我有一系列世界上所有国家：

$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？

$search_array = array('first' => 1, 'second' => 4);

if (array_key_exists('first', $search_array)) {

echo "The 'first' element is in the array";

}

?>

解决方法:

因为你正在读取文件,你可能会得到其他字符,尝试trim()：

if (array_key_exists(trim($in_data['country_select']), $countries)){

echo "Country Found";

}

else { echo "failed"; }

标签：html,php,arrays,forms,array-key-exists

来源： https://codeday.me/bug/20190729/1569373.html