Insert Into - php mysql - php

Question

HTML
<form action="inc/q/prof.php?pID=<?php echo $the_pID; ?>" method="post">
<select id="courseInfoDD" name="courseInfoDD" tabindex="1"><?php while($row3 = $sth3->fetch(PDO::FETCH_ASSOC)) {
echo "<option>".$row3['prefix']." ".$row3['code']."</option>"; }echo "</select>"; ?>
<input type="text" id="addComment" name="addComment" tabindex="3" value="Enter comment" />
<input type="hidden" name="pID" value="<?php echo $the_pID; ?>">
<input type="submit" name="submit" id="submit" />
</form>
PHP
$connect = mysql_connect("##", $username, $password) or die ("Error , check your server connection.");
mysql_select_db("###");
//Get data in local variable
if(!empty($_POST['courseInfoDD']))
$course_info=mysql_real_escape_string($_POST['courseInfoDD']);
if(!empty($_POST['addComment']))
$course_info=mysql_real_escape_string($_POST['addComment']);
if(!empty($_POST['pID']))
$the_pID=mysql_real_escape_string($_POST['pID']);
print_r($_POST);
echo $the_pID;
// check for null values
if (isset($_POST['submit'])) {
$query="INSERT INTO Comment (info, pID, cID) values('$the_comment','$the_pID','$course_info')";
mysql_query($query) or die(mysql_error());
echo "Your message has been received";
}
else if(!isset($_POST['submit'])){echo "No blank entries";}
else{echo "Error!";}
?>
?>
Table
commId int(11)
info text
date timestamp
reported char(1)
degree char(1)
pID int(11)
cID int(11)
It gives me "Error!" now, I try the db credentials and they are fine... ?? And the r_post() is still giving an error of Array()
Why isn't Array() accepting values? Anyone???

Answer 1

Like #user551841 said, you will want to limit your possibility of sql injection with his code.
You are seeing that error because you're code told it to echo that error if nothing was entered, which is the case upon first page load. You shouldn't need that until submit is done.
Edit: Sorry, I was assuming you are directly entering the page which needs the $_POST data without going through the form submit.
You also should do something along the lines of if(!isset($variable)) before trying to assign it to something less your server will spit out error of undefined variables.
if(!empty($_POST['courseInfoDD']))
$course_info=mysql_real_escape_string($_POST['courseInfoDD']);
do that to all of them.
Then you can check
if (!isset($user_submitted) && !isset($the_comment) && !isset($course_info) && !isset($the_pID) ){
echo "All fields must be entered, hit back button and re-enter information";
}
else{
$query="INSERT INTO Comment (info, pID, cID) values('$the_comment','$the_pID','$course_info')";
mysql_query($query) or die(mysql_error());
echo "Your message has been received";
}

Answer 2

Check that the hidden field "pID" has a value set from value=<?php echo $the_pID; ?>

Answer 3

Make sure that your data is valid before checking it.
For instance do
print_r($_POST);
and check if the keys and their data match up.
Also, as a side note, NEVER do what you're doing with :
$query="INSERT INTO Comment (info, pID, cID) values('$the_comment','$the_pID','$course_info')";
This is how mysql injection happens, either use prepared statements or
$course_info= mysql_real_escape_string($_POST['courseInfoDD']);

Answer 4

To answer to your question what is wrong here
you've got a huge gaping SQL-injection hole!!
Change this code
//Get data in local variable
$course_info=$_POST['courseInfoDD'];
$the_comment=$_POST['addComment'];
$the_pID=$_POST['pID'];
To this
//Get data in local variable
$course_info = mysql_real_escape_string($_POST['courseInfoDD']);
$the_comment = mysql_real_escape_string($_POST['addComment']);
$the_pID = mysql_real_escape_string($_POST['pID']);
See: How does the SQL injection from the "Bobby Tables" XKCD comic work?
For more info on SQL-injection.

Answer 5

i would change this line
if (isset($_POST['submit'])) {
to
if ($_POST) {
the sumbit button field will not always be posted, for example if you press return on keyboard instead of clicking on the submit button with the mouse.

Answer 6

Cleaner:
$submit = isset($_POST['submit']) ? true : false;
$comment = isset($_POST['comment']) ? trim($_POST['comment']) : '';
if ($submit && $comment) {
$query = 'INSERT INTO comments (comment) values("' . mysql_real_escape_string($comment) . '")';
//...
}
As you can see I place the escaping inside the query. And this is a good idea because sometimes you loose track of the complete code and this won't happen inside a query.