我有一个来自某个库的函数,该函数返回我想从另一个函数调用的Observable。我需要将该Observable传播到多个函数调用中。这是我的代码的结构:
extractSignature = (xml, signatureCount = 1) => {
const observ = this.generateDigest(signedContent, alg).pipe(map(digest => {
const sigContainer = {
alg: alg,
signature: signatureValue,
signedContent: signedContent,
digest: digest
};
console.log('sigContainer inside pipe: ');
console.log(sigContainer);
return sigContainer;
}));
return observ;
}
dissasemble(xml): Observable<SignatureContainerModel[]> {
const observables: Observable<any>[] = [];
for (let i = 1; i <= count; i++) {
const extractSigObservable = this.extractSignature(xml, i);
console.log('extractSigObs inside pipe: ');
console.log(extractSigObservable);
const observ = extractSigObservable.pipe(map(sigContainer => {
console.log('sigContainer inside pipe: ');
console.log(sigContainer);
const hashContainers: HashContainerModel[] = [];
const hashContainer: HashContainerModel = new HashContainerModel();
hashContainer.digestAlgorithm = sigContainer.alg;
hashContainer.bytes = sigContainer.digest;
hashContainers.push(hashContainer);
const signatureContainer: SignatureContainerModel = {
hashContainers: hashContainers,
signature: sigContainer.signature
};
console.log('observable inside pipe: ');
console.log(observ);
}));
observables.push(observ);
}
return forkJoin(observables);
}
verify() {
this.sigExec.dissasemble(this.fileContent).subscribe((signatureContainers: SignatureContainerModel[]) => {
// signatureContainers is [undefined] here
console.log('Sig Containers: ');
console.log(signatureContainers);
this.verifyHash(signatureContainers);
});
}
signatureContainers 变量在订阅中[未定义]。我不确定是什么问题,因为当我检查在map函数中编写的所有日志时,它们似乎都很好
答案 0 :(得分:2)
forkJoin
上的RXJS文档:
请注意,如果提供给
forkJoin
的内部可观察值有任何一个错误,则您将丢失其他任何可观察值的值,如果您没有正确地捕获内部可观察值的错误,则该值将已经或已经完成。如果仅关注所有内部可观测对象是否成功完成,则可以从外部捕获错误。 https://www.learnrxjs.io/learn-rxjs/operators/combination/forkjoin
您有可能在管道内部出错并且这些值丢失了。
此外,我注意到您没有从管道返回任何东西。这也可能是个问题。