如果我点击其中一个列表(满足切换功能条件),那么它应该显示或不显示下拉菜单。到目前为止,方法切换更改了chrome控制台内dropDownOpen的值,但视图中没有显示任何更改,代码:
控制台显示值已更改,并且未在模板中更改!
<ul>
<li v-for="(navHeading, index) in headerMenuItems" @click="toggle(navHeading.name)" :class="{'open': dropDownOpen}">
<icon :glyph="navMenuIconNames[index]" :viewBox="navMenuIconNames[index].viewBox" :height="25" :width="50"></icon>
<a class="navigation-title" href="#">{{ navHeading.name }}</a>
<ul class="drop-down-menu">
<li v-for="dropDownMenuItem in navHeading.children">
<a class="drop-down-menu-item-title">{{ dropDownMenuItem.name }}</a>
</li>
</ul>
</li>
</ul>
脚本:
data() {
const dropDownOpen = false
return { dropDownOpen }
},
methods: {
toggle: (dropDownName) => {
if (dropDownName === 'Meer') {
this.dropDownOpen = !this.dropDownOpen
}
console.log(this.dropDownOpen)
return this.dropDownOpen
}
}
CSS:
<style scoped>
.open .drop-down-menu {
display: block;
background-color: tomato;
}
.drop-down-menu {
display: none;
}
</style>
答案 0 :(得分:2)
this
没有指向vue实例,所以使用普通函数而不是胖箭头函数
methods: {
toggle: function(dropDownName) {
if (dropDownName === 'Meer') {
this.dropDownOpen = !this.dropDownOpen
}
console.log(this.dropDownOpen)
return this.dropDownOpen
}
请参阅warning为什么不应将=>
箭头函数与方法
在模板中添加v-show
这样的属性。
<ul v-show="dropDownOpen" class="drop-down-menu">
答案 1 :(得分:0)
不确切知道,但这有效:
methods: {
toggle(dropDownName) {
if (dropDownName === 'Meer') {
this.dropDownOpen = !this.dropDownOpen
}
return this.dropDownOpen
}
},
和模板:
<ul>
<li v-for="(navHeading, index) in headerMenuItems" @click="toggle(navHeading.name)">
<icon :glyph="navMenuIconNames[index]" :viewBox="navMenuIconNames[index].viewBox" :height="25" :width="50"></icon>
<span>{{ dropDownOpen }}</span>
<a class="navigation-title" href="#">{{ navHeading.name }}</a>
<ul v-show="dropDownOpen" class="drop-down-menu">
<li v-for="dropDownMenuItem in navHeading.children">
<a class="drop-down-menu-item-title">{{ dropDownMenuItem.name }}</a>
</li>
</ul>
</li>
</ul>