我试图创建一个可管道的运算符,并在不满足特定条件的情况下抛出该异常。但是,我无法抛出并捕获该错误。
这是我的管道:
// My custom pipeable
export function waitFor<T>(thisToComplete$: Observable<any>) {
return (beforeRunningThis$: Observable<T>) => {
return new Observable<T>(observer =>
thisToComplete$.pipe(first()).subscribe(x => {
if (x !== 'Success') {
console.log('Non success result encountered');
return throwError('Rando IO Error');
}
return beforeRunningThis$.subscribe(observer);
})
);
}
}
使用代码:
const preReq$ = timer(1000);
const dataReq$ = getData();
try {
dataReq$
.pipe(waitFor(preReq$), catchError(x => {
console.log('Code here reached');
return of('Error was handled either here')
}))
.subscribe(x => console.log(`I have result ${x.toString()}`));
} catch (e) {
console.log('Error was handled here');
}
但是上述控制台均未记录。
这里是stackblitz
答案 0 :(得分:1)
由于您使用的是Observable构造
observer.error
是你扔的方式
if (x !== 'Success') {
console.log('Non success result encountered');
observer.error('Rando IO Error');
}
答案 1 :(得分:1)
您需要摆脱管道运算符中的错误处理。基本上,您在这里遇到错误:
dataReq$
.pipe(waitFor(preReq$))
.subscribe(x => {
console.log(`I have result ${x.toString()}`);
}, error => {
console.log('Code here reached');
// handle error
});