I desperately need to have a constant website visitor that is online 24/7 on my website. Due to a js error. We need to keep an absolute minimum of one online site visitor. This is a temporary measure while we locate the error which so far has taken 3 js coders 3 days with no success in finding. We really need to keep the script thinking that a user is online.
Is there a bot setup that will simulate a website user that NEVER leaves a certain webpage?
If not. How can I set up the script so that I can fool it into thinking there is someone online? I really need this info... i have spent hours and another thread, with NO help on the repair. I really need to be able to keep the js rolling. I am on the verge of giving up on this project as impossible.
I am looking for a pop up to collect emails when a visitor is leaving my site, I don't which pop up software can provide that.I know how annoying pop up can be but I am sure my visitors will be ok with it, they are even asking for it.
After playing with all of the different jQuery interactions, widgets, visual effects, etc. (on the demo pages), I wonder of there is an online website builder that would allow me to easily create web pages with all of these functions, but without having to do any real programming ?
i'm trying to make the module for users online in my website i've used ajax to call database and update the persons table with the status of the users when he enters or leaves the page here is my code in the <body> tag :
I would like to add a prompt when a visitor of my website leaves the site or closes the browser.The prompt should be to optionally give his/her feedback about the site.
--> User can fill a simple form with comment & email id --> In this case, the comments should go to our mail id
--> The user can also leave the site without giving the feedback, saying NO to the prompt.
Is it possible to create a tracking code that would allow me to:
(1) a) Find out the identity of visitors to a certain website (a website unrelated to me and the visitors did not visit my website) b) The identity of those who recently searched for a certain keywords and/or...
(2) Find out what websites certain individuals have recently visited (or visiting in real-time)?
I want the lightbox window open only once a day for every unique visitor on my website. I understand I should use some sort of cookie implented, but I dont really understand how to.
I have a form with a textarea field. I want to validate the input from the textarea using javascript. Suppose I want to check that the user has not entered the string: "Hello World!"
To do this I am using the script: form["text"].value == "Hello World"
But this gives an "unterminated string constant error" because the browser converts this to: form["text"].value == "Hello World"
I'm getting an error message come up in IE6 when somenone subscribes to my newsletter from my home page (either from the drop down hover box form or the left hand form) here .
After entering a name and email address I get:
A runtime error has occurred....do you wish to debug? Line 4 Error: Unterminated string constant
When I click NO I then get a second small windows popup which says:
A runtime error has occurred....do you wish to debug? Error: Expected ')'
I've looked at lines 4 and 43 of my thanks.php page (the page subscribers get when they fill out the form but before they double optin via the email) and I can't find anything wrong. I guess the first message expects a semicolon and the second obviously a closing bracket. Code:
I am facing a problem in javascript. strdata is a variable of javascript and following is an assignment(hmm a Huge one but unfortunately its not mine code, I am just trying to rectify it) to this varibale.. Now I am facing a run time error(Unterminated string constant).....
in IE only (tested version 7) if var myWord = "English" then it works fine but if var myWord = "Modifier Chau00EEnes" then I get "Unterminated string constant" error.
What fix would you suggest to keep div.innerHTML = "" format?
make a scroller that could scroll continuously/constant both forward and backward (from left to right and right to left).
The instance shouldn't scroll unless holding down a link (arrows left&right) outside the scroller. (using onmousedown and stop onmouseup) and it would be great if it's not possible to scroll past the last and first item. (i don't want any empty space) But I guess I could do that with offset and exclude (:
I don't have any code for you, but if you want me to try some more first, I'll do that and send you if I got anything good out of it!
(I also tried the HoverScroll plugin, but since the arrows is implemented in the jquery/js-code I wasn't abled to tweak it as I wanted. That plugin is possibly the nearest already made scroller I've so far seen. However I want the buttons to be outside the scrollerarea, and I also want to be able to change the styling on the whole thing. and I want it to use onmousedown and onmouseup instead of onmouseover.)
I've tried with scrollable and ScrollTo/SerialScroll, but I'm not a code wizard so how to do this.. ;O These plugins seems really great, and I have used them in other projects, so I'm a bit familiar with how they work.
I'm having a hard time getting my head round it again. I know it could be more efficient in jQuery, but I'd be happy just to get it working, with an extra variable, in straight js.The function takes an array and progressively hides each element at constant interval, in this case 50ms:
function hide_50(arrayA,visibility,current) { if ( current == null ) current = 0; var arrayB=(typeof arrayA == 'string')? arrayA.split(',') : arrayA;
I'm trying to code a feature for my website using DHTML where the person viewing the website can rotate between viewing the positive/negative points of the website being reviewed. This example was in the publication called Using HTML 4 by Lee Anne Phillips. Code:
I have a ajax program on a page on website A. And I need to let it talk to a program on website B. I tried to put website B into the url of the parameter but it doesn't work.It gives me a "[objet XMLHttpRequest]" error.here is the code on website A:
