Possibility to use existing wrapper (parent) element. #51
Conversation
hideShowPassword.js
Outdated
| $(options.element).addClass(options.className).css(options.styles) | ||
| ); | ||
| this.element.css(options.innerElementStyles); | ||
| if (!options.useExistingParent) { |
There was a problem hiding this comment.
Instead of introducing useExistingParent to this already large set of options, could this be simplified by checking the value of the wrapper.element option and only performing the wrap if it's truthy? Then, if you wanted to omit the wrapper element, you could simply pass a falsy value for that option.
There was a problem hiding this comment.
Yeah it's good idea to use wrapper.element to either create a wrapper if it's string value e.g. <div> or just use it if it's jquery pointer to an DOM object. Let me redesign this next week. Thanks.
|
@erikjung redesigned. I used |
| ); | ||
| this.wrapperElement = this.element.parent(); | ||
| this.element.css(options.innerElementStyles); | ||
| if (!options.element.jquery) { |
There was a problem hiding this comment.
jQuery elements are also valid jQuery selectors. So instead of checking if the selector is a jQuery object, we could instead check to make sure it's not already the parent element before wrapping.
Example:
// Keep this separate, as you've already done
this.element.css(options.innerElementStyles);
// Create the wrapper element, using `options.element`
this.wrapperElement = $(options.element).addClass(options.className).css(options.styles);
// If this wrapper element doesn't already have the element as a descendent...
if (!this.wrapperElement.has(this.element).length) {
// ...then wrap the descendent and override the local reference
this.element.wrap(this.wrapperElement);
this.wrapperElement = this.element.parent();
}
|
Closing this PR due to lack of activity. |
Fixes #50