有一个类似于下面的数组:
$steps = array(0 => 'aaa', 1 => 'bbb', 2 => 'ccc', ......, 7 => 'hhh', 8 => 'iii', .....);
如何根据序列7来计算从密钥7到达密钥2所需的步数(密钥)?
答案 0 :(得分:2)
如果您的数字键从不丢失任何数字,则可以使用基本减法。
如果您需要考虑可能缺少的号码,或者密钥不是数字,则可以使用array_keys()
和array_search()
的组合:
$array = array(
0 => 'aaa',
1 => 'bbb',
3 => 'ccc',
'four' => 'ddd',
900 => 'eee',
13 => 'fff'
);
$from = 1;
$to = 900;
$keys = array_keys($array);
$from_index = array_search($from, $keys); // 1
$to_index = array_search($to, $keys); // 4
$steps = $to_index - $from_index;
// 3 steps: from 1-3, from 3-'four' and from 'four'-900
答案 1 :(得分:0)
I solved this problem by writing this code:
$tot_step = 0;
$steps = array(
0 => 'aaa',
1 => 'bbb',
2 => 'ccc',
3 => 'ddd',
4 => 'eee',
5 => 'fff',
6 => 'ggg',
7 => 'hhh',
8 => 'iii',
9 => 'jjj',
10 => 'aaa'
);
$from = "ddd";
$to = "bbb";
$from_index = array_search($from, $steps);
$to_index = array_search($to, $steps);
$last = $steps[(count($steps)-1)];
if ($from_index > 0) {
if ($to == $last)
$to_index = (count($steps)-1);
$arr_l = count($steps);
$mila = array();
for ($ind = $from_index; $ind <= ($arr_l-1); $ind++) {
if ($to == $last) {
if ($steps[$ind] != $last)
$mila[] = $steps[$ind];
} else {
$mila[] = $steps[$ind];
}
unset($steps[$ind]);
}
if (!empty($mila)) {
for ($i = (count($mila)-1); $i >= 0; $i--)
array_unshift($steps, $mila[$i]);
}
$to_new = array_search($to, $steps);
foreach ($steps as $key => $value) {
if ($key == $to_new)
break;
else
$tot_step++;
}
} elseif ($from_index == 0) {
if ($to_index == $from_index) {
$tot_step = (count($steps)-1);
} else {
foreach ($steps as $key => $value) {
if ($key == $to_index)
break;
else
$tot_step++;
}
}
}
echo $tot_step;
I hope it will be useful to someone