在 PHP 中循环并存储关联数组中的键值 [关闭]

Question

我正在尝试从我通过 SQL 查询创建的 PHP 关联数组中提取值

SELECT WEEKDAY(Date) AS weekday, 100 * SUM(Att_Type) / COUNT(Att_Type) AS percentage 
FROM Attendance 
WHERE ID=$ID AND WEEK(Date) = WEEK(NOW()) 
GROUP BY WEEKDAY(Date)

它被存储在变量$arr 中，其中包含我使用print_r 命令找到的内容：

Array ( [weekday] => 2 [percentage] => 100.0000 ) 
Array ( [weekday] => 1 [percentage] => 100.0000 ) 
Array ( [weekday] => 0 [percentage] => 66.6667 ) 
Array ( [weekday] => 3 [percentage] => 100.0000 )

我是 PHP 关联数组的新手，我想知道如何创建一个函数来获取工作日键的值，如果它是 0（星期一），则将百分比键的值存储在星期一其余的工作日值可变量等，最多 4 个（星期五）。

编辑：

我正在寻找的输出是将值保存到自变量的位置，因此在调用时它们会清楚地显示值，例如：

$monday = 66.6667
$tuesday = 100.0000
$wednesday = 100.0000
$thursday = 100.0000
$friday = 0 //when there is no value corresponding to that weekday in the array then it will be 0

因此它们可以在稍后的代码中回显。

Answer 1

$data = [['weekday' => 2, 'percentage' => 100.0000,], ['weekday' => 1, 'percentage' => 100.0000,], ['weekday' => 0, 'percentage' => 66.6667,], ['weekday' => 3, 'percentage' => 100.0000,]];    

$days = ['monday', 'tuesday', 'wednesday', 'thursday', 'friday', 'saturday', 'sunday'];

foreach ($days as $day) {
    ${$day} = 0;
}

如果您使用$results = mysqli_fetch_all($result, MYSQLI_ASSOC) 来获取结果数组，请使用此；

$results = mysqli_fetch_all($result, MYSQLI_ASSOC);

foreach($results as $arr) {
    ${$days[$arr['weekday']]} = $arr['percentage'];
    // $arr['weekday'] gets the weekday of the array;
    // $days[$arr['weekday']] gets the day index from the days array
    // ${$days[$arr['weekday']]} then creates a variable with the particular weekday as the variable name
}

或者，如果您使用的是while ($arr = mysqli_fetch_assoc($result))

while ($arr = mysqli_fetch_assoc($result)) {
    ${$days[$arr['weekday']]} = $arr['percentage'];
}