jQuery print value of variable

I am new to jQuery and I need some assistance with very simple code. So, I have a string variable that I want to print in the p with listprice class.

<script> 

$(document).ready( function() {
var string = "US $257.31";
$('.listprice开发者_高级运维).append(string);
}
</script>

Any ideas how can I achieve that?

Thank you,

H

You could do this

var string = "US $257.31";
$('.listprice').html(string);

http://jsfiddle.net/jasongennaro/akpn3/

Of course, this assumes there is nothing in the p as it would replace everything in there.

To add it on a separate line when the p contains content, do this

$('.listprice').append('<br />' + string);

http://jsfiddle.net/jasongennaro/akpn3/1/

EDIT

As per your comment

Do you know how can I do that with regular javascript? I think they are blocking jquery in there.

You could do this

var string = "US $257.31";

var a = document.getElementsByTagName('p');

a[0].innerHTML += string;

http://jsfiddle.net/jasongennaro/akpn3/2/

This finds the first p and adds the variable string.

You're missing a '.

$('.listprice).append(string);

should be

$('.listprice').append(string);

EDIT: Also missing a ) at the end.

Other than that it's fine.

You've it right except for some typos (and I added tag specifier to make the lookup bit faster):

$(document).ready( function() {
   var string = "US $257.31";
   $('p.listprice').append(string);
}

The above code will append US $275.31 to all elements with the listprice tag.

If you want to append it only to p tags with listprice, specify $('p.listprice').append(....

Oh, and remember to close your quotes after the identifier (listprice).

You aren't appending (adding an element to another). You just want to change the p text.

$(".listprice").text(string);

You Can Also Print Your Variable Value Between String Look like...

var filename = $('#resume').val();
$("#msg_show").text("Your resume "+filename+" has uploaded successfully. If you wish to upload a different document, please click the “Submit your Resume” button and select your file again.");