'showPopup' method is working wrong

[Ton That Hung]

Hi Syncfusion Team,
I'm using the DropdownList component and want to use a Button to show the popup of the DropdownList component. (When I click the button, the popup is opened). I am using showPopup property to implement. It's working well when I click the button slowly. But when I click quickly and many times in a short time, the popup doesn't show.

Link sample:https://stackblitz.com/edit/react-cdsqbq?file=index.js

How I can fix it.​

Thanks.

[UdhayaKumar Duraisamy]

Hi Ton,

We suspect that you are trying to open the popup even though the popup is in the opened state. So, we suggest you call the ShowPopup method when the popup is not opened as in the below code snippet.

function onClickButton() {

    if (!ref.current.isPopupOpen) {

      ref.current.showPopup();

    } else {

      ref.current.hidePopup();

    }

  }

 

Sample : https://stackblitz.com/edit/react-cdsqbq-c9amcw?file=index.js,index.html

Kindly try the above sample and let us know if this meets your requirement.

Regards,

Udhaya Kumar D

[Ton That Hung]

That's great. Thanks for your support.

[Buvana Sathasivam]

Most welcome. Please let us know if you need any further assistance.