时间日志监控系统PHP/MySQL

我的计时系统有问题。嗯，在我的系统里你可以看到员工的时间记录(例如。日期，签到，突破，闯入，退房)。我从生物计量仪上传csv文件后，将日期作为获取个人记录的主要来源。现在让我困扰的是，当我意识到有些员工在一个工作日内实行两班倒时，已经太晚了。例如:

在日期04/05/13(月/日/年)，员工在上午1:00入住，上午10:00退房。在同一天的晚上9点，该员工重新登记了他的两班制，因此该员工的退房日期和时间将是04/06/13上午6点。

现在我的系统记录中的输出将基于日期。所以在日期04/06/13中，将没有登记，因为它记录在日期04/05/13中，我在系统中所做的是，如果没有登记，它将不会显示记录的其余部分。就像这样:

| ACNo |   Date    |   CheckIn   |  Breakout  |   Breakin  | Checkout
|      |           |             |            |            |
| 1220 | 4/05/2013 | 01:00:18 AM | 5:46:58 AM | 6:22:41 AM | 9:42:19 AM
| 1220 | 4/05/2013 | 09:00:00 PM |            |            |
| 1220 | 4/06/2013 |             | 1:00:00 AM | 2:00:00 AM | 6:00:00 AM
this 3rd row will not appear in the system because of the condition that if there will be no
check-in it will not show the rest of the record.

上表的延续部分是计算员工的Tardiness, Over-Break, Over-Time和Under time(按行计算)。因此，如果总有一个员工要两班倒，我就无法看到他当天的总记录。我现在要做的是根据入住和退房的情况。顺便说一下，我的数据库表是CheckIn, CheckOut和break。

签到:

| CIn_ACNo |   CIn_Date  |   CIn_Time   | 
|          |             |              | 
|   1220   |  4/05/2013  | 01:00:18 AM  |  
|   1220   |  4/05/2013  | 09:00:00 PM  |

付款:

| COut_ACNo |  COut_Date  |   COut_Time  | 
|           |             |              |  
|   1220    |  4/05/2013  |  09:42:19 AM |
|   1220    |  4/06/2013  |  06:00:00 AM |

，下面是我用来获得结果的代码:

$sql1 = mysql_query("Select * from checkin where CIn_ACNo = '$id' order by CIn_Date asc");
$stop1 = mysql_num_rows($sql1);
if(!$stop1){
$checkin = null;
}
while($row1 = mysql_fetch_array($sql1)) {
$date1 = $row1['CIn_Date'];
$checkin = $row1['CIn_Time'];

//temporary change schedule
$tempsched = mysql_query("SELECT * FROM tempschedchange where TempACNo = '$ID' AND TempDate =   '$date1'");
$stop9 = mysql_num_rows($tempsched);
if(!$stop9){
    $tempin = null; $tempout = null; $tempdate = null;
}
while($row9 = mysql_fetch_array($tempsched)){
    $tempname = $row9['TempName'];
    $tempacno = $row9['TempACNo'];
    $tempdate = $row9['TempDate'];
    $tempin = $row9['TempSchedIn'];
    $tempout = $row9['TempSchedOut'];
}
//break out count
$sql2 = mysql_query("Select * from breaks where break_date = '$date1' AND break_ex = 'Out' AND break_acno = '$id' order by break_date asc");
$stop2 = mysql_num_rows($sql2);
    if(!$stop2){
        $breakoutt = "-";
    }
    while($row2 = mysql_fetch_array($sql2)) {
        $breakoutt = $stop2;
        $breakout = $row2['break_time'];
    }
//break in count
$sql3 = mysql_query("Select * from breaks where break_date = '$date1' AND break_ex = 'Out Back' AND break_acno = '$id' order by break_date asc");
$stop3 = mysql_num_rows($sql3);
    if(!$stop3){
        $breakinn = "-";
        $OBB="-";
    }
while($row3 = mysql_fetch_array($sql3)) {
    $breakinn = $stop3 + $stop2;
    $breakin = $row3['break_time'];
        if($breakinn > 2) {
            if($breakinn % 2){
                $breakinnn = $breakinn - 1;
                $breakinn = $breakinnn / 2;
            }else{
                $breakinn = $breakinn / 2;
            }
        }
        elseif($breakinn <= 2) {
            $breakinn = 1;
        }
}
//checkout
$sql4 = mysql_query("Select * from checkout where COut_Date = '$date1' AND COut_ACNo = '$id'");
$stop4 = mysql_num_rows($sql4);
    if(!$stop4){
        $checkout = "-";
    }
while($row4 = mysql_fetch_array($sql4)) {
    $date4 = $row4['COut_Date'];
    $checkout = $row4['COut_Time'];
}
//BREAKS!!!!!!!!!!!!!!!!!!!important
//break out 1
$sql5 = mysql_query("Select * from breaks where break_date = '$date1' AND break_acno = '$id' AND break_ex = 'Out' order by break_date desc limit 1");
$stop5 = mysql_num_rows($sql5);
    if(!$stop5) {
        $breakout1 = 0;
    }
    while($row5 = mysql_fetch_array($sql5)) {
        $breakout1 = $row5['break_time'];
        if($breakout == $breakout1){
        $breakout1 = 0;
        }
    }
//break in 1
$sql6 = mysql_query("Select * from breaks where break_date = '$date1' AND break_acno = '$id' AND break_ex = 'Out Back' order by break_date desc limit 1");
$stop6 = mysql_num_rows($sql6);
    if(!$stop6) {
        $breakin1 = 0;
    }
while($row6 = mysql_fetch_array($sql6)) {
    $date6 = $row6['break_date'];
    $breakin1 = $row6['break_time'];
    if($breakin == $breakin1){
        $breakin1 = 0;
        }
}
//break out 2
$sql7 = mysql_query("Select * from breaks where break_date = '$date1' AND break_acno = '$id' AND break_ex = 'Out' order by break_date desc limit 2");
$stop7 = mysql_num_rows($sql7);
    if(!$stop7) {
        $breakout2 = 0;
    }
    while($row7 = mysql_fetch_array($sql7)) {
        $breakout2 = $row7['break_time'];
        if($breakout1 == $breakout2){
        $breakout2 = 0;
        }
    }
//break in 2
$sql8 = mysql_query("Select * from breaks where break_date = '$date1' AND break_acno = '$id' AND break_ex = 'Out Back' order by break_date desc limit 2");
$stop8 = mysql_num_rows($sql8);
    if(!$stop8) {
        $breakin2 = 0;
    }
while($row8 = mysql_fetch_array($sql8)) {
    $date6 = $row8['break_date'];
    $breakin2 = $row8['break_time'];
    if($breakin1 == $breakin2){
        $breakin2 = 0;
    }
}

我需要一些帮助，如何创建一个条件，将获得签入和签出的值，而不参考日期。相反，它将基于员工的Check In和Check Out。比如，当有一个签入记录时，它会自动配对到一个签出。我不这么好在MySQL中使用join和什么只是简单的查询，你可以在我的代码中注意到。请问，你知道怎么做吗?请帮帮我，伙计们!非常感谢您阅读这篇很长的解释。任何帮助都非常感激!