Question

Angular2的

，有没有办法重构两个按钮元素的绑定表达式？

<input #c1="ngModel">
<input #c2="ngModel">
<input #c3="ngModel">
<input #c4="ngModel">
<input #c5="ngModel">
<input #c6="ngModel">
<button [disabled]='c1.invalid || c2.invalid || c3.invalid || c4.invalid || c5.invalid || c6.invalid'>b1</button>
<button [disabled]='c1.invalid || c2.invalid || c3.invalid || c4.invalid || c5.invalid || c6.invalid'>b2</button>

Answer 1

最好在这种情况下使用FORM，您只需要检查表单是否有效，以禁用按钮并相应地切换。像这样

<form [formGroup]="customForm">
  <input formControlName='c1'>
  <input formControlName='c2'>
  <input formControlName='c3'>
  <input formControlName='c4'>
  <input formControlName='c5'>
  <input formControlName='c6'>
  <button [disabled]='!customForm.valid'>b1</button>
  <button [disabled]='!customForm.valid'>b2</button>
</form>

角度2避免重复结合表达

1 个答案: