[英]How to merge these two arrays?
How to merge these two arrays... 如何合并这两个数组...
$arr1 = array(
[0] => 'boofoo',
[1] => '2'
);
$arr2 = array(
[0] => 'fbfb',
[1] => '6'
);
to get a third array as follows?: 得到第三个数组如下:
$arr3 = array(
[0] => 'boofoo',
[1] => '6'
);
That is, preserve strings that are longer and numeric values that are higher. 即,保留较长的字符串和较高的数值。
If both arrays have the same keys, you can do a simple foreach
and pick the better value according to their relationship regarding >
and strlen
respectively: 如果两个数组具有相同的键,则可以执行简单的foreach
并根据分别与>
和strlen
有关的关系来选择更好的值:
$arr3 = array();
foreach ($arr1 as $key => $val) {
if (array_key_exists($key, $arr2)) {
if (is_numeric($val)) {
$arr3[$key] = max($val, $arr2[$key]);
} else {
$arr3[$key] = strlen($val) > strlen($arr2[$key]) ? $val : $arr2[$key];
}
}
}
Before you ask: expr1 ? expr2 : expr3
在您问: expr1 ? expr2 : expr3
expr1 ? expr2 : expr3
is the conditional operator where the expressions expr2
or expr3
are evaluated based on the return value of expr1
. expr1 ? expr2 : expr3
是条件运算符 ,其中基于expr1
的返回值对表达式expr2
或expr3
进行求expr1
。 So if strlen($val) > strlen($arr2[$key])
is true, the conditional operation evaluates $val
, and $arr2[$key]
otherwise. 因此,如果strlen($val) > strlen($arr2[$key])
为true,则条件运算的结果$val
,否则$arr2[$key]
。
I think you can achieve via http://www.php.net/manual/en/function.array-walk.php . 我认为您可以通过http://www.php.net/manual/en/function.array-walk.php来实现。
I try and write some test for it soon and update post. 我会尝试为此编写一些测试并更新帖子。
PS: array_map is better(maybe walk not even good for this?) PS: array_map更好(也许步行甚至不行吗?)
<?php
function combine($arr1, $arr2) {
return array_map("callback", $arr1, $arr2);
}
function callback($item, &$item2) {
if (is_numeric($item)) {
return max($item, $item2);
}
if (strlen($item2) > strlen($item)) {
return $item2;
}
return $item;
}
class Tests extends PHPUnit_Framework_TestCase {
public function testMergingArrays() {
$arr1 = array(
'boofoo',
'2'
);
$arr2 = array(
'fbfb',
'6'
);
$arr3 = array(
'boofoo',
'6'
);
$this->assertEquals($arr3, combine($arr1, $arr2));
}
}
I got test in place and I could refactor it now. 我已经进行了测试,现在可以重构它了。 But this implementation does work. 但是这种实现确实有效。
您可以使用array_merge()
函数,然后再进行sort()
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.