php 如何仅在数组中保留特定的数组键/值？
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How to keep only specific array keys/values in array?
提问by Benn
I have a multidimensional array that I am searching trough for specific values. If those values are found I need to extract the indexes with those values ( make new array ) and remove all others.
我有一个多维数组，我正在搜索特定值的槽。如果找到这些值，我需要使用这些值提取索引（创建新数组）并删除所有其他值。
array_intersect worked fine on php 5.3 , now on 5.4 it complains Notice: Array to string conversion.
array_intersect 在 php 5.3 上工作正常，现在在 5.4 上它抱怨注意：数组到字符串的转换。
I found that array_intersect has an issue with multidimensional array on 5.4. https://bugs.php.net/bug.php?id=60198
我发现 array_intersect 在 5.4 上有一个多维数组的问题。 https://bugs.php.net/bug.php?id=60198
This is $options array I am searching trough
这是我正在搜索的 $options 数组
Array (
[index1] => html
[index2] => html
[index3] => slide
[index4] => tab
[index5] => Array
(
[0] => 123
)
)
code that works on php 5.3.x
适用于 php 5.3.x 的代码
$lookfor = array('slide', 'tab');
$found = array_intersect($options, $lookfor);
print_r($found);
Array
(
[index3] => slide
[index4] => tab
)
but in 5.4.x this trows the error mentioned above.
但在 5.4.x 中，这会引发上述错误。
What would be another way to do this without a loop please. and without suppressing the error.
请问没有循环的另一种方法是什么。并且没有抑制错误。
Thank you!
谢谢！
采纳答案by Amal Murali
array_intersect()isn't recursive. The function assumes the array is just one level deep and expects all the array elements to be scalars. When it finds a non-scalar value, i.e. a sub-array, it throws a Notice.
array_intersect()不是递归的。该函数假设数组只有一层深，并期望所有数组元素都是标量。当它找到一个非标量值，即一个子数组时，它会抛出一个通知。
This is vaguely mentioned in the documentation for array_intersect():
文档中array_intersect()含糊地提到了这一点：
Note: Two elements are considered equal if and only if: (string) $elem1 === (string) $elem2. In words: when the string representation is the same.
注意：两个元素被认为相等当且仅当： (string) $elem1 === (string) $elem2。换句话说：当字符串表示相同时。
One solution I could think of is to use array_filter():
我能想到的一种解决方案是使用array_filter()：
$lookfor = array('html', 'slide');
$found = array_filter($options, function($item) use ($lookfor) {
return in_array($item, $lookfor);
});
Note:This still performs a looping and isn't any better than a simple foreach. In fact, it might be slower than a foreachif the array is large. I have no idea why you're trying to avoid loops — I personally think it'd be more cleaner if you just use a loop.
注意：这仍然执行循环，并不比简单的foreach. 事实上，foreach如果数组很大，它可能比 a 慢。我不知道你为什么要避免循环——我个人认为如果你只使用循环会更干净。
Another solution I could think of is to remove the sub-arrays before using array_intersect():
我能想到的另一个解决方案是在使用之前删除子数组array_intersect()：
<?php
$options = array(
'index1' => 'html',
'index2' => 'html',
'index3' => 'slide',
'index4' => 'tab',
'index5' => array(123),
);
$lookfor = array('html', 'slide');
$scalars = array_filter($options,function ($item) { return !is_array($item); });
$found = array_intersect ($scalars, $lookfor);
print_r($found);
回答by zwacky
you could use array_filter()
你可以使用 array_filter()
$arr = array(
'index1' => 'html',
'index2' => 'html',
'index3' => 'slide',
'index4' => 'tab',
'index5' => array(0 => 123),
);
$with = array('html', 'slide');
$res = array_filter($arr, function($val) use ($with) {
return in_array($val, $with);
});
this will rerturn index1, index2 and index3.
这将返回 index1、index2 和 index3。
edit: just read your comment that your array will hold lots of entries. array_filterwill of course loop with a condition over them and create a new array.
编辑：只需阅读您的评论，即您的数组将包含大量条目。array_filter当然会在它们上面循环一个条件并创建一个新数组。
