PHP实现两种排课方式

吾爱主题阅读：131 2021-11-19 16:08:00 评论：0

两种排课方式：

固定每周的固定时间上课（例：共上20节，每周六、周日早上8点-10点上课。假如今天周六凌晨1点，那么排课也需要从今天开始）总共上几个周，每周上课时间比较个性化（例：共上三周，第一周周一周二早上8点-10点上课；第二周周三周四下午8点-10点上课；第三周周日中午11点-12点上课。）

第一种排课比较好实现，简要代码如下：

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 /** * 生成日期列表 * * @param int $startDate 开始日期时间戳格式 * @param array $timeList 课时计划列表 [ { "start_at": "09:09", //开课时间 "end_at": "10:09", //结束时间 "week_at": 1 //周几 }, { "start_at": "12:09", //开课时间 "end_at": "13:09", //结束时间 "week_at": 1 //周几 }, { "start_at": "09:09", "end_at": "10:09", "week_at": 5 } ] * @param int $amount 课时计划数量 * @param int $skipHoliday 跳过节假日 * * @return array */ public function generateDateList( $startDate , $timeList , $amount , $skipHoliday = 0) { // 计算开始日期是周几 $startDateWeek = intval ( date ( 'N' , $startDate )); //规范化课时数据 week_at 做key的三维数组 foreach ( $timeList as $item ) { $weekAt = $item [ 'week_at' ]; array_splice ( $item , 0, 0, $weekAt ); $key = array_shift ( $item ); $weeksTime [ $key ][] = $item ; $item = null; } unset( $timeList ); if ( empty ( $weeksTime )) { $this ->addError( '课时计划数据为空' ); return false; } //设置跳过假期,获取开始日期之后的节假日 if ( $skipHoliday ) { $holiday = new Holiday(); $holidayData = $holiday ->getHolidayList( $startDate ); $holiday = null; unset( $holiday ); } $nowTime = time(); $list = array (); for ( $weekStartTime = $startDate , $count = 0; $count < $amount ; $weekStartTime += 86400 * 7) { //$currentWeek :周几 foreach ( $weeksTime as $currentWeek => $weekTime ) { foreach ( $weekTime as $time ) { //算出对应的日期时间戳 $currentDateTime = $weekStartTime + (( $startDateWeek <= $currentWeek ? ( $currentWeek - $startDateWeek ) : (7 - $startDateWeek + $currentWeek )) * 86400); //对应的日期 = 开始时间 + ((开始时间对应周 <= 数据对应的周几？ (数据对应的周几 - 开始时间对应周) ：（7 - 开始时间对应周 + 数据对应的周几）) * 86400) //假期跳过排课 if ( $skipHoliday && ! empty ( $holidayData )) { $startUnix = $currentDateTime + $time [ 'start_at' ] * 3600; //开始时间 $endUnix = $currentDateTime + $time [ 'end_at' ] * 3600; //结束时间 $skip = false; //选择跳过节假日，且节假日与当前课程时间有重叠跳过 foreach ( $holidayData as $item ) { if (( $item [ 'start_at' ] < $endUnix && $item [ 'end_at' ] > $startUnix ) || ( $item [ 'start_at' ] === $startUnix && $item [ 'end_at' ] === $endUnix )) { $skip = true; continue ; } } if ( $skip ) { continue ; } } $currentDate = date ( 'Y/m/d' , $currentDateTime ); $startAt = strtotime ( $currentDate . $time [ 'start_at' ] . ':00' ); $endAt = strtotime ( $currentDate . $time [ 'end_at' ] . ':00' ); if ( $startAt < $nowTime || $endAt < $nowTime ){ $this ->addError( '上课时间不能小于当前时间' ); return false; } $list [] = [ 'date_at' => $currentDateTime , //日期 'week_at' => $currentWeek , //周几 'start_at' => $startAt , 'end_at' => $endAt ]; $count ++; if ( $count >= $amount ) { break 3; } } } } $weeksTime = null; unset( $weeksTime ); return $list ; }

（例子，只用来展示数据结构）假如总共5节课时，从6-25日开始排课，每周一、周六上课：

（例子，只用来展示数据结构）排课结果为：

第二种排课方式稍微麻烦一点，简要代码如下：

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 /** * 生成日期列表 * * @param int $startDate 开始日期时间戳格式 * @param array $taskList 任务列表 [ { "start_at": "09:09", //开始上课时间 "end_at": "10:09", //结束时间 "week_at": 1, //周几 "week_number": 1 //第几周 }, { "start_at": "09:09", "end_at": "10:09", "week_at": 2, "week_number": 1 }, { "start_at": "09:09", "end_at": "10:09", "week_at": 1, "week_number": 3 } ] * * @return array */ public function generateDateList( $startDate , $taskList ) { // 计算开始日期是周几 $startDateWeek = intval ( date ( 'N' , $startDate )); $list = []; $nowTime = time(); $weekSign = $week = 0; foreach ( $taskList as $key => $task ){ if ( $task [ 'week_number' ] > $weekSign && $task [ 'week_number' ] != $week ){ $weekSign = $task [ 'week_number' ] - $week ; } //计算每条数据对应的日期 $key == 0：确定第一周第一节课是在本周还是下一周 if ( $key == 0 || $task [ 'week_number' ] == $week ){ if ( $task [ 'week' ] >= $startDateWeek ){ $task [ 'date_at' ] = $startDate + (( $weekSign - 1) * 7 + ( $task [ 'week' ] - $startDateWeek )) * 86400; } else { $task [ 'date_at' ] = $startDate + (( $weekSign ) * 7 - ( $startDateWeek - $task [ 'week' ])) * 86400; } } else { if ( $task [ 'week' ] > $startDateWeek ){ $task [ 'date_at' ] = $startDate + (( $weekSign ) * 7 + ( $task [ 'week' ] - $startDateWeek )) * 86400; } else { $task [ 'date_at' ] = $startDate + (( $weekSign ) * 7 - ( $startDateWeek - $task [ 'week' ])) * 86400; } } $startDateWeek = intval ( date ( 'N' , $task [ 'date_at' ])); $week = $task [ 'week_number' ]; $startDate = $task [ 'date_at' ]; $dateAt = date ( 'Y/m/d' , $task [ 'date_at' ]); $startAt = strtotime ( $dateAt . '00:00:00' ); if ( $task [ 'start_at' ]){ $startAt = strtotime ( $dateAt . $task [ 'start_at' ] . ':00' ); } $endAt = strtotime ( $dateAt . '23:59:59' ); if ( $task [ 'end_at' ]){ $endAt = strtotime ( $dateAt . $task [ 'end_at' ] . ':00' ); } if ( $startAt < $nowTime || $endAt < $nowTime ){ $this ->addError( '上课时间不能小于当前时间' ); return false; } $task [ 'start_at' ] = $startAt ; $task [ 'end_at' ] = $endAt ; //生成课时数据 $list [] = [ 'date_at' => $task [ 'date_at' ], 'week_at' => $task [ 'week' ], 'start_at' => $startAt , 'end_at' => $endAt ]; } return $list ; }