Folks,
In PHP, anything inside single quotes is a string. Right ?
Example 1:
““
$type = gettype($_GET[‘page_no’]);
if(ISSET($_GET[‘page_no’]) && $type != ‘integer’)
{
echo ‘5a. Page Number is not INT!’;
}
The above code should be checking for the string ‘integer’ in the variable value $type. Right ?
So how come it’s checking for the value’s data type ? I can understand it checking for data type if the string word was inside double quotes.
What is the solution to this mystery ?
QUESTION 2:
Need to check if the value of $type is a number or not.
Look at these 2 examples. On one I use single equal sign while on the other double.
Which one is invalid and why. Answer that. It is important.
And yes, I know the differences between single quotes, double quotes, single equal sign, double equal sign and triple equal sign in PHP.
Should not both these following examples be TRUE where one checks if the $type variable has been assigned or not while the other example compares if the $type variable is INT or not ?
In my test, both seem to be showing that $type holds an INT value. Or did I misread the test result ?
Test both exames in your browser and tell me what you figure.
Example 2a:
““
$type = gettype($_GET[‘page_no’]);
if(ISSET($_GET[‘page_no’]) && $type !== “integer”)
{
echo ‘5b. Page Number is not INT!’;
}
Example 2b:
““
$type = gettype($_GET[‘page_no’]);
if(ISSET($_GET[‘page_no’]) && $type != “integer”)
{
echo ‘5c. Page Number is not INT!’;
}
QUESTiON 3:
Back to single quotes again.
Can you tell me what PHP is doing on the first example and what on the second ?
Example 3a:
““
$type = gettype($_GET[‘page_no’]);
if(ISSET($_GET[‘page_no’]) && $type !== “integer”)
{
echo ‘5b. Page Number is not INT!’;
}
Example 3b:
““
$type = gettype($_GET[‘page_no’]);
if(ISSET($_GET[‘page_no’]) && $type != “integer”)
{
echo ‘5c. Page Number is not INT!’;
Kindly provide your answers matching my questions numbers. Will be easy for me to understand your answers.
Thanks