基于以下内容的输出,我想知道是否有人可以帮助我:
$monthsOfTheYear = range(1, 12);
print_r ($monthsOfTheYear);
我理解上述过程,但是使用 for 循环重新创建此原则的最有效方法是什么?到目前为止,我想出了以下内容:
$months = ("January", "February", "March", "April", "May", "June", "July", "August", September", "October", "November", "December") {
for($months{0} = 1, $months{11} = 12, /*????*/) {
echo "/*????*/";
} }
如果可能的话,我想知道如何在循环代码中实现这一点,尽管我怀疑它可能需要更复杂的解决方案。
你想这样做吗?可能是,但我不确定你是否想要这个。请把你的问题说得更清楚一点。尝试添加示例输出。
$months = array("January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December");
for($i = 0;$i < 12; $i++){
echo $months[$i].'<br>';
}
for ($i = 0; $i < 12; $i++) {
echo $monthsOfTheYear[$i];
}
数十种可能的解决方案之一是
foreach(array("January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December") as $month) {
echo $month . "<br>";
}
你的意思是这样吗?
$months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"];
for ($i=0; $i<count($months); $i++) {
echo "$months[$i]'n";
}
请参阅 http://3v4l.org/T740C 中的输出