PHP:自动完成时间间隔
因此,我有一个具有以下结构值的数组:PHP:自动完成时间间隔,php,arrays,Php,Arrays,因此,我有一个具有以下结构值的数组: $values = array( "2012-01-01 00:15:00" => "0.01", "2012-01-01 00:30:00" => "0.34", "2012-01-01 00:45:00" => "0.51", ); 如您所见,值之间的间隔为15分钟。有时我会得到缺少的值,例如: $values = array( "2012-01-01 00:15:00" => "0.01",
$values = array(
"2012-01-01 00:15:00" => "0.01",
"2012-01-01 00:30:00" => "0.34",
"2012-01-01 00:45:00" => "0.51",
);
如您所见,值之间的间隔为15分钟。有时我会得到缺少的值,例如:
$values = array(
"2012-01-01 00:15:00" => "0.01",
"2012-01-01 00:30:00" => "0.34",
"2012-01-01 01:01:00" => "0.23",
);
以正确的顺序在数组中添加缺少的“2012-01-01 00:45:00”
,并添加标准值(如“0.00”)的最佳方法是什么(CPU密集度较低)
谢谢。下面的代码应该可以填补空白,如果您预计空白会超过30分钟,我可以稍微扩展一下
<?php
date_default_timezone_set('GMT');
$values = array(
"2012-01-01 00:15:00" => "0.01",
"2012-01-01 00:30:00" => "0.34",
"2012-01-01 01:31:00" => "0.23",
"2012-01-01 05:31:00" => "0.23",
"2012-01-02 01:31:00" => "0.23",
);
function process_values($values) {
$keys = array_keys($values);
$new_values = array();
$new_values[$keys[0]] = $values[$keys[0]];
for($i = 1; $i < count($keys); $i++) {
$timestamp = strtotime($keys[$i]);
$minute_difference = ($timestamp - strtotime($keys[$i - 1])) / 60;
$start_minutes = floor(date('i', strtotime($keys[$i - 1])) / 15) * 15;
for($k = 1; $k < floor($minute_difference / 15); $k++) {
$minutes = $start_minutes + $k * 15;
$formatted_date = date('Y-m-d H:' . str_pad($minutes % 60, 2, '0') . ':00', strtotime($keys[$i - 1]) + floor($minutes / 60) * 3600);
$new_values[$formatted_date] = '0.00';
}
$new_values[$keys[$i]] = $values[$keys[$i]];
}
return $new_values;
}
print_r(process_values($values));
这是你的答案,我已经测试过了,效果很好:
<?php
$values = array(
"2012-01-01 00:15:00" => "0.01",
"2012-01-01 00:30:00" => "0.34",
"2012-01-01 01:00:00" => "0.23",
);
$values2 = array();
$passFirst = 0;
$difference = 60*15;
foreach ($values as $k => $v) {
$tempTime = date('Y-m-d H:i:s', strtotime($k) -$difference);
if (!isset($values[$tempTime]) && $passFirst) {
$values2[$tempTime] = '0.00';
}
$values2[$k] = $v;
$passFirst = 1;
}
/* The next two lines are for debugging, remove or comment after that */
print('<pre>');
print_r($values2);
您能否先单独构建一个完整的日期数组,然后将其与上面的$values
数组中的键进行比较?请注意,我认为另一个解决方案只能以精确的15分钟增量工作,这就是他更改您最初提供的时间戳的原因是的,我注意到了“01:00:00”时间,但您的解决方案也会输出[2012-01-01 01:60:00]=>0[2012-01-01 01:45:00]=>0.43[2012-01-01 04:180:00]=>0
。谢谢。那真是太棒了。。。我已经更新了代码以使其正常工作,它甚至会在几天内失效,如提供的时间戳所示。只要试着运行代码,就会看到结果:)