thank you. That appears to have solved it.
----- Original Message -----
From: Santiago Zarate
To: Alexander
Sent: Saturday, January 27, 2007 7:26 PM
Subject: Re: I can't solve the error in the code
if(!input) { // an error here... should be !$input or even if(!add_to_table( $_REQUEST['picture'], $_REQUEST['thumbnail'] ))
echo 'Error\!';
}
Also you're calling a function that isn't created yet according with your code, also i dont see
Also in your function add_to_table, has some errors too... like you're NOT calling a db conection from it... and i guess it should... since i have to do it with classes... when it comes to functions... you should too...
for your write_form code... i think youre using your code wrong...
also check here: http://ve.php.net/manual/en/function.print.php
there might be more errors... lñemme do a fast debug here:
santiago@469:~$ nano php.php
santiago@469:~$ php -f php.php
Parse error: syntax error, unexpected $end in /home/santiago/php.php on line 69
that means there's an missing "}" somewhere... if youre on linux... install php cli and use it to debug... (Or use bluefish that is way more easy) and quanta to build your code... if you're on windows... well set your ini to E_ALL, then choose eitther using console for PHP and load the file... or use your browser and run the script until it runs fine without errors... if you're using a shared host... create a error function that logs everything into a file... might save you lots of time...
2007/1/27, Alexander <kurokawa@xxxxxxx>:
I'm writing a script for uploading into a database urls for a picture and its thumbnail, but for some reason when I test it out the page doesn't load, and not even the errors appear. Please help, here's the code:
<html>
<body>
<?php
$user = "userone";
$pass = "password";
$database = "mydatabase";
$link = mysql_connect("localhost", $user, $pass);
if(!$link) {
echo 'Error in linking';
}
$db = mysql_select_db($database, $link);
if(!$db) {
echo 'Error in DB';
}
else {
if ( !empty( $_REQUEST['picture'] ) && !empty( $_REQUEST['thumbnail'] ) ) {
$input = add_to_table( $_REQUEST['picture'], $_REQUEST['thumbnail'] );
if(!input) {
echo 'Error\!';
}
else {
echo 'Data enter successfully\!';
}
}
else {
write_form();
}
function add_to_table( $picture, $thumbnail ) {
$picture = mysql_real_escape_string($picture);
$thumbnail = mysql_real_escape_string($thumbnail);
$insert = 'INSERT INTO domains ( picture, thumbnail ) values( $picture, $thumbnail )';
$indata = mysql_query( $insert, $link );
}
function write_form() {
print <<<EOF
form method='post' action='{$_SERVER['PHP_SELF']}'>
<p>Enter url of large picture: <input type='text' name='picture' />
<p>Enter url of thumbnail: <input type='text' name='thumbnail' />
echo '<p><input type='submit' value='Add to Database' />'
</form>;
FORM;
}
?>
</body>
</html>
