Question

我有这样的条件：

net

真实场景中发生的事情是

对于Eg。 <td ng-repeat="sal in name.salaries" ng-if="name.avalue>sal.annual_value"> Value must be less than or equal to Balance Amount </td> 因为它不是parseInt，这就是它考虑name.value=1200 sal.annual_value=42。

的原因

我如何在AngularJS表达式中42 > 1200？

Answer 1

您可以在parseInt：

中使用ng-if

<td 
    ng-repeat="sal in name.salaries"
    ng-if="parseInt(name.avalue)>parseInt(sal.annual_value)">
        Value must be less than or equal to Balance Amount
</td>

或者您可以在控制器中创建一个检查此功能的功能：

<td 
    ng-repeat="sal in name.salaries"
    ng-if="isGreater(name.avalue, sal.annual_value)">
        Value must be less than or equal to Balance Amount
</td>

$scope.isGreater = function(val1, val2) {
    return parseInt(val1) > parseInt(val2);
};

Answer 2

我会使用范围函数进行验证：

{...}
scope.checkSalaries = function(avalue, annual_value) {
   return avalue > annual_value;
}
{...}

并在html中：

<td ng-repeat="sal in name.salaries"
    ng-if="checkSalaries(name.avalue,sal.annual_value)">
       Value must be less than or equal to Balance Amount
</td>

Answer 3

如果您想在视图中使用parseInt()方法，则可以在控制器中定义$scope.number = Number; $范围，然后在其中使用其内置方法<td ng-repeat="sal in name.salaries" ng-if="number.parseInt(name.avalue)>number.parseInt(sal.annual_value)"> Value must be less than or equal to Balance Amount</td>图。

e.g。

控制器：

public class Test {
int x =10;

public static void main(String[]args)
{
    Test t1 = new Test();
    System.out.println(t1.x);
}

查看：

<input *ngIf="fieldForm.value.type == 'phone'" pattern="^([0-9]{2}\s?){5}$" type="text" class="form-control" formControlName="field" [(ngModel)]="value">

Answer 4

您也可以尝试：

<td 
    ng-repeat="sal in name.salaries"
    ng-show="isGreater(name.avalue, sal.annual_value)">
        Value must be less than or equal to Balance Amount
</td>

$scope.isGreater = function(val1, val2) {
    return parseInt(val1) > parseInt(val2);
};

如何在angularjs表达式中parseInt

4 个答案: