未捕获的 ReferenceError：customerType 未定义（jQuery）

Question

单击触发 change_type 函数的单选按钮时出现以下错误：

Uncaught ReferenceError: customerType is not defined

我正在使用以下代码：

var date_selected = false;
var selected_year = false;
var current_customer = false;
var current_type = false;
var previous_value = 0;

var temp = <?php print_r($json); ?>;
var part = temp['Particulier'];
//console.log(part);
var type = part['Weekend'];
//console.log(type);
var yearArray = type['2016'];
//console.log(yearArray);

var dateSelect = document.getElementsByClassName('dateSelect');
var yearLabel = $("label[for='year']");
var year = document.getElementById('year');
var month = document.getElementById('month');
var date = document.getElementById('date');

$(dateSelect).hide();
yearLabel.hide();

function change_customer(value) {
    current_customer = value;

    var customerType = temp[value];
    console.log(customerType);
}

function change_type(value) {
    $(dateSelect).show();
    yearLabel.show();
    current_type = value;
    if(previous_value == 0) {
        $(dateSelect).show();
    }

    if(current_type == 'Weekend') {
        year.selectedIndex = 0;
        month.style.visibility = 'hidden';
        date.style.visibility = 'hidden';
    } else if(current_type == 'Midweek') {
        year.selectedIndex = 0;
        month.style.visibility = 'hidden';
        date.style.visibility = 'hidden';
    } else if(current_type == 'Week') {
        year.selectedIndex = 0;
        month.style.visibility = 'hidden';
        date.style.visibility = 'hidden';
    }
    previous_value = value;

    var dateType = customerType[value];
    console.log(dateType);
}

如果需要任何其他信息，请询问，新手，以及 jQuery。 提前致谢！

Answer 1

问题是因为您在change_customer() 函数内声明了customerType 数组，所以它超出了change_type() 函数的范围。您需要在两者的范围内声明它。试试这个：

var customerType = []; // declare empty to prevent unexpected 'access by index' errors

function change_customer(value) {
    customerType = temp[value]; // remove the 'var'
}

function change_type(value) {
    var dateType = customerType[value]; // is now in scope
}

请注意，我在上面的示例中只包含了相关代码。

Answer 2

您在函数change_customer 中定义了customerType，因此在该函数之外无法访问它。要使 customerType 在所有函数中都可访问，您需要将其设为全局。

var customerType = []; // In global scope(outside of all functions)
.
.
.
function change_customer(value) {
    current_customer = value;

    customerType = temp[value];
    console.log(customerType);
}

Answer 3

function change_customer(value) {
    current_customer = value;

    var customerType = temp[value];
    console.log(customerType);
}

function change_type(value) {
    $(dateSelect).show();
    yearLabel.show();
    current_type = value;
    if(previous_value == 0) {
        $(dateSelect).show();
    }

    if(current_type == 'Weekend') {
        year.selectedIndex = 0;
        month.style.visibility = 'hidden';
        date.style.visibility = 'hidden';
    } else if(current_type == 'Midweek') {
        year.selectedIndex = 0;
        month.style.visibility = 'hidden';
        date.style.visibility = 'hidden';
    } else if(current_type == 'Week') {
        year.selectedIndex = 0;
        month.style.visibility = 'hidden';
        date.style.visibility = 'hidden';
    }
    previous_value = value;

    var dateType = customerType[value];
    console.log(dateType);
}

在上面的代码中你已经在里面初始化了customerType变量 change_customer 函数，而您正试图在超出其范围的 change_type 函数中访问它 范围，所以你的选择是让它像你做一些变量一样全球化。

Answer 4

你不要在change_type 函数内的任何地方调用你的change_customer 函数。因此customerType 永远不会被实例化。

尝试像这样添加你的函数：

function change_customer(value) {
    var current_customer = value;

    var customerType = temp[value];
    return customerType;
}

function change_type(value) {
    var customerType = change_customer(value);
    $(dateSelect).show();
    yearLabel.show();
    ......
}

虽然，您的 temp 数组在哪里，键为 value？

Answer 5

您的 change_type 函数在尚未声明时尝试使用 customerType。您需要声明 customerType 以及其余的全局变量，而不是在 change_customer 函数的范围内:)