Question

根据以下内容的输出，我想知道是否有人可以帮助我：

  $monthsOfTheYear = range(1, 12);
  print_r ($monthsOfTheYear);

我理解上面的过程，但是用for循环重建这个原理的最有效方法是什么？到目前为止，我已经提出了以下内容：

  $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，但我怀疑它可能需要更复杂的解决方案。

Answer 1

你是不是想这样做。可能但是我不确定你是否想要这个。请让你的问题更清楚一点。试着添加样本输出。

$months = array("January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December");
for($i = 0;$i < 12; $i++){
 echo $months[$i].'<br>';
}

Answer 2

for ($i = 0; $i < 12; $i++) {
    echo $monthsOfTheYear[$i];
}

Answer 3

许多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>";
}

Answer 4

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

的输出

创建一个for循环，以1为增量回显一系列数字

4 个答案: