AngularJS Typeahead：如何在selectTraining中获取选中对象？

需求 #

使用AngularJS的typeahead组件，需实现两个目标：

• 在selectTraining(...)方法中获取选中的完整trainingEntry对象
• 保持ngModel绑定为training.topic

原代码 #

<input 
    type="text" 
    class="form-control" 
    data-ng-model="training.topic"
    typeahead="trainingEntry.topic as trainingEntry.topic + ' ' + trainingEntry.location + ' ' + trainingEntry.trainingTyp + ' von ' + trainingEntry.from + ' bis ' + trainingEntry.to for trainingEntry in findTrainingsOfInstitutionByTerm($viewValue) | limitTo:5" 
    typeahead-on-select="selectTraining(trainingEntry)" />

解决方法 #

将typeahead-on-select的参数替换为$item，$item代表选中的完整trainingEntry对象，同时原有的ngModel绑定逻辑不受影响（因为typeahead表达式中trainingEntry.topic as ...已经指定了绑定到ngModel的值为trainingEntry.topic）。

修改后代码 #

<input 
    type="text" 
    class="form-control" 
    data-ng-model="training.topic"
    typeahead="trainingEntry.topic as trainingEntry.topic + ' ' + trainingEntry.location + ' ' + trainingEntry.trainingTyp + ' von ' + trainingEntry.from + ' bis ' + trainingEntry.to for trainingEntry in findTrainingsOfInstitutionByTerm($viewValue) | limitTo:5" 
    typeahead-on-select="selectTraining($item)" />

内容的提问来源于stack exchange，提问作者quma