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73 lines
2.2 KiB
73 lines
2.2 KiB
2 months ago
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import Vue from 'vue'
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import Vuex from 'vuex'
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Vue.use(Vuex)
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let lifeData = {};
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try {
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// 尝试获取本地是否存在lifeData变量,第一次启动APP时是不存在的
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lifeData = uni.getStorageSync('lifeData');
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} catch (e) {
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}
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// 需要永久存储,且下次APP启动需要取出的,在state中的变量名
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let saveStateKeys = ['vuex_user', 'vuex_token'];
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// 保存变量到本地存储中
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const saveLifeData = function(key, value) {
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// 判断变量名是否在需要存储的数组中
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if (saveStateKeys.indexOf(key) != -1) {
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// 获取本地存储的lifeData对象,将变量添加到对象中
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let tmp = uni.getStorageSync('lifeData');
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// 第一次打开APP,不存在lifeData变量,故放一个{}空对象
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tmp = tmp ? tmp : {};
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tmp[key] = value;
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// 执行这一步后,所有需要存储的变量,都挂载在本地的lifeData对象中
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uni.setStorageSync('lifeData', tmp);
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}
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}
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const store = new Vuex.Store({
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// 下面这些值仅为示例,使用过程中请删除
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state: {
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// 如果上面从本地获取的lifeData对象下有对应的属性,就赋值给state中对应的变量
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// 加上vuex_前缀,是防止变量名冲突,也让人一目了然
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// 如果vuex_version无需保存到本地永久存储,无需lifeData.vuex_version方式
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// vuex_version: '1.0.1',
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vuex_token: lifeData.vuex_token ? lifeData.vuex_token : '',
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vuex_user: lifeData.vuex_user ? lifeData.vuex_user : {},
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vuex_dept: '',
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vuex_sex: [{
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label: '男',
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value: 0
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}, {
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label: '女',
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value: 1
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}]
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},
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mutations: {
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$uStore(state, payload) {
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// 判断是否多层级调用,state中为对象存在的情况,诸如user.info.score = 1
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let nameArr = payload.name.split('.');
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let saveKey = '';
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let len = nameArr.length;
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if (nameArr.length >= 2) {
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let obj = state[nameArr[0]];
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for (let i = 1; i < len - 1; i++) {
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obj = obj[nameArr[i]];
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}
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obj[nameArr[len - 1]] = payload.value;
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saveKey = nameArr[0];
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} else {
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// 单层级变量,在state就是一个普通变量的情况
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state[payload.name] = payload.value;
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saveKey = payload.name;
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}
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// 保存变量到本地,见顶部函数定义
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saveLifeData(saveKey, state[saveKey])
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}
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}
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})
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export default store
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