[英]jQuery cookie value doesn't increase
I want to display an alert when a visiter visits certain pages. 我想在访客访问某些页面时显示警报。
I tried this code : 我尝试了这段代码:
$(document).ready(function () {
var visited = 0;
if ($.cookie('visited')) {
visited = $.cookie('visited');
}
if (visited == 3) {
alert('test');
} else {
visited++;
var date = new Date();
date.setTime(date.getTime() + (10 * 1000));
$.cookie('visited', visited, {expires: 1});
return false;
}
});
It works, but when I load my page 4 times, the value of visited
stays at 3 and the alert function always display. 它可以工作,但是当我将页面加载4次时,“
visited
的值保持为3,并且始终显示警报功能。
visited
value does not increase and the cookie does not set a new value. visited
值不会增加,并且cookie不会设置新值。
The other point is I want to set an additional parameter to my cookie: I want to display my alert function only in certain pages. 另一点是,我想为cookie设置一个附加参数:我只想在某些页面中显示警报功能。
here's how your code works. 这是您的代码的工作方式。
When visited = 0; increment visited by 1;
when visited = 1; increment visited by 1;
when visited = 2; increment visited by 1;
when visited = 3; display alert.
now your visited is always 3. Therefore, the value doesn't change. 现在您访问的总是3。因此,值不变。
You don't need the else statement, try this: 您不需要else语句,请尝试以下操作:
$(document).ready(function () {
var visited = 0;
if ($.cookie('visited')) {
visited = $.cookie('visited');
}
if (visited == 3) {
alert('test');
}
visited++;
var date = new Date();
date.setTime(date.getTime() + (10 * 1000));
$.cookie('visited', visited, {expires: 1});
return false;
});
And think to indent your code 并考虑缩进代码
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