ngx-bootstrap工具提示中的isDisabled属性

Question

我已经实现了ngx-bootstrap的工具提示，工具提示在正常情况下工作正常。但我想在有条件的基础上显示工具提示。

当我包含isDisabled属性时，我在以下任何一种情况下都看不到工具提示。我错过了什么吗？

<span tooltip="{{toolTipText}}" triggers="" #pop="bs-tooltip" placement="bottom" isDisabled="true"></span>
<button type="button" (click)="toggleSelect()" (mouseenter)="pop.show()" (mouseleave)="pop.hide()">
    <span class="pull-left">Click Here!</span>
</button>

（要么）

<span tooltip="{{toolTipText}}" triggers="" #pop="bs-tooltip" placement="bottom" isDisabled="false"></span>
<button type="button" (click)="toggleSelect()" (mouseenter)="pop.show()" (mouseleave)="pop.hide()">
    <span class="pull-left">Click Here!</span>
</button>

Answer 1

您需要使用[bracket syntax]将输入参数评估为javascript而不是静态值。

意思是你需要做[isDisabled]="false/true"instead的isDisabled="false/true"。做后面的例子实际上会传入字符串'true'或'false'

这是一个有效的PLUNKER

<div style="margin: 100px;">
  <button type="button" class="btn btn-default btn-secondary"
    tooltip="isDisable = true"
    [isDisabled]="true"
    placement="top">
    isDisable = true
  </button>
</div>
<div style="margin: 100px;">
  <button type="button" class="btn btn-default btn-secondary"
    tooltip="isDisable = false"
    [isDisabled]="false"
    placement="top">
    isDisable = false
  </button>
</div>