如何在 Angular JS 的弹出窗口外单击时隐藏 uib-popover？

Question

<button class="btn btn-default blue btn-lg h-ask btn-block" popover-placement="bottom" uib-popover-template="popover_home.templateUrl">ASK</button>

The above is the code for angular js popover, in the html template of the Popover two input fields also there, so when user submits the html form inside popover i will have control over the popover, but i need to manually close when user opens this popover and does not do any things on the popover(eg: clicking outside the popover).

Any idea about this? I used "popover-trigger='focus'",but it doesnt allows me to write into the input field (whenever hover the popover it closes)

Answer 1

Don't know which version of angular-ui boostrap you are using, but I'm using the latest (1.1.1) and adding this do the trick:

popover-trigger="outsideClick"

You can see in this demo.

popover close when click outside

Hope it helps!

Answer 2

For those who don't read comments:-) Based on @kuanyui comment if it dosen't work with the accepted answer try to add simple cotes '

popover-trigger=" 'outsideClick' "