当弹出窗口悬停 (ngx-bootstrap) 时，如何让弹出窗口保持活动状态？

Question

当弹出窗口悬停 ngx-bootstrap/popover angular 2 包时，如何让弹出窗口保持活动状态？

感谢您的回复

Answer 1

工具提示是你需要的：

https://valor-software.com/ngx-bootstrap/#/tooltip

<button
    class="btn btn-primary"
    type="button"
    tooltip="This text will be shown when the button is hovered">
    Hover me!
</button>

Answer 2

您可以参考ngx-popover

    import { PopoverModule } from 'ngx-bootstrap/popover';// or import { PopoverModule } from 'ngx-bootstrap';

@NgModule({
  imports: [PopoverModule.forRoot(),...]
})
export class AppModule(){}

模板

<button type="button" class="btn btn-primary"
        popover="Vivamus sagittis lacus vel augue laoreet rutrum faucibus.">
  Live demo
</button>

组件

import { Component } from '@angular/core';

@Component({
  selector: 'demo-popover-basic',
  templateUrl: './basic.html'
})
export class DemoPopoverBasicComponent {}

当弹出窗口悬停 (ngx-bootstrap) 时，如何让弹出窗口保持活动状态？

how can i keep popover alive while the popover is being hovered for (ngx-bootstrap)?

ngx-bootstrap

angular