Question

我创建了一个预先输入

<input type="text" placeholder="Search" onfocus="this.placeholder=''" onblur="this.placeholder='Search'" ng-model="selected" typeahead="Customer as Customer.name for Customer in typeaheadSrc | filter:$viewValue" />

以下数据，

$scope.typeaheadSrc = [{
id:1,
name:'data1'
},
{
id:2,
name:'data2'
}];

上面的代码完全正常，但我想动态传递列名。在上面的代码我给了'Customer.name'，我不想硬编码列名。我在范围变量中获得列名，现在我想在此处指定它。我尝试按以下方式执行，但它无法正常工作

$scope.columndata = 'id';  

 <input type="text" placeholder="Search" onfocus="this.placeholder=''" onblur="this.placeholder='Search'" ng-model="selected" typeahead="Customer as Customer.columndata for Customer in typeaheadSrc | filter:$viewValue" />

而不是名字字段我正在分配columndata，但它不起作用。请告诉我怎么做。

Answer 1

您可以尝试：

<input type="text" placeholder="Search" onfocus="this.placeholder=''" onblur="this.placeholder='Search'" ng-model="selected" typeahead="Customer as Customer[{{columndata}}] Customer in typeaheadSrc | filter:$viewValue" />

此处 columndata 将是范围变量

具有angularjs中动态字段名称的typeahead

1 个答案: