Question

如何使用javascript历史记录推送状态删除所有获取变量？

它应该适用于以下情况：

http://example.com/slug1/slug2/index.php?myvar=1&myvar2=4
http://example.com/index.php?myvar=1&myvar2=4
http://example.com/slug1/slug2/?myvar=1&myvar2=4
http://example.com/slug2/?myvar=1&myvar2=4

之后应该是这样的：

http://example.com/slug1/slug2/index.php
http://example.com/index.php
http://example.com/slug1/slug2/
http://example.com/slug2/

也许像这样的函数：

function removeGetVariablesFromUrl() {
// Do stuff
}

它不应该返回带有更改的地址，它应该更改地址字段中的实际URL而不重新加载页面。

Answer 1

以下是在javascript中执行此操作的动态方法：

[Mon Apr 11 21:36:15 2016] [notice] child pid 8781 exit signal Floating point exception (8)

[Mon Apr 11 21:36:15 2016] [notice] child pid 8783 exit signal Floating point exception (8)

[Mon Apr 11 21:36:15 2016] [notice] child pid 8784 exit signal Floating point exception (8)

[Mon Apr 11 21:36:15 2016] [notice] child pid 8787 exit signal Floating point exception (8)

[Mon Apr 11 21:36:15 2016] [notice] child pid 8788 exit signal Floating point exception (8)

[Mon Apr 11 21:36:15 2016] [notice] child pid 8789 exit signal Floating point exception (8)

[Mon Apr 11 21:36:15 2016] [notice] child pid 8793 exit signal Floating point exception (8)

[Mon Apr 11 21:36:15 2016] [notice] child pid 8794 exit signal Floating point exception (8)

这会将您带到当前网址并删除任何内容，包括＆＃39;？＆＃39;符号（表示$ _GET变量）

Answer 2

这很简单：

function removeGetVariablesFromUrl(my_url) {
    my_url = "An url without get variables";
    history.pushState({}, 'The title', my_url);
}

Javascript删除具有历史推送状态的所有get变量

2 个答案: