如何在一个数组的键之间找到差异

Question

这是我的代码。 例如。

<?php 
  $testing="70, 60, 60, 30";
  $test=explode(",",$testing);
  $total=$test[1]+$test[2]+$test[3];
  echo $total-$test[0];
?> 

现在我想要那种类型的函数，它自动添加数组的所有最后一个键，除了该数组的第一个键。最后，数组的所有最后一个键的总和从数组的第一个键中减去。 任何人都知道我是怎么做到的。

Answer 1

如下所示： -

    ExecuteQueryDynamic(@"DECLARE @Discontinued bit={0}; DECLARE @ProductID Int={1}; 
                          DECLARE @CategoryID Int={2}; 
                          SELECT DISTINCT * FROM Products 
                          WHERE Discontinued = @Discontinued AND ProductId = @ProductID 
                          AND CategoryID = @CategoryID;                     
                         ", true, 5, 2).Dump();

输出： - https://eval.in/822897

Answer 2

<?php 
$total=0;
$testing="70, 60, 60, 30";
$test=explode(",",$testing);
for($i=0;$i<count($test);$i++){
if($i==0){}
else{$total +=$test[$i];}
}
echo $total-$test[0];
?> 

Answer 3

有许多方法可以使用单行代码（有时是两行）来实现这样一个简单的目标，它可以与输入数组中的任意数量的值一起使用：

解决方案＃1

// Extract the first value from the array
$first = array_shift($test);
// Subtract the first value from the sum of the rest
echo(array_sum($test) - $first);

解决方案＃2

// Change the sign of the first value in the array (turn addition to subtraction)
$test[0] = -$test[0];
// Add all values; the first one will be subtracted because we  changed its sign
echo(array_sum($test));

解决方案＃3

// Sum all values, subtract the first one twice to compensate
echo(array_sum($test) - 2 * $test[0]));
// There is no second line; that's all

解决方案＃4

// Compute the sum of all but the first value, subtract the first value
echo(array_sum(array_chunk($test, 1)) - $test[0]);
// There is no second line; two function calls are already too much

解决方案＃5

留给读者练习。
有用的阅读：PHP array functions的文档。

Answer 4

希望这会有所帮助。

   <?php 
     $testing="70, 60, 60, 30";
     $test=explode(",",$testing);
     $total=0;
     foreach ($test as $key =>$value) {
       if($key === 0) {
           $total = $total - $value;
       } else  {
          $total = $total + $value;
       }
      }
    echo $total;
    ?>