php判断数组中是否存在指定的键(key)

php中有两个函数用来判断数组中是否包含指定的键，分别是array_key_exists和isset

array_key_exists语法如下

array_key_exists($key, $array)

如果键存在返回true

isset函数语法如下

isset($array[$key])

如果键存在返回true

演示代码

<?php 
$array = array("Zero"=>"PHP", "One"=>"Perl", "Two"=>"Java");
print("Is 'One' defined? ".array_key_exists("One", $array)."\n");
print("Is '1' defined? ".array_key_exists("1", $array)."\n");
print("Is 'Two' defined? ".isset($array["Two"])."\n");
print("Is '2' defined? ".isset($array[2])."\n");
?>

返回结果如下：

Is ‘One’ defined? 1
Is ’1′ defined?
Is ‘Two’ defined? 1
Is ’2′ defined?