PHP array_key_exists不起作用;数组不是多维的

$countries = array(
  "GB" => "United Kingdom","US" => "United States","AF" => "Afghanistan","AL" => "Albania","DZ" => "Algeria","AS" => "American Samoa","AD" => "Andorra","AO" => "Angola","AI" => "Anguilla","AQ" => "Antarctica","AG" => "Antigua And Barbuda","AR" => "Argentina","AM" => "Armenia","AW" => "Aruba","AU" => "Australia","AT" => "Austria","AZ" => "Azerbaijan","BS" => "Bahamas","BH" => "Bahrain","BD" => "Bangladesh","BB" => "Barbados","BY" => "Belarus","BE" => "Belgium","BZ" => "Belize","BJ" => "Benin","BM" => "Bermuda","BT" => "Bhutan","BO" => "Bolivia","BA" => "Bosnia And Herzegowina","BW" => "Botswana","BV" => "Bouvet Island",);

所有国家都是如此;我100％肯定每个国家都正确列出.

我有一个申请表,将结果存储在服务器上存储的文件中.目前,该应用程序的评论页面是一个基本的文本版本,我现在正在将它放入一个模拟表单中,以便我的客户有一个更具视觉吸引力的方法来审查应用程序.

因此,名为$in_data的数组存储来自文件的结果.该阵列的结构如下：“emergency_medical_insurance”=> “value_user_entered”.每个键都是它所形成的HTML元素的名称,值是用户输入的值.

表单上的国家/地区选择列表返回该国家/地区的双字母代码.所以我想要做的是搜索价值为$in_data [‘country_select’]的$country,然后返回该国家/地区的名称.

echo $in_data [‘country_select’];返回’CA’加拿大的字母代码和我输入的测试国家/地区.

echo $countries [‘CA’];返回’加拿大’

if (array_key_exists($in_data['country_select'],$countries)){ 
      echo "Country Found";
} 
else { echo "failed"; }

没有回报.

if (array_key_exists('CA',$countries)){ 
      echo "Country Found";
} 
else { echo "failed"; }

也没有任何回报.当我什么都不说时,我什么都不说,不是空的,不是真的,不是虚假的;只是甚至没有运行.

我的问题很简单;下面的代码(取自官方的PHP手册)如何与我的代码完全相同,工作,但我的代码甚至不返回任何东西？

<?php
$search_array = array('first' => 1,'second' => 4);
if (array_key_exists('first',$search_array)) {
    echo "The 'first' element is in the array";
}
?>