_error());
$row_d = mysql_fetch_array($results_d);
extract($row_d);
$director = $people_fullname;
}

function get_leadactor() {
global $movie_leadactor;
global $leadactor;

$query_a = "SELECT people_fullname" .
		   "FROM people" .
		   "WHERE people_id='$movie_leadactor'";
$results_a = mysql_query($query_a)
	or die(mysql_error());
$row_a = mysql_fetch_array($results_a);
extract($row_a);
$leadactor = $people_fullname;
}


while ($row = mysql_fetch_array($result)) {
$movie_name = $row['movie_name'];
$movie_director = $row['movie_director'];
$movie_leadactor = $row['movie_leadactor'];
//get director's name from people table
get_director();
//get lead actor's name from people table
get_leadactor();
$movie_details .=<<<EOD
	<tr>
		<td>$movie_name</td>
		<td>$director</td>
		<td>$leadactor</td>
	</tr>
EOD;
}


$movie_details .=<<<EOD
	<tr>
		<td>Total :$num_movies Movies</td>
	</tr>
EOD;

$movie_footer ="</table>";

$movie =<<<MOVIE
		$movie_header
		$movie_details
		$movie_footer
MOVIE;

echo "There are $num_movies movies in our database";
echo $movie;
?>

Error: Parse error: parse error in C:\Users\plagel\Desktop\local\table2.php on line 90

[Plagel]

So..after hours of tweaking I've come nowhere..a friend has tipped me that it has to do with the quotations within my database query used when calling the string..

 

It's really amazing to me how far the internet has come, lolol

 

Here is my new error message:

 

Notice: Use of undefined constant e_all - assumed 'e_all' in C:\Users\plagel\Desktop\local\table2.php on line 3

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'people_id='1'' at line 1

 

and the code:

<?php
ini_set ("display_errors", "1"); 
error_reporting(e_all);
$link = mysql_connect("localhost", "root", "pass")
or die(mysql_error());
mysql_select_db("moviesite")
or die(mysql_error());
$query = "select movie_name, movie_director, movie_leadactor " .
	 "from movie";

$result = mysql_query($query, $link)
or die(mysql_error());
$num_movies = mysql_num_rows($result);

$movie_header=<<<eod
<h2><center>movie review database</center></h2>
<table width="70%" border="1" cellpadding="2" cellspacing="2" align="center">
	<tr>
		<th>movie title</th>
		<th>year of release</th>
		<th>movie director</th>
		<th>movie lead actor</th>
		<th>movie type</th>
	</tr>

eod;

function get_director() {
global $movie_director;
global $director;
$query_d = "select people_fullname " .
		   "from people " .
		   "where people_id='$movie_director'";
$results_d = mysql_query($query_d)
	or die(mysql_error());
$row_d = mysql_fetch_array($results_d);
extract($row_d);
$director = $people_fullname;
}

function get_leadactor() {
global $movie_leadactor;
global $leadactor;

$query_a = "select people_fullname" .
		   "from people" .
		   "where people_id='$movie_leadactor'";
$results_a = mysql_query($query_a)
	or die(mysql_error());
$row_a = mysql_fetch_array($results_a);
extract($row_a);
$leadactor = $people_fullname;
}


while ($row = mysql_fetch_array($result)) {
$movie_name = $row['movie_name'];
$movie_director = $row['movie_director'];
$movie_leadactor = $row['movie_leadactor'];
//get director's name from people table
get_director();
//get lead actor's name from people table
get_leadactor();
$movie_details.=<<<eod
	<tr>
		<td>$movie_name</td>
		<td>$director</td>
		<td>$leadactor</td>
	</tr>
eod;
}


$movie_details.=<<<eod
	<tr>
		<td>total :$num_movies movies</td>
	</tr>
eod;

$movie_footer ="</table>";

$movie=<<<movie
		$movie_header
		$movie_details
		$movie_footer
movie;

echo "there are $num_movies movies in our database";
echo $movie;
?>

[Buddski]

your error_reporting is using an invalid var.

error_reporting(E_ALL); // note the case

and your query_a is missing a space

$query_a = "select people_fullname " .
"from people " .
"where people_id='$movie_leadactor'";

[Plagel]

You rock

[ignace]

Don't tell us we know. That's why were here