4
我是一名仍在學習如何編寫代碼的初級程序員。 我有這個數組中我的PHP將得到的數組與數據庫中的數據進行比較PHP
//data inside the array are days name
$days = array();
for($date = $from_date; $date <= $to_date; $date->modify('+1 day')) {
array_push($days,strtolower($date->format('l')));
}
從這個陣列,將有名單已經由用戶選擇天(星期一,星期二,等等)
然後,我有此表在我的數據庫
work_scheme
的work_scheme由表
//field_name => data
Monday => Working Day
Tuesday => Working Day
Wednesday => Working Day
Thursday => Working Day
Friday => Working Day
Saturday => Half Day
Sunday => Off Day
的,這是我的禾該數據從數據庫
$working_days = array();
if(count($work_scheme) > 0){
foreach($work_scheme as $r){
$working_days[0] = array(
"monday" => $r['monday']
);
$working_days[1] = array(
"tuesday" => $r['tuesday']
);
$working_days[2] = array(
"wednesday" => $r['wednesday']
);
$working_days[3] = array(
"thursday" => $r['thursday']
);
$working_days[4] = array(
"friday" => $r['friday']
);
$working_days[5] = array(
"saturday" => $r['saturday']
);
$working_days[6] = array(
"sunday" => $r['sunday']
);
}
}
檢索所以現在我怎麼能在數據庫中,我從用戶的活動所獲得的數組比較我的表rking_days陣列?
我有這個下面的代碼,但它不能正常工作
for($i = 0; $i < count($days); $i++){
for($x = 0; $x < count($working_days); $x++){
$total_days = 0;
if($days[$i] == $working_days[$x]){
echo "hello world";
}
}
}
我注意到$ working_days [$ X]不會回到我一天的名字,而是將返回我要麼工作日,半天或休息日 如何比較從$ days()到$ working_days天的日期名稱?
所以如果我們說選定的日子是星期五,星期六和星期天,我該如何編寫能夠返回我1.5天的代碼?
Working day = 1
Half day = 0.5
Off day = 0