如何断开getter中派生对象与Vuex存储状态的连接?
我有一个像这样的Vuex getter:如何断开getter中派生对象与Vuex存储状态的连接?,vuex,Vuex,我有一个像这样的Vuex getter: sectionsAndSubSectionsJoined(state) { var sections = state.aBMudbone.sections; var subsections = state.aBMudbone.subSections; var combinedSections = []; _.forEach(sections, function (section)
sectionsAndSubSectionsJoined(state) {
var sections = state.aBMudbone.sections;
var subsections = state.aBMudbone.subSections;
var combinedSections = [];
_.forEach(sections, function (section) {
var subSectionsFiltered = _.filter(subsections, ['sectionId', section.sectionId]);
var sectionAndSubSections = section;
sectionAndSubSections.children = subSectionsFiltered;
combinedSections.push(sectionAndSubSections);
});
return combinedSections
}
在控制台中,我得到以下错误:
[vuex] do not mutate vuex store state outside mutation handlers.
显示此错误的行为:
sectionAndSubSections.children = subSectionsFiltered;
据我所知,即使节
,子节
和节和子节
都是新对象,因为它们是从存储状态创建的,它们受到一些绑定的约束,因此在我尝试添加到此状态时会发出警告
如何断开此新状态与存储状态的连接,以避免收到此警告?