Php 将两个日期之间的所有日期插入具有相反数据的Mysql数据库

我试图在两个日期之间将行插入数据库

这是我的桌子

用户将填写一份简短表格，说明开始和结束日期以及第一周的值（0或1）

$WorkDays

为0或1（0为周一、周二、周三，1为周四、周五、周六）

我可以使用以下命令插入第一周罚款：

$mon=$Start;              
$tue=date('Y-m-d', strtotime($Start. ' + 1 days'));
$wed=date('Y-m-d', strtotime($Start. ' + 2 days'));
$thu=date('Y-m-d', strtotime($Start. ' + 3 days'));
$fri=date('Y-m-d', strtotime($Start. ' + 4 days'));
$sat=date('Y-m-d', strtotime($Start. ' + 5 days'));

if($WorkDays==1){$monWork=1; $tueWork=1; $wedWork=1; $thuWork=0; $friWork=0; $satWork=0; }
if($WorkDays==2){$monWork=0; $tueWork=0; $wedWork=0; $thuWork=1; $friWork=1; $satWork=1; }      

然后我每天都会插入查询mon-sat

然后，我需要相同的脚本来查看开始和结束之间的所有日期，并将其插入到相同的数据库中，但每天必须具有最后一周的oposite值

例如

第一周：周一=1，周二=1，周三=1，周四=0，周五=0，周六=0
第二周：周一=0，周二=0，周三=0，周四=1，周五=1，周六=1

以下是我尝试过的：

//Get the date 1 after the start date as first week already inserted
   $SevenDays=date('Y-m-d', strtotime($Start. ' + 7 days'));  
    $startTime = strtotime( $SevenDays);
    $endTime = strtotime( $End );

//Loop through each day between the start and end date
    for ( $i = $startTime; $i <= $endTime; $i = $i + 86400 ) { 
    $given_date =  date( 'Y-m-d', $i );
    $given_date_day =  date( 'l', $i );

//Select the same day the week before
    $result1512=mysql_query("SELECT * FROM RosteredPatternHours where PatternID='$PatternID' and WorkingDate='$SevenDaysGiven'")or die('Error 84' . mysql_error());
        for ($ii = 1; $ii <= mysql_num_rows($result1512); $ii++) {
        $row = mysql_fetch_array($result1512);  
        $Day = $row ['Day'];
        $Working = $row ['Working'];

//Set Working to opposite of last week
        if($Working==1){$WorkingDay=0;}
        if($Working==0){$WorkingDay=1;}

//获取开始日期后的日期1，因为已插入第一周
$SevenDays=日期（'Y-m-d'，strottime（'Start.'+7天'）；
$startTime=strotime（七天）；
$endTime=strottime（$End）；
//在开始日期和结束日期之间循环每天
对于（$i=$startTime；$i）
//Get the date 1 after the start date as first week already inserted
   $SevenDays=date('Y-m-d', strtotime($Start. ' + 7 days'));  
    $startTime = strtotime( $SevenDays);
    $endTime = strtotime( $End );

//Loop through each day between the start and end date
    for ( $i = $startTime; $i <= $endTime; $i = $i + 86400 ) { 
    $given_date =  date( 'Y-m-d', $i );
    $given_date_day =  date( 'l', $i );

//Select the same day the week before
    $result1512=mysql_query("SELECT * FROM RosteredPatternHours where PatternID='$PatternID' and WorkingDate='$SevenDaysGiven'")or die('Error 84' . mysql_error());
        for ($ii = 1; $ii <= mysql_num_rows($result1512); $ii++) {
        $row = mysql_fetch_array($result1512);  
        $Day = $row ['Day'];
        $Working = $row ['Working'];

//Set Working to opposite of last week
        if($Working==1){$WorkingDay=0;}
        if($Working==0){$WorkingDay=1;}