javascript - 如何在异常捕获中获取数据库引用？

Question

出现异常时需要关闭并删除当前indexeddb 代码行如下所示，

export async function getCurrUser(window) {
    ...

    let openRequest = indexedDB.open('userData',2);
    openRequest.onsuccess =  e => {
       let db = e.target.result;//this is not accessable in the catch block
        try{}
        catch(e){
            //close db and delete it. but I cannot get db reference here
            //I can't db.close() here
        }
    };
}

有什么有效的方法可以在异常捕获中获取db吗？

Answer 1

您可以简单地使用finally(始终执行)并用 bool 值控制它:

export async function getCurrUser(window) {
...

let openRequest = indexedDB.open('userData',2);
let error = false;
openRequest.onsuccess =  e => {
   let db = e.target.result;//this is not accessable in the catch block
    try{}
    catch(e){
         error = true;
        //print exception
    }finally{
      if (error){
      db.close();
      }
   }
};
}

了解更多信息:https://www.w3schools.com/js/js_errors.asp