如何检查是否存在多个数组键

Question

我有各种各样的数组,将包含

story & message

or just

story

How would I check to see if an array contains both story and message? array_key_exists() only looks for that single key in the array.

Is there a way to do this?

Answer 1

这是一个可扩展的解决方案,即使您要检查大量密钥:

<?php

// The values in this arrays contains the names of the indexes (keys) 
// that should exist in the data array
$required = array('key1', 'key2', 'key3');

$data = array(
    'key1' => 10,
    'key2' => 20,
    'key3' => 30,
    'key4' => 40,
);

if (count(array_intersect_key(array_flip($required), $data)) === count($required)) {
    // All required keys exist!
}

Answer 2

If you only have 2 keys to check (like in the original question), it's probably easy enough to just call array_key_exists() twice to check if the keys exists.

if (array_key_exists("story", $arr) && array_key_exists("message", $arr)) {
    // Both keys exist.
}

However this obviously doesn't scale up well to many keys. In that situation a custom function would help.

function array_keys_exists(array $keys, array $arr) {
   return !array_diff_key(array_flip($keys), $arr);
}

Answer 3

令人惊讶的array_keys_exist是不存在？!在此期间,留出一些空间来为这个常见任务找出单行表达式.我在想一个shell脚本或另一个小程序.

注意:以下每个解决方案都使用[…]php 5.4+中提供的简明数组声明语法

array_diff + array_keys

if (0 === count(array_diff(['story', 'message', '…'], array_keys($source)))) {
  // all keys found
} else {
  // not all
}

(给Kim Stacks提示)

这种方法是我发现的最简短的方法.array_diff()返回参数1中不存在于参数2中的项目数组.因此,空数组表示找到了所有键.在PHP 5.5中,您可以简化0 === count(…)为简单empty(…).

array_reduce + unset

if (0 === count(array_reduce(array_keys($source), 
    function($in, $key){ unset($in[array_search($key, $in)]); return $in; }, 
    ['story', 'message', '…'])))
{
  // all keys found
} else {
  // not all
}

更难阅读,易于改变.array_reduce()使用回调迭代数组以获得值.通过提供我们感兴趣的键的$initial值,$in然后删除源中找到的键,如果找到所有键,我们可以期望以0个元素结束.

结构很容易修改,因为我们感兴趣的键非常适合底线.

array_filter&in_array

if (2 === count(array_filter(array_keys($source), function($key) { 
        return in_array($key, ['story', 'message']); }
    )))
{
  // all keys found
} else {
  // not all
}

编写比array_reduce解决方案更简单,但编辑略有琐事.array_filter也是一个迭代回调,允许您通过在回调中返回true(复制项到新数组)或false(不复制)来创建过滤数组.gotchya是你必须改变2你期望的项目数.

这可以更持久,但边缘可读:

$find = ['story', 'message'];
if (count($find) === count(array_filter(array_keys($source), function($key) use ($find) { return in_array($key, $find); })))
{
  // all keys found
} else {
  // not all
}

Answer 4

在我看来,到目前为止最简单的方法是:

$required = array('a','b','c','d');

$values = array(
    'a' => '1',
    'b' => '2'
);

$missing = array_diff_key(array_flip($required), $values);

打印:

Array(
    [c] => 2
    [d] => 3
)

这也允许检查哪些键完全丢失.这可能对错误处理很有用.

Answer 5

另一种可能的解决方案：

if (!array_diff(['story', 'message'], array_keys($array))) {
    // OK: all the keys are in $array
} else {
   // FAIL: some keys are not
}

Answer 6

上述解决方案很聪明,但速度很慢.使用isset的简单foreach循环的速度是array_intersect_key解决方案的两倍多.

function array_keys_exist($keys, $array){
    foreach($keys as $key){
        if(!array_key_exists($key, $array))return false;
    }
    return true;
}

(对于1000000次迭代,344ms vs 768ms)