您需要通过引用传递变量:

You need to pass the variable by reference:

function AmISet(&$fieldName) {
    if (isset($fieldName)) {
        echo "I am set\n";
    } else {
        echo "I am not set\n";    
    }
}

测试用例:

$fieldName = 'foo';
AmISet($fieldName); // I am set

unset($fieldName);
AmISet($fieldName); // I am not set

但是，此函数实际上没有用，因为它只会输出一个字符串.您可以创建一个接受变量的函数，如果变量存在则返回(来自

However, this function is not useful as it is, because it will only output a string. You can create a function that accepts a variable and return if it exists (from this post):

function issetor(&$var, $default = false) {
    return isset($var) ? $var : $default;
}

现在可以像这样使用了:

Now it can be used like so:

echo issetor($fieldName); // If $fieldName exists, it will be printed