如何在php中比较三个数字时找到最高组合

There are three numbers in set A {3,4,7} and in set B {2,4,7}.

It is not possible to get the result true because the the first numbers in a and b are not same.

But I need to get the result as true by comparing other two number and leaving the first number.

How is it possible to do it in PHP?

try this it will helps you.. it will display the 1 => for more number of combinations as same and 0 => for more of combination as different . in the below code the more number of combination as different so its returns 0.

<?php
$A = array(2,3,7,5,6);
$B = array(4,3,7,8,9);

$flagTrue = 0;
$flagFalse = 0;
for($i=0; $i < count($A); $i++)
{
    if($A[$i] == $B[$i])
    {
        $flagTrue=$flagTrue+1;
    }
    else 
    {
        $flagFalse=$flagFalse+1;
    }
}

$var_is_greater_than_two = ($flagTrue >= $flagFalse) ? 1 : 0;
echo $var_is_greater_than_two;
?>

<?php

$a = array(3,4,7);
$b = array(2,4,7);

echo $a === $b ? 'TRUE' : 'FALSE';

echo PHP_EOL;

array_shift($a);
array_shift($b);

echo $a === $b ? 'TRUE' : 'FALSE';
?>

Shows:

FALSE TRUE

UPD:

If you need to extract values from strings, then:

$strA = '3,4,7';
$strB = '2,4,7';

$a = explode(',', $strA);
$b = explode(',', $strB);

array_shift($a);
array_shift($b);

echo $a === $b ? 'TRUE' : 'FALSE';

Should work.

<?
$A = array(2,3,7);
$B = array(4,3,7);

$isTrue=1;
for($i=1; $i < count($A); $i++) if($A[$i]!=$B[$i]) $isTrue=0;

echo $isTrue;
?>

EDIT:

If you want to return true if exactly two elements are the same, then the code would be:

$common=0;
for($i=0; $i < count($a); $i++) if($a[$i]==$b[$i]) $common++;
if($common==2) $isTrue=1;

 <?php

  $a = array(3,4,7);
  $b = array(2,4,7);

  for($i=0;$i<3;$i++)
 {
      if($a[$i] > $b[$i]
       echo true;
 }

 ?>

will give true if a is greater.

function compareSets($a, $b) {
    $result = TRUE;
    $diffArray = array_diff($a, $b);
    foreach ($diffArray as $key => $value) {
        if ($key > 0) {
            $result = FALSE;
            break;
        }
    }
    return $result;
}