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需求:
Hi All,
I have a requirement which goes like this:
User logs in to the application navigates to one particular page and suppose if the user is idle with out posting any requests, after some time if user tries to click on some thing and say if the session times out, then the page saying session timed out gets displayed.
Now, i want to popup a window which would notify user saying that the application would time out in next x minutes where x + idle time = session time out value.Is this fessible?
If yes, how can i acheive this....ie say i am on Page A where i dont click anything causing idle time...where should i start writing my logic....and to do this i think i should start some parallel thread which would start the timer to keep track of the elapsed and idle times to throw popup.
Thanks,
Sushma.
注意以上是时间到后,自动弹出窗口,自动执行
Hi,
Here are my responses:
Now, i want to popup a window which would notify user saying that the application would time out in next x minutes where x + idle time = session time out value.Is this fessible?
>>There is a profile option "ICX:Session Timeout",which is nothing but Maximum idle time for the Oracle Applications user session
(specified in minutes). You can put a js function in onload api of oabody bean, which will take value from the profile option, say value is x1 minutes, and suppose x minutes before of timeout , u need to throw an js alert,then ur pseudo code should be
String s="function update(){alert('The session will be timed out in"+<String variable comtaining value of(x)>+" ');}";
pageContext.putJavaScriptFunction("update","
//put value of (x1-x)*1000 in the second argument of api
String javaS = "javascript:setTimeout(\"update();\",<(x1-x)*1000>);";
bodyBean.setOnLoad(javaS);
Note, i haven't tested , but this should work fine, also check syntaxes.
--Mukul
建议是采用javascript来实现
暂时记录,有待于验证 |
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