PHP syntax error - using elseif [duplicate] - php

Question

This question already has answers here:
PHP parse/syntax errors; and how to solve them
(20 answers)
Closed 5 years ago.
This is the code that I wrote and it's not working. it's suppose to ask the user a odd number and check is it correct.
Code:
$guess = $_POST['guess'];
$submit = $_POST['submit'];
if(isset($submit)){
if ($guess/2 %0){
echo "the number you are guessing is not odd";
}elseif{
($guess<$rand)
echo "Your Guess is Smaller than the secret number";
}elseif{
($guess>$rand)
echo "Your Guess is bigger than the secret number";
}else{
($guess==$rand)
echo "you guessed currectly. now try anouthe number!";}
else
header("Location: index.php");
exit();}
?>

Answer 1

Can you try this?
You placed the '()' in your elseif wrong.
<?php
$rand = rand(1, 99);
$guess = $_POST['guess'];
$submit = $_POST['submit'];
if(isset($submit))
{
if($guess / 2 % 0)
{
echo "the number you are guessing is not odd";
}
elseif($guess < $rand)
{
echo "Your Guess is Smaller than the secret number";
}
elseif($guess > $rand)
{
echo "Your Guess is bigger than the secret number";
}
elseif($guess == $rand)
{
echo "you guessed currectly. now try anouthe number!";
}
}
else
{
header("Location: index.php");
exit();
}
?>
I have not tested this code, so i need your feedback. Edit: You've confirmed this worked.
I'd like to provide you the manual about elseif:
http://php.net/manual/en/control-structures.elseif.php
Please consider easier/cleaner coding. Personally I like to use ':' instead of '{}', it's less code and easier to read when you use HTML mixed with PHP like:
<?php
$rand = rand(1, 99);
$guess = $_POST['guess'];
$submit = $_POST['submit'];
if(isset($submit)):
if($guess / 2 % 0):
echo "the number you are guessing is not odd";
elseif($guess < $rand):
echo "Your Guess is Smaller than the secret number";
elseif($guess > $rand):
echo "Your Guess is bigger than the secret number";
elseif($guess == $rand):
echo "you guessed currectly. now try anouthe number!";
else:
header("Location: index.php");
exit();
endif;
?>
And don't forget to check the $_POST data.
Same goes for array, but thats a side note:
$arr = array(1 => 'hi', 2 => 'hello'); // old
$arr = [1 => 'hi', 2 => 'hello']; // new

Answer 2

That's not the correct syntax for an if-else construct in php.
The elseif part needs to have a condition right after it (before the opening brace), while else doesn't expect a condition at all.
if ($guess/2 %0){
echo "the number you are guessing is not odd";
} elseif ($guess<$rand) {
// ....
} else {
echo "you guessed currectly. now try anouthe number!";
}
Of course, before your else you have to be sure that the if and elseifs match all the "wrong" cases.