The pages I am attempting to step through using HttpUnit have implemented a cookie counter mechanism to deal with users using the back button. The Javascript looks like this:
<SCRIPT LANGUAGE="JavaScript">
var curCookie = "tx=" + i;
document.cookie=curCookie;
</SCRIPT>
where i is incremented on each subsequent page.
This being said:
The back end code depends on this cookie for a couple of things and it appears I have no access to it thru HttpUnit.
The first element the WebResponse.getDOM() returns is <html> the <SCRIPT. element is not included in the DOM.
Up at the web conversation level I can print cookies, add cookies but trying to add the same cookie with a different value does not work, and clearing the web conversation clobbers my context. I want more fine grained control over cookies and it looks like it is not there. Any suggestions. I am going to hack the WebConversation etc to give me more control over cookies and I should be good to go.
Any Help is appreciated
Thanks
Dean
If you would like to refer to this comment somewhere else in this project, copy and paste the following link:
The pages I am attempting to step through using HttpUnit have implemented a cookie counter mechanism to deal with users using the back button. The Javascript looks like this:
<SCRIPT LANGUAGE="JavaScript">
var curCookie = "tx=" + i;
document.cookie=curCookie;
</SCRIPT>
where i is incremented on each subsequent page.
This being said:
The back end code depends on this cookie for a couple of things and it appears I have no access to it thru HttpUnit.
The first element the WebResponse.getDOM() returns is <html> the <SCRIPT. element is not included in the DOM.
Up at the web conversation level I can print cookies, add cookies but trying to add the same cookie with a different value does not work, and clearing the web conversation clobbers my context. I want more fine grained control over cookies and it looks like it is not there. Any suggestions. I am going to hack the WebConversation etc to give me more control over cookies and I should be good to go.
Any Help is appreciated
Thanks
Dean