[webkit-reviews] review denied: [Bug 29961] Position of "Close" button in docked mode doesn't match the position of window "Close" button in detached mode : [Attachment 40440] propsed change

[bugzilla-daemon at webkit.org]

Timothy Hatcher <timothy at hatcher.name> has denied Mikhail Naganov
<mnaganov at chromium.org>'s request for review:
Bug 29961: Position of "Close" button in docked mode doesn't match the position
of window "Close" button in detached mode
https://bugs.webkit.org/show_bug.cgi?id=29961

Attachment 40440: propsed change
https://bugs.webkit.org/attachment.cgi?id=40440&action=review

------- Additional Comments from Timothy Hatcher <timothy at hatcher.name>


> +    // Place "Close" button in docked mode on the same side as the window
"Close" button in detached mode.
> +    var closeElement;
> +    if (InspectorController.platform().substr(0, 3) == "mac") {
> +	   closeElement = document.getElementById("close-left");
> +    } else {
> +	   closeElement = document.getElementById("close-right");
> +    }
> +    var closeButton = document.createElement("button");
> +    closeButton.id = "close-button";
> +    closeButton.addEventListener("click", this.close, true);
> +    closeElement.appendChild(closeButton);

You can do more this in CSS. Using the .platfor-mac and .platform-win classes.

So attach the listener to both elements and let CSS decided what one to show
and hide. And have the button in both.