合并PHP中的开放时间

时间：2015-04-14 08:54:50

标签： php arrays

我正在从我的数据库中循环一些开放时间，并且我已经能够将时间戳合并为基于一周中7天的数组。例如

array(7) {
  [1]=>
  string(28) "12:00 - 14:00, 16:00 - 23:30"
  [2]=>
  string(13) "12:00 - 23:30"
  [3]=>
  string(13) "12:00 - 23:30"
  [4]=>
  string(13) "12:00 - 23:30"
  [5]=>
  string(13) "12:00 - 23:30"
  [6]=>
  string(13) "12:00 - 23:30"
  [7]=>
  string(13) "12:00 - 23:30"
}

我想知道如何将上面的数组合并到这样的数组中：

array(2) {
  ['Mon']=>
  string(28) "12:00 - 14:00, 16:00 - 23:30"
  ['Tue - Sun']=>
  string(13) "12:00 - 23:30"
}

我的冲突是我有时会有这样的开放时间：

array(7) {
  [1]=>
  string(13) "12:00 - 23:30"
  [2]=>
  string(28) "12:00 - 14:00, 16:00 - 20:00"
  [3]=>
  string(28) "12:00 - 14:00, 16:00 - 20:00"
  [4]=>
  string(13) "12:00 - 23:30"
  [5]=>
  string(13) "12:00 - 23:30"
  [6]=>
  string(13) "12:00 - 23:30"
  [7]=>
  string(1) "-" // Means closed
}

结果应该是这样的：

array(2) {
  ['Mon']=>
  string(13) "12:00 - 23:30"
  ['Tue & Wed']=>
  string(28) "12:00 - 14:00, 16:00 - 20:00"
  ['Thu - Sat']=>
  string(13) "12:00 - 23:30"
  ['Sun']=>
  string(6) "Closed"
}

到目前为止，这是我的功能：

function mergeOpeningHours($arr) {

    $days = array();

    $day_name = array('Mon', 'Tue', 'Wed', 'Thu', 'Fri', 'Sat', 'Sun');

    // Firstly we merge the days timestamps
    foreach ($arr as $day => $data) {

        if (isset($days[$data['day']])) {
            // then the day has already been defined and we procedd by applying the days following timestamps
            $days[$data['day']] .= ', ' . $data['open'] . ' - ' . $data['close'];
        } else {
            $days[$data['day']] = $data['open'] . ' - ' . $data['close'];
        }

    }

    $final_days = array();

    $d = 1;
    $dd = $days[$d];

    for ($i=1; $i < 7; $i++) { 

        if ($days[$d+1] == $dd) {

            $final_days[$day_name[$d] . ' - ' . $day_name[$d + 1]] = $dd;

        } else {

            $dd = $days[$i+1];

        }

        $d++;

    }


}

出现了这样的事情：

array(5) {
  ["Wed - Thu"]=>
  string(13) "12:00 - 23:30"
  ["Thu - Fri"]=>
  string(13) "12:00 - 23:30"
  ["Fri - Sat"]=>
  string(13) "12:00 - 23:30"
  ["Sat - Sun"]=>
  string(13) "12:00 - 23:30"
  ["Sun - "]=>
  string(13) "12:00 - 23:30"
}

非常感谢任何建议 - 谢谢！

**解决方案**

感谢Stackoverflow社区的大量帮助，我设法将所需的功能集合在一起。

function mergeOpeningHours($arr) {

    $days = array();

    // Firstly we merge the days timestamps
    foreach ($arr as $day => $data) {

        if (isset($days[$data['day']])) {
            // then the day has already been defined and we procedd by applying the days following timestamps
            $days[$data['day']] .= ', ' . $data['open'] . ' - ' . $data['close'];
        } else if (empty($data['open']) || empty($data['close']) || $data['open'] == NULL || $data['close'] == NULL) {

            $days[$data['day']] = '-';

        } else {
            $days[$data['day']] = $data['open'] . ' - ' . $data['close'];
        }

    }

    $week = array (1 => 'Mon', 2 => 'Tue', 3 => 'Wed', 4 => 'Thu', 5 => 'Fri', 6 => 'Sat', 7 => 'Sun');
    $newarray = $finalarray = $valArray = array();

    $i = $e = 0;
    foreach ($days as $key => $val) {
        if (isset($newarray[$e][$val]) && count($newarray[$e][$val]) > 0 && ((int)$key - 1 == end($newarray[$e][$val]))) {
            $newarray[$e][$val][$i] = $key;
            $i++;
        } else {

            $e++;
            $i = 0;
            $newarray[$e][$val][$i] = $key;
            $i++;

        }
    }

    foreach ($newarray as $k => $v) {
        foreach ($v as $nk => $nv) {
            $cnt = count($nv);
            if ($cnt > 2) {     
                $finalarray[$week[array_shift($nv)].' - '.$week[array_pop($nv)]] = $nk;
            } elseif ($cnt == 2) {
                $finalarray[$week[array_shift($nv)].' & '.$week[array_pop($nv)]] = $nk;
            } else {
                $finalarray[$week[$nv[0]]] = $nk;
            } 
        }   
    }


    return $finalarray;

}

它可以采用这样的数组：

array(9) {
[0]=>
array(4) {
  ["id"]=>
  string(1) "1"
  ["day"]=>
  string(1) "1"
  ["open"]=>
  string(5) "12:00"
  ["close"]=>
  string(5) "23:30"
}
[1]=>
array(4) {
  ["id"]=>
  string(1) "2"
  ["day"]=>
  string(1) "2"
  ["open"]=>
  string(5) "12:00"
  ["close"]=>
  string(5) "14:00"
}
[2]=>
array(4) {
  ["id"]=>
  string(1) "3"
  ["day"]=>
  string(1) "3"
  ["open"]=>
  string(5) "12:00"
  ["close"]=>
  string(5) "14:00"
}
[3]=>
array(4) {
  ["id"]=>
  string(1) "4"
  ["day"]=>
  string(1) "4"
  ["open"]=>
  string(5) "12:00"
  ["close"]=>
  string(5) "23:30"
}
[4]=>
array(4) {
  ["id"]=>
  string(1) "5"
  ["day"]=>
  string(1) "5"
  ["open"]=>
  string(5) "12:00"
  ["close"]=>
  string(5) "23:30"
}
[5]=>
array(4) {
  ["id"]=>
  string(1) "6"
  ["day"]=>
  string(1) "6"
  ["open"]=>
  string(5) "12:00"
  ["close"]=>
  string(5) "23:30"
}
[6]=>
array(4) {
  ["id"]=>
  string(1) "7"
  ["day"]=>
  string(1) "7"
  ["open"]=>
  string(0) ""
  ["close"]=>
  string(0) ""
}
[7]=>
array(4) {
  ["id"]=>
  string(1) "8"
  ["day"]=>
  string(1) "2"
  ["open"]=>
  string(5) "16:00"
  ["close"]=>
  string(5) "20:00"
}
[8]=>
array(4) {
  ["id"]=>
  string(1) "9"
  ["day"]=>
  string(1) "3"
  ["open"]=>
  string(5) "16:00"
  ["close"]=>
  string(5) "20:00"
}
}

并将其转换为：

array(4) {
  ["Mon"]=>
  string(13) "12:00 - 23:30"
  ["Tue & Wed"]=>
  string(28) "12:00 - 14:00, 16:00 - 20:00"
  ["Thu - Sat"]=>
  string(13) "12:00 - 23:30"
  ["Sun"]=>
  string(1) "-"
}

感谢所有的帮助和支持..

2 个答案:

答案 0 :(得分：1)

尝试这样的事情：

$week = array (1 => 'Mon', 2=> 'Tue', ....);
$newarray = $finalarray = $valArray = array();
$i = 0;
foreach ($array as $key => $val) {
    if (count($newarray[$val]) > 0 && ((int)$key - 1 == end($newarray[$val]))) {
        $newarray[$i][] = $key;
        $valArray[$i] = $val;
        $i = $i +1;
    }   
}

foreach ($newarray as $k => $v) {
    $cnt = count($v);
    if ($cnt > 2) {     
        $finalarray[$week[array_shift($v)].' - '.$week[array_pop($v)]] = $valArray[$k];
    } elseif ($cnt == 2) {
        $finalarray[$week[array_shift($v)].' & '.$week[array_pop($v)]] = $valArray[$k];
    } else {
        $finalarray[$week[$v[0]]] = $valArray[$k];
    }    
}

答案 1 :(得分：0)

我得到了这个：

$array=array("
12:00 - 14:00, 16:00 - 23:30",
"12:00 - 23:30",
"12:00 - 23:30",
"12:00 - 23:30"
,"12:00 - 23:30"
,"12:00 - 23:30"
,"12:00 - 23:30"
,"12:00 - 14:00"

);
$days=array(0=>'Mon',1=>'Tuesday',2=>'Wed',3=>'Thu',4=>'Fri',5=>'Sat',6=>'Sun');
//var_dump($array);
//var_dump($days);
$previousvalue=0;
    for($i=0;$i<7;$i++){
        if($i!=0){
            if($array[$i]==$previousvalue){
                echo $days[$i].'=>'.$array[$i].'<br>';
            }else{

            }
        }
        $previousvalue=$array[$i];
    }

此代码仅回显与之前代码相同的日期。因此，我基本上将其他日子分开，与相互联系的日子是连续的并且有相同的时间。这可能会有所帮助。现在我需要的是将具有相同值的日期合并为：连续的第一天 - 连续的最后一天。我会在找到的东西后立即编辑。

注意：

我的代码从索引0开始，而你的代码从1开始

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