[$Name data not making into database]

Let's see your HTML form source code. I'm guessing you forgot to change the original form to name='editor_name'.

 

ps: and you don't need those session lines about editor_name

[$Name data not making into database]

are you getting the "The answer was submitted successfully" message?

 

add echo $sql and post the command here

[Form values into an indexed array]

No, you will just echo the same thing $n times. Take a look at this

 

<?php
// display the number of input rows for invites. 
for ( $c=0, $c<$n, $c++) {
echo "


\n";
} 
?>

 

View your html source to see the generated code.

You will have name='whole_name[0]'

Also, I removed id= because unless you're doing stuff with client side java script you don't need it, and if you do it must be a unique value (to be valid html) so you'd have to append $c to the id value

[PHP Question - Join function?]

take a look at url_encode

 

$string = '&fn='. url_encode($firstname). '&ln='. url_encode($lastname);

[file uploading on mac]

Do you get values displayed for Sent: Sixe: and Type:

 

also see the use of basename and move_uploaded_file

 

if (move_uploaded_file( $_FILES['file']['tmp_name'],  "/usr/local/zend/apache2/htdocs/" . basename($_FILES['file']['name']) ) 
   echo "File is valid, and was successfully uploaded.\n";
else 
   echo "File upload failed!\n";

[Form values into an indexed array]

close, just added brackets [] is used for checkboxes, because you have text fields you shoud include an index inside the brackets.

 

//assuming you have a for loop like
for ($row=0; $row<5; $row++)
{
 echo "... name=\"whole_name[$row]\" ...";
}

// fyi html also likes single quotes, this just looks cleaner

 echo "... name='whole_name[$row]' ...";

// it will return

$_POST['whole_name'][0]

[$Name data not making into database]

my guess is hat name is a reserved word. you can either rename it to something like $student_name or try changing your sql statement to enclose field names in accents

 

$sql="INSERT INTO Responses (`name`,`Answer1`,`Answer2`,`Answer3` ...

[strtoupper question]

einsteinsboi is close, $name would be better, but $new_name is probably correct.

 

if ($new_name != "Jimmy Two Guns") {

$this->name = strtoupper($new_name);

}

[PHP OOP Inserting records into db]

I think it's because you don't have quotes around the data fields in the sql statement. Using ',' may seem like you have the quotes, but that only creates a comma, you have to have "','"

 

function newSchool()
{
  $sql_command = "INSERT INTO schools (school, country, city, state, zip) VALUES ('" . $this->school . "','" . $this->country . "','" . $this->city . "','" . $this->state . "','" . $this->zip . "')";
// by placing this in a variable you can test with
 echo $sql_command;

 $sql = mysql_query ($sql_command);
 return $sql; 
}