Question

我正在尝试计算从脚本返回的数组的结果。我有两种情况可以找回选项a

Array ( [Id] => 1779 [SupplierId] => 1809 [SupplierName] => cccccc)

第二个选项ib

Array ( 
[0] => Array ( [Id] => 2020 [SupplierId] => 1809 [SupplierName] => vvv) 
[1] => Array ( [Id] => 2058 [SupplierId] => 1809 [SupplierName] => bbb) 
[2] => Array ( [Id] => 2063 [SupplierId] => 1809 [SupplierName] => xx) 
)

如果我确实相信我回来的情况3 我如何计算选项A中我回到1，而在选项B中我会回来3？

Answer 1

您可以创建如下函数：

function countRes($arr) {
    return is_array(end($arr)) ? count($arr) : 1;
}

$arr1 = array("Id" => 1779, "SupplierId" => 1809, "SupplierName" => "cccccc");

$arr2 = array(array("Id" => 2020, "SupplierId" => 1809, "SupplierName" => "vvv"),
              array("Id" => 2058, "SupplierId" => 1809, "SupplierName" => "bbb"),
              array("Id" => 2063, "SupplierId" => 1809, "SupplierName" => "xx"));

echo countRes($arr1); // 1
echo countRes($arr2); // 3

它检查最后一个元素是否是一个数组，并返回结果数组中的数组数。否则返回1，因为结果数组本身包含数据。

Answer 2

首先，评论。方法应始终返回相同类型的对象。如果您可以修改脚本代码以始终返回数组数组，请执行此操作。

如果你做不到，那么hacky解决方案就是有序的。

你可以使用php的typeof运算符来查找第一个数组中的元素是整数还是更多数组：

http://php.net/manual/es/function.gettype.php

Answer 3

$array1 = array ( 'Id' => 1779, 'SupplierId' => 1809, 'SupplierName' => 'cccccc');

$array2 = array(
        array ( 'Id' => 1779, 'SupplierId' => 1809, 'SupplierName' => 'cccccc'),
        array ( 'Id' => 1779, 'SupplierId' => 1809, 'SupplierName' => 'cccccc'),
        array ( 'Id' => 1779, 'SupplierId' => 1809, 'SupplierName' => 'cccccc')
    );

// sub arrays with key
$array3 = array( 
    'first'=>array ( 'Id' => 1779, 'SupplierId' => 1809, 'SupplierName' => 'cccccc'), 
    'second'=>array ( 'Id' => 1779, 'SupplierId' => 1809, 'SupplierName' => 'cccccc'), 
    'third'=>array ( 'Id' => 1779, 'SupplierId' => 1809, 'SupplierName' => 'cccccc') 
    );


echo arrayCount($array3);


function arrayCount($array){
    foreach($array as $key=>$value){
        if(is_array($value)){
            return count($array);
        }else{
            return 1;
        }
    }
}

php数组计入动态结果

3 个答案: