如何用jQuery定义函数参数

Question

我有一些语法问题......这是我的小脚本:

$('.basket_details, .advanced_search_panel, .producers_major_panel').hover(function () {
    mouse_is_inside = true;
}, function () {
    mouse_is_inside = false;
});
$("body").mouseup(function () {
    if (mouse_is_inside) {
        $('.advanced_search_panel, .producers_major_panel').fadeOut('slow');
        $('.basket_details').slideUp('slow');
    }
});

一切都很好,但我的Chrome控制台尖叫,我有语法错误,并mouse_is_inside没有定义,如何纠正这个错误？

Answer 1

尝试关闭:

(function() {

  var mouse_is_inside = false;

$('.basket_details, .advanced_search_panel, .producers_major_panel').hover(function () {
    mouse_is_inside = true;
}, function () {
    mouse_is_inside = false;
});
$("body").mouseup(function () {
    if (mouse_is_inside) {
        $('.advanced_search_panel, .producers_major_panel').fadeOut('slow');
        $('.basket_details').slideUp('slow');
    }
});

})();

这将允许您在函数范围内使用'mouse_is_inside'.它也阻止你需要使用全局.