php PHP根据值将数组拆分为两个数组
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PHP splitting array into two arrays based on value
提问by Austin
I have a PHP array that I am trying to split into 2 different arrays. I am trying to pull out any values that contain the word "hidden". So one array would contain all the values that do not contain the word "hidden". The other array would contain all the values that do contain the word "hidden". I just can't figure out how to do it though.
我有一个 PHP 数组，我试图将其拆分为 2 个不同的数组。我正在尝试提取包含“隐藏”一词的任何值。因此，一个数组将包含所有不包含“隐藏”一词的值。另一个数组将包含所有包含“隐藏”一词的值。我就是不知道该怎么做。
The original array is coming from a form post that contains keys and values from a bunch of check boxes and hidden inputs. so the actual post value looks something like this:
原始数组来自一个表单帖子，其中包含来自一堆复选框和隐藏输入的键和值。所以实际的帖子值看起来像这样：
Group1 => Array([0] => item1,[1] => item2hidden,[2] => item3,[3] => item4,[4] => item5hidden)
so to simplify it:
所以为了简化它：
$myArray = Array(item1, item2hidden, item3, item4, item5hidden)
final output
最终输出
$arr1 = (item1, item3, item4)
$arr2 = (item2hidden, item5hidden)
Anyone know how to do something like this?
有谁知道如何做这样的事情？
采纳答案by Vlad Preda
This should do the trick:
这应该可以解决问题：
$myArray = array('item1', 'item2hidden', 'item3', 'item4', 'item5hidden');
$secondaryArray = array();
foreach ($myArray as $key => $value) {
if (strpos($value, "hidden") !== false) {
$secondaryArray[] = $value;
unset($myArray[$key]);
}
}
It moves all the entries that contain "hidden" from the $myArrayto $secondaryArray.
它将所有包含“隐藏”的条目从 移动$myArray到$secondaryArray。
Note:It's case sensitive
注意：区分大小写
回答by Glavi?
You can use array_filter()function:
您可以使用array_filter()功能：
$myArray = array('item1', 'item2hidden', 'item3', 'item4', 'item5hidden');
$arr1 = array_filter($myArray, function($v) { return strpos($v, 'hidden') === false; });
$arr2 = array_diff($myArray, $arr1);
回答by Viacheslav Kondratiuk
$myArray = Array('item1', 'item2hidden', 'item3', 'item4', 'item5hidden');
$arr1 = array();
$arr2 = array();
foreach ($myArray as $item) {
if (strpos($item, "hidden") !== false) {
$arr1[] = $item;
} else {
$arr2[] = $item;
}
}
This solution checks if 'hidden' present at current item, if no, move to $arr1else to $arr2
此解决方案检查当前项目中是否存在“隐藏”，如果不存在，则移至$arr1其他$arr2
回答by Nikolay Krasnov
You can use array_filter:
您可以使用 array_filter：
function filtreHiddens($e) {
if (isset($e['hidden']) && $e['hidden']) return true;
else return false;
}
function filtreNotHiddens($e) {
if (isset($e['hidden']) && !$e['hidden']) return true;
else return false;
}
$arrayToFiltre = array(
array('hidden' => true, 'someKey' => 'someVal'),
array('hidden' => false, 'someKey1' => 'someVal1'),
array('hidden' => true, 'someKey2' => 'someVal3'),
);
$hidden = array_filter($arrayToFiltre, 'filtreHiddens');
$notHidden = array_filter($arrayToFiltre, 'filtreNotHiddens');
print_r($hidden);
print_r($notHidden);
回答by fresbeeplayer
Maybe it's just me, but I would go for the clarity of regular expressions...
也许这只是我，但我会为了正则表达式的清晰度而去......
foreach($myArray as $item) {
if (preg_match("/hidden$/i", $item)) {
array_push($arr2, $item);
} else {
array_push($arr1, $item);
}
}
