FRInteractives:
well done jregz but you did'nt get question 4 correctly.

FRInteractives:
well done jregz but you did'nt get question 4 correctly.

FRInteractives:
Question 1

Q: What is wrong with this query:
"SELECT *
FROM table WHERE id = $_POST
[ 'id' ]" ?

spikesC:


Question 1 is also wrong.



You cannot have the single quotes of an array index when the array is in a double quote. It is a syntax error.

Therefore,

"SELECT *
FROM table WHERE id = $_POST
[ id ]"


Pls post reasonable questions, this is really too basic

FRInteractives:
Question 2

Q: What is the preferred way to write this if
statement, and why?

if( 5 == $someVar ) or
if( $someVar == 5 )

Due to right-to-left associativity of the operator being used, if($someVar == 5) is preferred.

Craigston:

FRInteractives:
Question 5 (OOP)
Q: What is the difference between p u b l i c, p r o t e c t e d and p r i v a t e in a class definition?

I can't put these in exactly correct statements now but I know declaring an object property or a method as public (default) grants outside code access to it and allows sub-classes to inherit it; declaring it as protected denies outside code access to it but allows sub-classes to inherit it and declaring it as private grants only the class methods access to it: outside code cannot access it and sub-classes cannot inherit it

FRInteractives:
Question 4
Q: Consider these lines of codes:
//first code
$myVar = "I am Something";
$myVar .= "So Special";
echo "Do you Know $myVar";
And
//second code
$myVar = "I am Something";
$myVar .= "So Special";
echo 'Do you Know $myVar';
Now if you run these two codes in a Browser what will be the output?

The first code, by Variable Substitution, will output the string 'Do you know I am something so special'; the second code outputs a string literal (using apostrophes instead of quotes) so the output will be 'do you know $myVar'.

FRInteractives:
Question 3
Q: Given this code:
function doSomething( &$arg )
{
$return = $arg;
$arg += 1;
return $return;
}
$a = 3;
$b = doSomething( $a );
...what is the value of $a and $b after the
function call and why?

Since the parameter '$arg' is passed by reference, the function accesses its value directly. The expressions in the function will operate directly on its value without making any copy. Therefore, the function 'doSomething' will return 4 and will also change the value of $a to 4.

Craigston:

FRInteractives:
You Nailed it Bro. Keep it up. I'll be posting some more difficult questions soon

FRInteractives:
Question 1

Q: What is wrong with this query:
"SELECT *
FROM table WHERE id = $_POST
[ 'id' ]" ?

Craigston:



Since the parameter '$arg' is passed by reference, the function accesses its value directly. The expressions in the function will operate directly on its value without making any copy. Therefore, the function 'doSomething' will return 4 and will also change the value of $a to 4.

Craigston:
That's great. I'm a beginner so the basics are still fresh in my skull. I hope the harder trivia will spur me on to learn fast. I just finished the basics of objects and arrays, and I'm now learning SQL (MySQL). After that, I'll integrate it with PHP in a small project and learn JS too. Holla.