Cannot use isset() on the result of an expression (you can use "null !== expression" instead)

if (isset($array[2])){

抛出错误 

Cannot use isset() on the result of an expression (you can use "null !== expression" instead)





if (null !== $array[2]){

抛出提示 

Notice:  Undefined offset: 2 in /var/www/alt.php on line 144





对了，这家伙是数组，目的是检测该数组中某个 key 值是否存在，而不是检测某个变量是否有，应该用 array_key_exists() 

if(array_key_exists('2', $array)){

果然不再有错误