show data only if currentmonth and currentyear are larger than - php

Question

Let say currentmonth is 06 and currentyear 2017
i have dropdown of year and month ..
if i select year 2017 and month of July which is larger then currentmonth
how do i show there is not data available yet
my current code..and its not working..i'm a newbie and not sure im doing this right or not...
$mystartingyear=2017;
$mystartingmonth=01;
$currentmonth=date("m");
$currentyear=date("Y");
if ($currentyear >= $mystartingyear && $currentmonth >= $mystartingmonth)
{
echo "Show Data";
}
else
{
echo "No Data yet";
}
i also tried it like this
$mystartingyear=2017;
$mystartingmonth='01';
$currentmonth=12;
$currentyear=2017;
//$currentmonth=date("m");
//$currentyear=date("Y");
if ($currentyear >= $mystartingyear && $currentmonth >= $mystartingmonth)
{
echo "Show Data";
}
else
{
echo "No Data yet";
}
it always display "show data"

Answer 1

Edit:
Integers from 01 to 07 are ok just as long as you don't do 010 (with 3 integers) since that will be represented as 8. But as soon as you start hitting 08 and 09 for the months of August and September (which may be what's in your unknown dropdown), you will have problems.
So it's best you quote it.
Consult "Footnotes".
Original answer:
The leading zero in 01 for:
$mystartingmonth = 01;
^^
is treated as an octal.
It needs to be quoted:
$mystartingmonth = '01';
Octals references:
http://php.net/manual/en/language.types.integer.php
https://en.wikipedia.org/wiki/Octal
Footnotes:
If your current code is failing you, then something else is failing you.
You mention in your question that you're using this from a dropdown and that code wasn't included, nor how it's being passed.
Use PHP's error reporting, set to catch and display:
http://php.net/manual/en/function.error-reporting.php
Verify your conditional statement's logic also; both conditions must be met.

Answer 2

There is missing semicolon after first echo i.e. code should be
echo "Show Data";
Then every thing should work fine,
The date(); function in php returns the date in string and when you compare a string with integer it is implicitly converted into equivalent integer variable,if you want for sure you can use intvar(); function which converts variable to equivalent integer.But in your case it is not necessary.For more about i recommend you to read php manual .