jQuery is just a javascript library that makes some extra stuff available when writing javascript – so there is no reason to use jQuery for declaring variables. Use “regular” javascript: EDIT: As Canavar points out in his example, it is also possible to use jQuery to get the form value: given that the text box … Read more
Think about what means. || means “or.” The negation of this is (by DeMorgan’s Laws): In other words, the only way that this could be false is if a state equals 10, 15, and 19 (and the rest of the numbers in your or statement) at the same time, which is impossible. Thus, this statement will always be true. State 15 … Read more
I don’t believe there’s a way to query the mouse position, but you can use a mousemove handler that just stores the information away, so you can query the stored information. But almost all code, other than setTimeout code and such, runs in response to an event, and most events provide the mouse position. So your code that needs to know where … Read more
Place your widget.js after core.js, but before any other jquery that calls the widget.js file. (Example: draggable.js) Precedence (order) matters in what javascript/jquery can ‘see’. Always position helper code before the code that uses the helper code.
Place the following in your jQuery mouseover event handler: To set both color and size at the same time: You can set any CSS attribute using the .css() jQuery function.
Did you load jQuery in head section? Did you load it correctly? This code assumes jquery.js is in scripts directory. (You can change file name if you like) You can also use jQuery as hosted by Google: As per your comment: Apparently, your web server is not configured to return jQuery-1.6.1.js on requesting /webProject/jquery-1.6.1.js. There may be numerous reasons for this, such as wrong file … Read more
Try like this… or unobtrusively using just
How would I create an object in jQuery, and then proceed to make a couple of different instances of this object I.e Create an object named box which holds a variable called color. And then make a couple of instances of this object with different stored colours.
The trouble is that you can not return a value from an asynchronous call, like an AJAX request, and expect it to work. The reason is that the code waiting for the response has already executed by the time the response is received. The solution to this problem is to run the necessary code inside … Read more
You’re thinking too complicated. It’s actually just $(‘#’+openaddress).