Javascript 函数返回值未定义

Question

我有以下代码。函数 get_last_catergory_value 的返回值始终是 undeifned。我搜索了 stackoverflow 但无法调试问题。

当我在 return 语句之前显示返回的值时，它显示了正确的值。但是当它从函数返回时，它是 undefined。你能帮我解决这个问题吗？

function fetch_product(brand) {
	var brand = brand;

	//get last category id
	var last_category = 'subcategory10';

	var last_category_value = get_last_catergory_value(last_category);
	alert(last_category_value); //undefined
}

function get_last_catergory_value(last_category) {

	if($('.' + last_category).find(':selected').val() == 'none') {

		last_category_number = last_category.substring(11);
		last_category_number = parseInt(last_category_number);
		last_category_number--;

		last_category = last_category.substring(0, 11);
		last_category = last_category + last_category_number;
		get_last_catergory_value(last_category);  //recall the function with values 'subcategory9', 'subcategory8' and so on...
	} else {
		var value = $('.' + last_category).find(':selected').val();

		alert(value); //Gives the correct value here
		return value;
	}
}

如果这是一个微不足道的问题，请原谅我。提前致谢。

Answer 1

get_last_catergory_value(last_category) 的 if block 中缺少 return 语句

 function get_last_catergory_value(last_category) {

if($('.' + last_category).find(':selected').val() == 'none') {

    last_category_number = last_category.substring(11);
    last_category_number = parseInt(last_category_number);
    last_category_number--;

    last_category = last_category.substring(0, 11);
    last_category = last_category + last_category_number;
    return get_last_catergory_value(last_category);
} else {
    var value = $('.' + last_category).find(':selected').val();

    alert(value); //Gives the correct value here
    return value;
}
}