[英]Increment variable with number in variable + PHP
我有一些像這樣的變量:
$_8 = 0; $_9 = 0; $_10 = 0; $_11 = 0; $_12 = 0; $_13 = 0; $_14 = 0; $_15 = 0; $_16 = 0; $_17 = 0; $_18 = 0; $data = array();
然后我做以下事情:
$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] = ${"_" . $hours}+1;
}
}
這些是我從數據庫中提取的日期:
當我輸出$ data時,我得到了這個:
array (size=2)
8 => int 1
9 => int 1
但通常情況下,鍵9的值應為2.所以他不增加,他只是加1而沒有。 有人可以幫我這個嗎?
你不需要像$ _8這樣的變量
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;
}
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