脚本宝典收集整理的这篇文章主要介绍了PHP array_key_exists不起作用;数组不是多维的,脚本宝典觉得挺不错的,现在分享给大家,也给大家做个参考。
$countries = array( "GB" => "UnITed Kingdom","US" => "United states","AF" => "Afghanistan","AL" => "Albania","DZ" => "Algeria","AS" => "american Samoa","AD" => "Andorra","AO" => "Angola","ai" => "Anguilla","AQ" => "Antarctica","AG" => "Antigua And Barbuda","AR" => "argentina","AM" => "Armenia","AW" => "Aruba","AU" => "Australia","AT" => "Austria","AZ" => "Azerbaijan","BS" => "Bahamas","BH" => "Bahrain","BD" => "Bangladesh","BB" => "BarbaDOS","BY" => "Belarus","BE" => "Belgium","BZ" => "Belize","BJ" => "Benin","BM" => "Bermuda","BT" => "Bhutan","BO" => "Bolivia","BA" => "Bosnia And Herzegowina","BW" => "Botswana","BV" => "Bouvet Island",);
所有国家都是如此;我100%肯定每个国家都正确列出.
我有一个申请表,将结果存储在服务器上存储的文件中.目前,该应用程序的评论页面是一个基本的文本版本,我现在正在将它放入一个模拟表单中,以便我的客户有一个更具视觉吸引力的方法来审查应用程序.
因此,名为$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"; } ?>
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