Question

我正在尝试编写一个脚本来从工作中使用的数据库中提取计划数据。

员工的工作时间每天都存储为像这样的字符串

000000000000000000000000000000001111111111111111111111111111111111111111000000000000000000000000

每个角色代表15分钟。第一个角色空间是凌晨12:00，第二个是凌晨12:15，依旧......

员工上面的例子工作时间为8:00 - 6:00。

我为每个角色位置创建了一个这样的数组。

$time[0] = "12:00";
$time[1] = "12:15";
$time[2] = "12:30";
$time[3] = "12:45";
$time[4] = "1:00";
$time[5] = "1:15";
$time[6] = "1:30";
$time[7] = "1:45";
$time[8] = "2:00";

我可以像这样显示员工的时间

echo $time[strpos($string, '1')] . "-" . $time[strpos($string, '0', strpos($string, '1'))];

但如果某人有分班，我无法弄清楚如何使这项工作，例如9:30 - 2:00 / 4:00 - 7:00

000000000000000000000000000000000000001111111111111111110000000011111111111110000000000000000000

对不起，如果我的英语很差。

由于

Answer 1

如果您需要像这样处理拆分班次，您可以在一般情况下解决它，迭代每个字符。

这样的事情：

<?PHP
$date = '2010-09-24'; // or whatever.

$string = '00000000001111100000001111...';

//convert your string in to an array
$data = explode('',$string); 

$time = strtotime("$date 00:00:00"); //midnight on whatever day $date is

$workstate = 0; // zero means not working, 1 means working.
foreach($data as $index=>$value){
    // employee started working
    if ($value && ! $workstate){ 
        $state = 1;
        echo date('Y-m-d H:i:s',$time) .'-';
    }
    // employee stopped working
    if ($workstate && ! $value){
        $state = 0;
        echo date('Y-m-d H:i:s',$time) . PHP_EOL;
    }
    $time = strtotime('+15 minutes', $time); increase time by 15 minutes
}

以上代码未经过测试或其他任何内容，但至少应该说明算法。

它应该生成员工工作的次数列表，每行一班。

编辑：您可能需要处理边缘情况，例如员工是在23：45-24：00工作

Answer 2

谢谢，有几个修复，它完美无缺！

<?php

$date = '10/01/2010';

$string = '000000000000000000000000000000000000001111111111111111110000000011111111111110000000000000000000';

//convert your string in to an array
$data = str_split($string); 

$time = strtotime("$date 00:00:00"); //midnight on whatever day $date is

$workstate = 0; // zero means not working, 1 means working.
foreach($data as $index=>$value){
    // employee started working
    if ($value && !$workstate){ 
        $workstate = 1;
        echo date('h:i:s',$time) .'-';
    }
    // employee stopped working
    if ($workstate && !$value){
        $workstate = 0;
        echo date('h:i:s',$time) . PHP_EOL;
    }
    $time = strtotime('+15 minutes', $time); #increase time by 15 minutes
}

?>

Answer 3

这适用于我的两个示例字符串：

date_default_timezone_set('UTC');

$shift_strings = array(
    '000000000000000000000000000000001111111111111111111111111111111111111111000000000000000000000000',
    '000000000000000000000000000000000000001111111111111111110000000011111111111110000000000000000000'
);

$initial_time = 0; // 12AM
$now = 0; // counter
$increment = 15 * 60; // each digit = 15 minutes, but store as seconds
$offset = 0;
foreach ($shift_strings as $string) {
    echo $string, "\n";

    $shifts = preg_split('/(0+)/', $string, -1, PREG_SPLIT_DELIM_CAPTURE |PREG_SPLIT_NO_EMPTY);

    foreach($shifts as $shift) {
        $start = strftime('%I:%M%P', $now * $increment);
        $now += strlen($shift);
        $end = strftime('%I:%M%P', $now * $increment);
        switch(substr($shift, 0, 1)) {
            case '0':
                echo "Offshift $start - $end\n";
                break;
            case '1':
                echo "Onshift  $start - $end\n";
                break;
            default:
                echo "Someone dun goofed, digit is ", $substr($shifts, 0, 1), "?\n";
        }
    }
    echo "\n";
}

输出是：

000000000000000000000000000000001111111111111111111111111111111111111111000000000000000000000000
Offshift 12:00am - 08:00am
Onshift  08:00am - 06:00pm
Offshift 06:00pm - 12:00am

000000000000000000000000000000000000001111111111111111110000000011111111111110000000000000000000
Offshift 12:00am - 09:30am
Onshift  09:30am - 02:00pm
Offshift 02:00pm - 04:00pm
Onshift  04:00pm - 07:15pm
Offshift 07:15pm - 12:00am

PHP - 需要一些字符串函数的帮助

3 个答案: