how do I indicate a javascript generated ID using javascript/jQuery?

I'm counting the number of p tags in a certain div with individual IDs where the p tags and the IDs are randomly generated by php and in varying numbers but in a patterned ID (otherUser1, otherUser2, otherUser3, etc.)

for (var i = 1; i <= numUsers; i++) {
    var tagId = '#otherUser' + i + '';
    document.write(tagId);
    var userName = $(tagId).html();
    document.write(userName);
}

whe开发者_如何学Pythonn I document write the userName variable above, it gives me null because just inserting the variable tagId in the jQuery brackets doesn't do the job. How do I do this properly? - I don't mind either using javascript or jQuery, either is fine.

I have tried this using jQuery. This is very well working with my localhost. But, still I don't this will solve your issue or not.

I have implemented javaScript as :

            function submit(){
                var numUsers=3;
                for (var i = 1; i <= numUsers; i++) {
                    var tagId = '#otherUser' + i;
                    var userName = $(tagId).text().toString();
                    $('#user').append("Username" + i + " : " + userName + "<br/>");
                }
            }

And that your HTML is :

        <div id="otherUser1">userName1</div>
        <div id="otherUser2">userName2</div>
        <div id="otherUser3">userName3</div>

        <input type="button" value="Click Me!!" onclick="submit();"/>
        <div id="user"></div>

This will display the DIVs currently. But, in place of this HTML, there would be your PHP auto-generated DIVs, that will not be displayed later.

I have also created this fiddle. I don't know why this is not working even if being the same !!!

Please correct me if I am wrong anywhere.

Thanking You.

EDIT : perhaps the jsfiddle not supporting the functions, (so removed function line from javaScript) ! So, updated the fiddle & that is working properly.