以下代码无法使用。尽管该函数正在调用
$name1 = "apple";
$gender1 = "orange";
$country1 = "pepe";
function profile1($name1, $gender1, $country1)
{
echo $name1 . "\n";
echo $gender1 . "\n";
echo $country1 . "\n";
}
profile1();
答案 0 :(得分:0)
函数参数$name1
,$gender1
和$country1
与在函数外部声明的变量是分开的且不同的。在不提供任何参数的情况下调用该函数时,参数的值将是不确定的。
您可以:
profile1( $name1, $gender1, $country1 );
或(不推荐!):
function profile1()
{
global $name1, $gender1, $country1;
echo $name1 . "\n";
echo $gender1 . "\n";
echo $country1 . "\n";
}
profile1();