Simple number check problem

[apotd]

Hi,

 

I'm totally new to PHP and I'm afraid I don't understand a lot yet. However, I need to make this: when you fill in a number in a form, it should be displayed whether the number is odd or even. I tried the following code, which is obviously not working and I have no idea on how to fix this.

 

<?php 
$var = $_POST['var'];

for ($i = 0; $i < $var; $i++) {
  if ($i % 2 == 0) {
    print("even");
  } else {
   print("odd");
  }
}
?>

[Pikachu2000]

What isn't working about it? It seems fine to me.

[apotd]

For example, I fill in the number 30. The result I get is: "evenoddevenoddevenoddevenoddevenoddevenoddevenoddevenoddevenoddevenoddevenoddevenoddevenoddevenoddevenodd".

[Pikachu2000]

Well, yeah. That's what it's written to do. What output are you trying to get from it?

[apotd]

I just want it to say whether the number is odd or even, nothing more. What did I do to make it write all this stuff?

[litebearer]

Your script says check each number for zero til it reaches your number and print even/odd for each number.

 

no need for the for loop - just do the number its-self

[apotd]

My bad, of course. I used the following code instead, which works:

 

<?php 
$number = $_POST['number'];

if($number % 2) {
print("odd");
} else {
    print("even");
}

?>

 

The thing is, I need to perform a for loop on the result (I won't be actually printing it in the end), so I was sort of confused by that. One more thing, is it possible just to put the for loop at the place where the "print" is now situated, or am I thinking totally wrong here?

[litebearer]

Not clear as to what you are asking. Can you provide an example

[apotd]

I'm sorry. I know I sound confusing. Fact is I am confused. But, I found out I had to use a while loop. Tried completing my code but I'm already having problems on the first part. I just want to do a calculation on the number if it is odd, and then make it print the outcome (end1) as long as the number is bigger than one. However, I get an error on the "$number / 2 = $end1;" line.

 

<?php 
$number = $_POST['number'];

if($number % 2) {
while ($number > 1){	
	$number / 2 = $end1;
	print($end1);
}
} else {
...
}

?>

[litebearer]

try changing this...

$number / 2 = $end1;

to this

$end1 = $number / 2;

 

[litebearer]

In fact, get rid of the while

 

if(($number % 2 == 0) && ($number>1)) {
	$end1 = $number /2;
	echo $end1;
}else{
echo "odd";
}

[apotd]

When I try that (your second post), it just divides the number in two. I need the script to keep dividing it in two till the number equals one. That's why I thought I should use a while loop to keep it going till that point. I also want it to print every step. But I don't see how I can do this without a loop. Is it wrong to use "$end1 = $number / 2; " as you first said? Because this actually removed the error, but it just prints an awful lot of 0's onto my page.

[litebearer]

you while loop is endless. try...

 

while ($number > 1){	
echo $number . "<br />";
$number = $number / 2;
}

[apotd]

Works like a charm, thank you!

 

One more thing. I cannot find any serious information on an else statement with a while loop, so I guess this is not possible. However, for an iteration of the code I wanted to create something like this:

 

while ($number % 2){	
   $number / 2;
- i know this is wrong but - } else {
   $number / 4;
    }

 

So that, as long as the number is even, it will divide itself by 2, otherwise by 4, and keeps doing this (so I was again thinking of the while loop), but I have no clue on how to do this to be honest.

[litebearer]

while loops do not use ELSE in and of themselves - Also you must be careful that you allow some event to occur during a while loop that will eventually make the loop end (or else you will have it looping endlessly).

 

 

[apotd]

Would there be some workaround solution for this? For example another if statement within the while loop? Previously you said I should use this

while ($number > 1){	
echo $number . "<br />";
$number = $number / 2;
} 

for the particular code part. But I don't really see what made the loop end, since all that was added was the . "<br />, if I am not mistaken.

[MMDE]

<?php
if(is_numeric($_POST['var'])){
if($_POST['var']%2==0){
	echo 'even';
}else{
	echo 'odd';
}
}
?>

 

There are probably far more effective ways, but I'm not exactly sure how % does it, like looking at the number in binary numbers and see if it ends with 0 or 1, perhaps using one of the bitwise operators.

http://php.net/manual/en/language.operators.bitwise.php

You could also compare the number divided by 2 and the number divided by 2 and rounded to nearest whole number.

[litebearer]

But I don't really see what made the loop end, since all that was added was the . "<br />, if I am not mistaken.

while ($number > 1){	
echo $number . "<br />";
$number = $number / 2;
}

 

the while loop says "Do this as long as $number is greater than 1".

inside the while loop, we say "replace the value of $number with whatever the value of $number/2 is".

ie.

pretend $number has an initial value of 30.

first time thru the loop $number is 30 - we echo it - then we replace 30 with 15

second time thru we echo 15 - then replace 15 with 7.5

third time - we echo 7.5 - then replace 7.5 with 3.75

fourth time - echo 3.75 - then replace 3.75 with 1.875

fifth time - echo 1.875 - then replace 1.875 with .9375

Since $number is now .9375, it is less then 1 (the condition we set in our WHILE) so the while loop ends

 

Clear?