Question

如果我有这样的代码：

let result;
try {
    result = doSomethingThatCanThrow();
} catch (e) {
    throw new Error(`Thrown error because <some info> ${e}`)
}

// do something with result if didn't throw

这是JS中的有效模式吗？我不喜欢使用let。我更喜欢const但我不能在这种情况下使用它，因为它不会在try-catch块之外可用。

我可以将整个代码放在try块中并删除let但是在我看来它会更难阅读：

try {
    const result = doSomethingThatCanThrow();
    // do something with result if didn't throw
} catch (e) {
    throw new Error(`Thrown error because <some info> ${e}`)
}

请记住，它总能增长。

此外，我总是可以跳过重新抛出，但我想为错误添加更多信息。

哪个选项更好，为什么？还有其他可能性吗？

Answer 1

您可以使用finally

(async(doSomethingThatCanThrow) => {

  let result;
  try {
    result = await doSomethingThatCanThrow();
  } catch (e) {
    throw new Error(`Thrown error because <some info> ${e}`)
  } finally {
    if (result !== undefined) {
      // do something with result if didn't throw
      console.log(result);
    } else {
      console.log(result);
    }
  }
  // do stuff 
})(() => Promise.reject(123))
.catch(err => console.error(err));

Answer 2

尝试测试条件，看看变量结果是否为空！

＆＃13;

var result
try {
    result = "This time it worked!";
} catch (e) {
    throw new Error(`Thrown error because <some info> ${e}`)
}

if (result) {
    console.log(result)
}

＆＃13;

var result
try {

} catch (e) {
  }

// do something with result if didn't throw

if (!result) {
    // var result undefined
    // Result not thrown!
    console.log("Failed")
}

＆＃13;

Answer 3

在这种情况下，我认为避免let的唯一真实方法（除了建议使用var作为其他响应，否则不要这样做）是将try-catch提取到函数中并返回直接用try块得到结果，即

const doSomethingThatCanThrowAndRethrowWithUpdatedMessageIfFails = () => {
  try {
    return doSomethingThatCanThrow();
  } catch (e) {
    throw new Error(`Thrown error because <some info> ${e}`);
  }
}

const result = doSomethingThatCanThrowAndRethrowWithUpdatedMessageIfFails();

尝试抓住 - 重新抛出 - 正确的模式

3 个答案: