[RESOLVED] Database connection & query.... Notice: Undefined index:

riddellchris

I'm having trouble with the following code

<?php 
$link = mysql_connect("localhost", "username", "password")
	or die(mysql_error());
mysql_select_db("moviesite")
	or die (mysql_error());

/* Function to calculate if a movie made a profit loss or broke even */
function calculate_differences($takings, $cost) {
	$difference = $takings - $cost;

if ($difference < 0) {
		$difference = substr($difference, 1);
		$font_colour = 'red';
		$profit_or_loss = "$" / $difference . "m";
}	elseif ($difference > 0) {
	$font_colour = 'green';
	$profit_or_loss = "$" . $difference . "m";
} else {
	$font_colour = 'blue';
	$profit_or_loss = "Broke_even";
}
return "<font colour=\"$font_colour\">" . $profit_or_loss . "</font>";
}

// Function to get the director's name from the people table //
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 to get the director's name from the people table //
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;
}

$movie_id=$_GET['movie_id'] ;  //here//

$query="SELECT * FROM movie WHERE movie_id='$movie_id'";

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

$movie_table_headings=<<<EOD
	<tr>
		<th>Movie Title<th>
		<th>Year of Release<th>
		<th>Movie Director<th>
		<th>Movie Lead Actor<th>
		<th>Movie Running Time<th>
		<th>Movie Health<th>
	</tr>
EOD;

while ($row = mysql_fetch_array($result)) {
	$movie_name = $row['movie_name'];
	$movie_director = $row['movie_director'];
	$movie_leadactor = $row['movie_leadactor'];
	$movie_year = $row['movie_year'];
	$movie_running_time = $row['movie_running_time']." mins";
	$movie_takings = $row['movie_takings'];
	$movie_cost = $row['movie_cost'];

//get director's name from people table
get_director($movie_director);

//get lead actor's name from people table
get_leadactor($movie_leadactor);
}

$movie_health = calculate_differences($movie_takings, $movie_cost); //here//
$page_start =<<<EOD

<html>
<head>
<title>Details and Review for: $movie_name</title>
</head>
<body>   //here//
EOD;  

$movie_details =<<<EOD
<table width="70%" border="0" cellspacing="2"
				cellpadding="2" align="center">
	<tr>
		<th colspan="6"><u><h2>$movie_name: Details</h2></u></th>
	</tr>
	$movie_table_headings  //here//
	<tr>
		<td witdh="33%" align ="center">$movie_name</td>
//here//		<td align="center">$movie_year</td>                          

//here//		<td align="center">$movie_director</td>
//here//		<td align="center">$movie_leadactor</td>
//here//		<td align="center">$movie_running_time</td>
		<td align="center">$movie_health</td>
	</tr>
</table>
<br>
<br>
EOD;

$page_end =<<<EOD
</body>
</html>
EOD;

$detailed_movie_info =<<<EOD
	$page_start
	$movie_details
	$page_end
EOD;

echo $detailed_movie_info;
mysql_close();

?>

The errors that it give me are:

Notice: Undefined index: movie_id in C:\apache\htdocs\movie_review_website\movie_details.php on line 56

Notice: Undefined variable: movie_takings in C:\apache\htdocs\movie_review_website\movie_details.php on line 90

Notice: Undefined variable: movie_cost in C:\apache\htdocs\movie_review_website\movie_details.php on line 90

Notice: Undefined variable: movie_name in C:\apache\htdocs\movie_review_website\movie_details.php on line 97

Notice: Undefined variable: movie_name in C:\apache\htdocs\movie_review_website\movie_details.php on line 106

Notice: Undefined variable: movie_name in C:\apache\htdocs\movie_review_website\movie_details.php on line 109

Notice: Undefined variable: movie_year in C:\apache\htdocs\movie_review_website\movie_details.php on line 110

Notice: Undefined variable: movie_director in C:\apache\htdocs\movie_review_website\movie_details.php on line 111

Notice: Undefined variable: movie_leadactor in C:\apache\htdocs\movie_review_website\movie_details.php on line 112

Notice: Undefined variable: movie_running_time in C:\apache\htdocs\movie_review_website\movie_details.php on line 113

Basically these variables are the column headings from the table "movie" in the database "moviesite". I assume its not connecting properly for some reason even though i have got it to with other pages OR i'm doing something wrong with the query and the treatment of the results. My other concern is that i've stuffed up brackets or something stupid somewhere because of the strange placement of some of the errors but I have looked over this with my editor as best as possible.

Anyway I'm stuck and haven't got the answers. Any help would be much appreciated!

krotkruton

Try changing these lines (around line 56):

$movie_id=$_GET['movie_id'] ;  //here//

$query="SELECT * FROM movie WHERE movie_id='$movie_id'";

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

To this:

if (isset($_GET['movie_id'])) {
  $movie_id=$_GET['movie_id'] ;  //here//
} else {
  echo "movie_id not set in $_GET";
}

$query="SELECT * FROM movie WHERE movie_id='$movie_id'";

$result = mysql_query($query, $link)
    or die("Failed to get movie information for " . $movie_id . ": " .mysql_error());

You'll probably find that $_GET['movie_id'] hasn't been set, which is screwing up your query because you WHERE clause has movie_id set equal to an empty string. I didn't check the rest of the file because that jumped out at me, but if you're still having problems, let me know what the results of the new code are.

riddellchris

Thanks so much for your help!

Firstly i'm still learning so what do you mean by being "set"... and this is also related to the new error which adds to the others and is ontop i.e. is the first line of errors.

movie_id not set in Array

does this mean? it's there and a column in the db with numbers in it... is the column the array?

Sxooter

When you call another page from a form, php takes the incoming data and creates an array called either $GET[] or $POST[]. This array has all the vars that were setup by the calling form. so, if you have a form like this:

then the page you call with that form will have an array passed into it called $_POST and it will have an element movie_id that you can access like this:

$movie_id=$_POST['movie_id']

If you call it with method="get" then you'll have an element like $_GET['movie_id'].

To see all the elements being passed into a page, you can use the print_r function:

print_r($POST);
print_r($GET);

riddellchris

Thanks for your help!

When i try this all i get is

Array()

.... I assume this means that I'm not succesfully passing anything onto this page for some reason? I did play around with where this code was added on the page incase it made a difference but it always returned the same thing? What to do?

Sxooter

Are you using print() or print_r() ???

riddellchris

print_r()

Sxooter

You should get an output something like this:

Array
(
[a] => 1
=> 2
)

did you try it for the right array? i.e. $GET or $POST?

riddellchris

i did both and got nothing empty arrays? Not cool hey! Any ideas?

Sxooter

OK, have you got phpinfo(); turned on? Try going to a page like:

http://myserver.mydomain/phpinfo.phtml?key=value&key2=value2 (or html, or php, or whatever extensions your php server uses.)

And see what it says. There's all kinds of useful info on that page, and one of them is things like $GET or $POST or whatever they're called on your machine.

See if you can get phpinfo(); to give you a useful page for troubleshooting. Oh and what version php are you running?

riddellchris

i'm not sure quite what you mean by this

"And see what it says. There's all kinds of useful info on that page, and one of them is things like $GET or $POST or whatever they're called on your machine."

What should I be seeing or doing with the information in that file?

I used Crtl+F for both get and post and only found a few things but i'm pretty sure none of them were applicable they didn't really have anything useful. I'm running php 5.2.3.

I'm really sorry, the new error I got came kotkruton's code... thus

if (isset($_GET['movie_id'])) { 
  $movie_id=$_GET['movie_id'] ;  //here//

is not being set off... is this what you were talking about? I don't know? I hope so.
Anyway what could be wrong with my setup that would be stopping the $GET or $POST from working properly?

thanks for your help

riddellchris

i'm not sure quite what you mean by this

"And see what it says. There's all kinds of useful info on that page, and one of them is things like $GET or $POST or whatever they're called on your machine."

What should I be seeing or doing with the information in that file?

I used Crtl+F for both get and post and only found a few things but i'm pretty sure none of them were applicable they didn't really have anything useful. I'm running php 5.2.3.

I'm really sorry, the new error I got came kotkruton's code... thus

if (isset($_GET['movie_id'])) { 
  $movie_id=$_GET['movie_id'] ;  //here//

is not being set off... is this what you were talking about? I don't know? I hope so.
Anyway what could be wrong with my setup that would be stopping the $GET or $POST from working properly?

thanks for your help

Sxooter

OK, so I took a day or two off this to think it out more clearly. I'm not sure we're communicating real well here. What you need to do is figure if either your web server isn't processing GET / POST data or if you're doing it wrong. Towards that end, let's try a little experiment. Create a file called gettest.php and in it put in JUST this:

<?php
print_r($_GET);
?>

and save it. Then call it, like so:

http://localhost/dir/to/gettest.php?key=value

and see what you get for output.

If that works, then your server is set up to handle $_GET data and pass it on. Which means that elsewhere you're doing it wrong. If your server is misconfigured, and that phpinfo(); thing I was talking about might help. for that create a page called phpinfo.php and put just this in it:

<?php
phpinfo();
?>

and call it like this:

http://localhost/dir/to/phpinfo.html?key=value

and see where you find _GET on that page (or not).

riddellchris

Firstly you are a bloody legend! Thanks so much for all your help, simply amazing!

Ran the little test and I am pretty sure that I got the outcome we wanted...

Array ( [key] => value )

Hence the problem isn't the server! I've been over it a million times and I can't find any errors with my knowledge but clearly i'm not a superstar at this.

tsinka

Hi,

I read the posts several times and didn't find an answer to the question that came to my mind first ...

How does the request look like, how do you call the script exactly. Post the html form or the (exact) url you use to call that script.

There must be another reason since your little test script worked.

There are some additional more or less interesting questions but this one is the first one ...

Thomas

riddellchris

AAAAAAAAAAAAAAAAAAAAH!

that's it! It wasn't getting put up properly from the preceding page into the url and so it wasn't getting read!

Thanks to everyone for all your help!