PHP从一个函数返回两个值答案

【问题标题】：PHP return two values from a functionPHP从一个函数返回两个值
【发布时间】：2020-03-19 21:39:34
【问题描述】：

我在控制器中调用了这样的函数：

$all_calendars = Appointment::organizeAppointment($daysView['currentDay']['appointments'], $openingHours);

经过长时间的计算，包括几个查询，它返回一个像这样的集合：

return $appointments;

$appointments 的内容如下所示：

appointments:1 [▼
  "Employee1" => array: [▼
    "08:00" => ""
    "08:15" => ""
    "08:30" => ""
    "08:45" => ""
    "09:00" => Appointment {#11949 ▶}
    "09:15" => "Ja"
    "09:30" => "Ja"
    "09:45" => "Ja"
    "10:00" => Appointment {#11950 ▶}
    "10:15" => "Ja"
    "10:30" => "Ja"
    "10:45" => "Ja"
    "11:00" => ""
    "11:15" => ""
    "11:30" => ""
    "11:45" => ""
    "12:00" => Appointment {#11952 ▶}
    "12:15" => "Ja"
    "12:30" => "Ja"
    "12:45" => "Ja"
    "13:00" => Appointment {#11953 ▶}
    "13:15" => "Ja"
    "13:30" => "Ja"
    "13:45" => "Ja"
    "14:00" => Appointment {#11954 ▶}
    "14:15" => "Ja"
    "14:30" => "Ja"
    "14:45" => "Ja"
    "15:00" => Appointment {#11955 ▶}
    "15:15" => "Ja"
    "15:30" => "Ja"
    "15:45" => "Ja"
    "16:00" => Appointment {#11956 ▶}
    "16:15" => "Ja"
    "16:30" => "Ja"
    "16:45" => "Ja"
    "17:00" => ""
  ],
    "Employee2" => array: [],
    "Employee3" => array: [],
    "Employee4" => array: [],
    ...
]

现在我想知道每个员工有多少约会，这只是进一步添加所有约会的一步：

$totalPerEmployee[$name] = count(array_filter($all_calendars));

在上面显示的情况下，employee1 将有 7 个。

但我不想调用同一个进程，而且我不能将此值添加到每个员工子集合的末尾，因为它会破坏前端的“foreach”。

所以我决定创建这个值并将其存储在第二个数组中，如下所示：

totals:1 [▼
  "Employee1" => array: [▼
    "total" => 7
  ],
    "Employee2" => array: [
    "total = 4
],
    "Employee3" => array: [],
    "Employee4" => array: [],
    ...
]

我的问题是我现在如何从一个函数返回两个值？

我在 PHP 中这样说：

return ($appointments and $totals);  ///<--- of course this is not valid PHP

只需一次通话。

【问题讨论】：

标签： php function return

【解决方案1】：

返回一个包含 2 个项目的数组。返回的原始数据和新数据

return (['appointments' => $appointments, 'totals' => $totals]);

【讨论】：

我确信它会非常简单！谢谢。

【解决方案2】：

您也可以使用 stdClass 返回一个对象。

$returnObj = new stdClass(); 
$returnObj->appointments = $appointments;
$returnObj->totals= $totals;

return $returnObj;

【讨论】：