我试图用Laravel创建一个简单的考勤应用程序。我对逻辑方面有点困惑。 我想要做的是将用户的时间日志与正确的时间表相匹配。
因此,在我们的办公室,我们可以有2到3次休息,所以典型的工作班次将如下所示:
Time In - 06:00 AM
Break Out 1 - 07:15 AM
Break In 1 - 07:30 AM
Break Out 2 - 11:30 AM
Break In 2 - 12:00 PM
Time Out - 03:00 PM
现在,用户通常只使用生物识别设备并在"中选择"或" out"。
我的问题是这样的:
05:58 AM, in
07:17 AM, out
07:31 AM, in
05:58 AM, out
系统如何判断哪个时间日志(07:17 AM)属于哪个时段(Break Out 1)?
非常感谢您的帮助,如果我的问题不清楚,我很抱歉。
这是我到目前为止所做的示例代码。它有效,但我认为它真的很混乱:
$day_shift = [
"time_in" => date('H:i:s', strtotime("06:00 AM")),
"break_out_1" => date('H:i:s', strtotime("07:15 AM")),
"break_in_1" => date('H:i:s', strtotime("07:30 AM")),
"break_out_2" => date('H:i:s', strtotime("11:30 AM")),
"break_in_2" => date('H:i:s', strtotime("12:00 PM")),
"break_out_3" => '',
"break_in_3" => '',
"time_out" => date('H:i:s', strtotime("03:00 pM")),
];
foreach ($day_shift as $key => $userScheduleTime) {
if ($userScheduleTime !== "") {
$time = $logDate->date." ".$userScheduleTime;
} else {
$time = "";
}
$schedules[] = array('time' => $time, 'type' => $types[$i%2],'name'=>$key);
$i++;
}
// logTimes is a collection of time logs
foreach ($logTimes as $logTime) {
$logs[] = $logTime->toArray();
}
$cloneLogs = $logs;
$lastlog = end($cloneLogs);
if ($lastlog["log_type"] == "login") {
$dayOut = "";
} else {
$dayOut = $lastlog["log_time"];
}
$lastout = null;
$size = count($logs);
do {
if ($logs[$size-1]['log_type'] == 'logout') {
$lastout = $logs[$size-1];
break;
}
$size--;
} while ($size > 0);
$lastlog = null;
for ($i = 0; $i < count($schedules); $i++) {
$logTime = current($logs);
if ($lastlog == $logTime['log_type']) {
next($logs);
}
// checks the first log, calculates how late the person is
if ($i == 0) {
if ($logTime['log_type'] == "login") {
$timein = $schedules[0]['time'];
if (strtotime(date("Y-m-d H:i", strtotime($logTime['log_time']))) >
strtotime(date("Y-m-d H:i", strtotime($timein)))) {
$lates[$logTime['log_time']] = true;
$lates["totallate"] += getDifferenceInMinutes($logTime['log_time'], $timein);
}
$lastlog = $logTime["log_type"];
next($logs);
}
}
if ($schedules[$i]['type']==$logTime['log_type'] && $i!=0 && $lastlog !== $logTime["log_type"]) {
$nextSched = isset($schedules[$i+1])?$schedules[$i+1]:$schedules[$i];
$j = 1;
while ($nextSched['time']=="" && ($i+$j)<=count($schedules)) {
$nextSched = $schedules[$i+$j];
$j++;
}
if (strtotime($nextSched['time'])>strtotime($logTime['log_time']) && $logTime["log_time"] != $dayOut) {
// checks and calculates if the user has undertime
if (strtotime($nextSched['time']) > strtotime($logTime['log_time']) && $nextSched['type']=='login') {
if (strtotime($logTime['log_time']) < strtotime($schedules[$i]['time'])) {
$lates[$logTime['log_time']] = true;
$lates["totalunder"] += getDifferenceInMinutes($logTime['log_time'], $schedules[$i]['time']);
}
}
// checks and calculates if the user has overbreaks
if (date('H:i', strtotime($schedules[$i]['time'])) <
date('H:i', strtotime($logTime['log_time'])) &&
$logTime['log_type'] == 'login') {
$lates[$logTime['log_time']] = true;
$lates["totalover"] += getDifferenceInMinutes($schedules[$i]['time'], $logTime['log_time']);
}
$lastlog = $logTime["log_type"];
next($logs);
}
if ($i+1==count($schedules)) {
next($logs);
}
if ($logTime["log_time"] == $dayOut && $dayOut !== null) {
$timeOut = $schedules[count($schedules)-1]["time"];
if (strtotime(date("Y-m-d H:i", strtotime($logTime["log_time"]))) <
strtotime(date("Y-m-d H:i", strtotime($timeOut)))) {
$lates[$logTime["log_time"]] = true;
$lates["totalunder"] += getDifferenceInMinutes($logTime['log_time'], $timeOut);
$lastlog = $logTime["log_type"];
}
break;
}
}
}