PHP Code:

<?

include("tags.php");

$tname = $_GET['tname'];
$ttype = $_GET['ttype'];
$caption = $_GET['caption'];
$tutorial = $_GET['tutorial'];
$author = $_GET['author'];

if(!$tname) {
exit("You need to input a title for the tutorial");
}
if(!$ttype) {
exit("You need to input a type of tutorial");
}
if(!$caption) {
exit("You need to input a caption");
}
if(!$author) {
exit("You need to input an author for the tutorial!");
} else {

$tname = addslashes($tname);
$caption = addslashes($caption);
$tutorial = addslashes($tutorial);
$tutorial = tags($tutorial);

mysql_connect("localhost", "lalala", "lalalala") or die("Could not connect to DB because: ".mysql_error());
mysql_select_db("lalala") or die("Could not select the DB because: ".mysql_error());

$query = "INSERT INTO tutorials VALUES ('', '".$tname."', '".$ttype."', '".$caption"', '".$tutorial."', '".$author."')";
mysql_query($query) or die("Could not complete query because: ".mysql_error());

?>

I get an error on line 34...

Parse error: parse error, unexpected T_CONSTANT_ENCAPSED_STRING in /home/xingeh/public_html/test/insert.php on line 34

Can anyone help please?

At a first glance, it looks like you missed a full stop at $caption

PHP Code:

$query = "INSERT INTO tutorials VALUES ('', '".$tname."', '".$ttype."', '".$caption"', '".$tutorial."', '".$author."')"; 


Ok great, I missed that one after checking it all... Dam, thanks anyway man. This file is used when the form is submitted to it. The only problem is that the error checking works no matter what, the data isn't get sent to the file, any help on this one?

Form > POST > php

php > process > insert

That's not happening though?

Put a Curlybrace on the line before PHP closes. }, you forgot to close off your if...else statement.

Ahha, this has been sorted now! I had put $_GET instead of $_POST, sorry for the post guys! I started a new topic for it, rather pointless... Thanks for the help though!
Also I did forget to close the IF statement, noticed that too, thanks mate!

OK, another small problem... I get this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/xingeh/public_html/test/viewtutorial.php on line 11

The code is:

PHP Code:

<?

$id = $_GET['id'];

mysql_connect("localhost", "lalala", "lala") or die("Could not connect to DB because: ".mysql_error());
mysql_select_db("lalalalala") or die("Could not select the DB because: ".mysql_error());

$query = "SELECT * FROM tutorials WHERE id = ".$id." LIMIT 1";
mysql_query($query) or die("Could not do query because: ".mysql_error());

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

$id = $row['id'];
$tname = $row['tname'];
$ttype = $row['ttype'];
$tutorial = $row['tutorial'];
$author = $row['author'];
$caption = $row['caption'];

echo("Tutorial $id");
echo("<br>");
echo("$tname");
echo("<br><br>");
echo("$tutorial");

}
mysql_close();
?>

Whats wrong with that?

heh heh, you're trying to fetch this "SELECT * FROM tutorials WHERE id = ".$id." LIMIT 1";
instead of the actual result of the query.


you need to put something like this in

PHP Code:

$result = mysql_query("SELECT * FROM tutorials WHERE id = ".$id." LIMIT 1");
while($row=mysql_fetch_array($result)) { 

ect ect 


But isn't that just the same? Making the variable, carrying out the query, doing the fetch function... Same both ways isn't it?

Hehe, I have sorted this problem now again lol, and another has arise ...

This time it is:

Could not do query because: Unknown column 'PHP' in 'where clause'

PHP Code:

<?

$cat = $_GET['cat'];

mysql_connect("localhost", "lala", "lala") or die("Could not connect to DB because: ".mysql_error());
mysql_select_db("lala") or die("Could not select the DB because: ".mysql_error());

$query = mysql_query("SELECT * FROM tutorials WHERE ttype = ".$cat."")
or die("Could not do query because: ".mysql_error());

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

$id = $row['id'];
$tname = $row['tname'];
$ttype = $row['ttype'];
$tutorial = $row['tutorial'];
$author = $row['author'];
$caption = $row['caption'];

$tname = stripslashes($tname);
$caption = stripslashes($caption);
$tutorial = stripslashes($tutorial);

echo("$ttype");
echo("<br>");
echo("$tname");
echo("<br><br>");
echo("$caption");
echo("<br><br>");

}
mysql_close();
?>

That just means the query returned 0 results. I wouldn't suggest making the script die if the query fails. You should designate a specific output if the query returns 0 results rather than making the entire script die.

OK, I will try that, but the thing is, in the database.. $ttype is PHP for 2 items in the database (because there are only 2, im just testing it), so why does it give an error?