vue.js - 如何在单击按钮时关闭 Vuesax PopUp 屏幕？

Question

我在 VueSax PopUp 屏幕中创建了一个表单，并且在单击按钮时提交了该表单，但是一旦提交 from，PopUp 就不会关闭。请帮助编写代码。

单击提交按钮后，它会触发下面的方法。

  createRecord() {

        this.$vs.loading();


          jwt.createLeads(this.newRecord).then((response) => {

            this.$vs.loading.close();

            this.$store.dispatch("userManagement/upsertToState", { type: "Leads", data: response.data.leads });

            this.newRecord.leadtype = this.newRecord.first_name = this.newRecord.last_name = this.newRecord.email_id = this.newRecord.phone_number = this.newRecord.location = this.newRecord.postcode = this.newRecord.lead_address = this.newRecord.gender = this.newRecord.lead_source = this.newRecord.required_sub = this.newRecord.teach_mode = this.newRecord.location_pref = this.newRecord.lead_frequency = this.newRecord.session_duration = this.newRecord.pref_days = this.newRecord.pref_time_slot = this.newRecord.tutor_pref = this.newRecord.lead_owner = this.newRecord.notes = ''

           alert("Lead Creation Success");
           location.reload();
        }).catch((error) => {
            console.log(error)

            this.$vs.loading.close();


            this.$vs.notify({
                title: 'Error',
                text: 'There was an error creating the Lead',
                iconPack: 'feather',
                icon: 'icon-alert-circle',
                color: 'danger'
            });
        });

    },

Answer 1

在 API 部分的 vuesax 文档中，我可以看到“active.sync”属性通过更改此属性的值来确定弹出窗口是否处于事件状态(可见)，您可以关闭弹出窗口。 Vuesax Documentation