萌新求教！Vue this.$nextTick() 之后再执行，导致渲染不准确各位有什么建议？

Question

问题this.$nextTick(()=>{this.$refs.multipleTable.toggleAllSelection()this.isChecked=false})注意：this.$refs.multipleTable.toggleAllSelection()必须放在this.$nextTick()里面执行才能生效如上所示我希望先执行this.$refs.multipleTable.toggleAllSelection()，再this.isChecked=false，但是经过测试发现this.isChecked=false总是在前执行导致页面渲染不正确弱鸡解决法this.$nextTick(()=>{this.$refs.multipleTable.toggleAllSelection()setTimeout(()=>{this.isChecked=false},100);})这样可以解决这个问题，但是总感觉不太好尝试过封装asyncawit但是无效如下：constnextTickAsync=()=>{returnnewPromise((resolve,reject)=>{this.$nextTick(()=>{this.$refs.multipleTable.toggleAllSelection();resolve(true)})})}constisCheckedAsycn=()=>{returnnewPromise((resolve,reject)=>{this.$nextTick(()=>{this.isChecked=falseresolve(true)})})}(async()=>{awaitnextTickAsync()awaitisCheckedAsycn()})()求问解决方法？如何先执行完this.$nextTick()在执行后续的this.isChecked=false

Answer 1

this.$refs.multipleTable.toggleAllSelection()
this.$nextTick(()=>{
this.isChecked=false
})
                            

Answer 2

注意：this.$refs.multipleTable.toggleAllSelection()必须放在this.$nextTick()里面执行才能生效