Question

by Angular2，有没有什么办法来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

更好的是在这种情况下使用的形式，你只需要检查表单是否有效不是禁用按钮，并相应地进行切换。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>