Closed Bug 624723 Opened 14 years ago Closed 14 years ago

Get visited sites using getComputedStyle and CSS

Categories

(Core :: CSS Parsing and Computation, defect)

1.9.2 Branch
x86
Windows XP
defect
Not set
critical

Tracking

()

RESOLVED DUPLICATE of bug 147777

People

(Reporter: simon.roesler, Unassigned)

References

()

Details

User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US) AppleWebKit/534.10 (KHTML, like Gecko) Chrome/8.0.552.224 Safari/534.10 Build Identifier: Mozilla/5.0 (Windows; U; Windows NT 5.1; de; rv:1.9.2.13) Gecko/20101203 Firefox/3.6.13 With using the Javascript code getComputedStyle and the CSS-Visited Tag it´s possible to check wheater an user has visited a specific website. The Code to Check: if(window.getComputedStyle(document.getElementById("check"),null).getPropertyValue("display") == "inline") And the CSS for this: div.history a { display: none;} div.history a:visited { display: inline;} Later I saw that it should already be fixed but it isn´t. Reproducible: Always Steps to Reproduce: 1. Visit a site 2. enter the site adress in the textbox at the included site 3. Click Check and see what happen Actual Results: It says weather a site was visited or not. Expected Results: There shouldn´t be an ability of checking this with javascript. I read, that this is already solved but with my code (I don´t know what i did different) it´s working in my firefox.
Component: General → Style System (CSS)
Product: Firefox → Core
QA Contact: general → style-system
Version: unspecified → 1.9.2 Branch
> I read, that this is already solved It's solved in Firefox 4.
(In reply to comment #1) > It's solved in Firefox 4. In bug 147777.
Status: UNCONFIRMED → RESOLVED
Closed: 14 years ago
Resolution: --- → DUPLICATE
Group: core-security
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