Question

我有一些像这样的变量：

$_8 = 0; $_9 = 0; $_10 = 0; $_11 = 0; $_12 = 0; $_13 = 0; $_14 = 0; $_15 = 0; $_16 = 0; $_17 = 0; $_18 = 0; $data = array();

然后我执行以下操作：

$vistoday = $app['db']->fetchAll('SELECT `Datum Bezoek 1` FROM `psttodo-uit` WHERE CAST(`Datum Bezoek 1` AS DATE) = CURRENT_DATE AND PB = 1');

foreach($vistoday as $v){
    $date = strtotime($v['Datum Bezoek 1']);
    $hours = (int)date('h', $date);
    $minutes = (int)date('i', $date);

    if($minutes > 30)
    {
        $hours = $hours + 1;
        $count = ${"_" . $hours} + 1;
        $data[$hours] = $count;
    }
    else
    {
        $data[$hours] = ${"_" . $hours}+1;
    }
}

这些是我从数据库中提取的日期：

2014-02-27 08:25:34
2014-02-27 08:50:34
2014-02-27 08:55:34

当我输出$ data时，我得到了这个：

array (size=2)
   8 => int 1
   9 => int 1

但通常键9的值应为2.所以他不增加，他只是加1而没有。有人可以帮我吗？

Answer 1

Yoг不需要像$ _8

这样的变量

foreach($vistoday as $v){
    $date = strtotime($v['Datum Bezoek 1']);
    $hours = (int)date('h', $date);
    $minutes = (int)date('i', $date);

    $hours = round($hours + $minutes / 61); // if $minutes > 30 ceil else floor
    $data[$hours] = isset($data[$hours]) ? $data[$hours] + 1 : 1;
}

在变量+ PHP中使用数字增加变量

1 个答案: