我有一个指令,通过添加小数点作为用户类型来过滤货币。
我的问题是它不允许有负面影响。如果允许用户输入一个“-”符号,并让它始终出现在数字的开头,或者键入a '+‘并让它消失,那就太好了。
需要:
-看看我的柱塞。它有input = number,但不允许负数:
app.directive('format', ['$filter', function ($filter) {
return {
require: 'ngModel', //there must be ng-model in the html
link: function (scope, elem, attr, ctrl) {
if (!ctrl) return;
ctrl.$parsers.unshift(function (viewValue, modelValue) {
var plainNumber = viewValue.replace(/[^-+0-9]/g,'');
// use angular internal 'number' filter
plainNumber = $filter('number')(plainNumber / 100, 2).replace(/,/g, '');
// update the $viewValue
ctrl.$setViewValue(plainNumber);
// reflect on the DOM element
ctrl.$render();
// return the modified value to next parser
return plainNumber;
});
}
};
}]);发布于 2016-07-12 17:18:04
这将完成您希望它做的事情:
var app = angular.module('App',[]);
app.controller('MainCtrl', function ($scope) {
});
app.directive('format', ['$filter', function ($filter) {
return {
require: 'ngModel', //there must be ng-model in the html
link: function (scope, elem, attr, ctrl) {
if (!ctrl) return;
ctrl.$parsers.unshift(function (viewValue, modelValue) {
var plainNumber = viewValue.replace(/[^-+0-9]/g,'');
var newVal = plainNumber.charAt(plainNumber.length-1);
var positive = plainNumber.charAt(0) != '-';
if(isNaN(plainNumber.charAt(plainNumber.length-1))){
plainNumber = plainNumber.substr(0,plainNumber.length-1)
}
//use angular internal 'number' filter
plainNumber = $filter('number')(plainNumber / 100, 2).replace(/,/g, '');
if(positive && newVal == '-'){
plainNumber = '-' + plainNumber;
}
else if(!positive && newVal == '+'){
plainNumber = plainNumber.substr(1);
}
plainNumber.replace('.', ',');
//update the $viewValue
ctrl.$setViewValue(plainNumber);
//reflect on the DOM element
ctrl.$render();
//return the modified value to next parser
return plainNumber;
});
}
};
}]);只需从以下位置删除type="number“:
<input ng-model="amount" format="number" />https://stackoverflow.com/questions/38334580
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