【发布时间】:2019-12-04 08:15:11
【问题描述】:
在发送如下请求时,而不是使用传统的try/catch 来处理错误
let body;
try {
const response = await sendRequest(
"POST",
"/api/AccountApi/RefundGetStatus",
JSON.stringify(refundParams),
undefined,
headers
);
body = response.body;
} catch (error) {
log("error", "An error occurred sending a request", { error });
throw createError("An error has occurred", DefaultErrors.API_ERROR);
}
我在想是否可以像下面这样重写错误处理来获得相同的结果。
const response = await sendRequest(
"POST",
"/api/AccountApi/RefundGetStatus",
JSON.stringify(refundParams),
undefined,
headers
).catch((err: any) => {
log("error", "An error occurred sending a request", { err });
throw createError("An error has occurred", DefaultErrors.API_ERROR);
});
上面的代码块是否能够处理未定义的错误以及JSON.parse 错误?
【问题讨论】:
-
不,这不是错误捕获的工作原理。
标签: javascript typescript error-handling