javascript - 如何保持循环运行并发送所有消息？

Question

我正在尝试发送超过 1000 条消息。问题发生在 twilio catch 中:当错误返回时，循环中断并且不再继续前进。如何保持循环运行并确保所有消息的发送？

try {
        const targets: TargetSms[] = data.targets;
        const rbd = data.rbd;
        const idMessage = data.idMessage;

        const messageRef =  admin.firestore().collection(`/RBD/${rbd}/messages`).doc(idMessage);
        await messageRef.set({ serverResponse: true }, { merge: true });


        let countSeg = 0;
        for (const target of targets) {
            if (target.messagePayload) return target;
            const textMessage = {
                body: target.plainMsg,
                to: target.targetNumber,  // Text to this number
                from: twilioNumber, // From a valid Twilio number
            };

            const payloadMessage  = await client.messages
                .create(textMessage)
                .then(mess => mess)
                .catch(err => {
                    console.warn('ocurrio un error al enviar mensaje', err)
                    target.messageError = err;
                    return null;
                });

                if(payloadMessage){
                    countSeg = countSeg + parseInt(payloadMessage.numSegments);
                    target.messagePayload = JSON.parse(JSON.stringify(new MessagePayload(payloadMessage)));
                }
            await admin.firestore().collection(`/RBD/${rbd}/targets`).doc(target.id).set(target);
            await timeout(100);
        };

        await messageRef.set({totalSegments:countSeg},{merge:true});
        await admin.firestore().doc(`/RBD/${rbd}`).set({config: {"countSms": admin.firestore.FieldValue.increment(countSeg)}},{merge:true});


        return JSON.stringify({ suss: true, message: 'Mensajes enviados' })

    } catch (error) {
        return JSON.stringify({ suss: false, message: error })
    }

Answer 1

您需要稍微重新格式化您的代码，例如:

for (const target of targets) {
    if (target.messagePayload) return target;

    const textMessage = {
        body: target.plainMsg,
        to: target.targetNumber,  // Text to this number
        from: twilioNumber, // From a valid Twilio number
    };

    let payloadMessage;
    try {
        payloadMessage  = await client.messages.create(textMessage);
    } catch (err) {
        console.warn('ocurrio un error al enviar mensaje', err)
        target.messageError = err;
        // continue;  // Remove this line to continue the execution of finally block
    } finally {
        if (payloadMessage) {
            countSeg = countSeg + parseInt(payloadMessage.numSegments);
            target.messagePayload = JSON.parse(JSON.stringify(new MessagePayload(payloadMessage)));
        }

        await admin.firestore().collection(`/targets`).doc(target.id).set(target);
        await timeout(100);
    }
};

通过这种方式，您可以摆脱“旧的”then 语法，因为 async - wait 允许您放置 await 代码在 try - catch block 中拦截来自 await 调用的错误，如果成功，后面的所有内容都将被执行。

还可以使用继续来代替返回null。

也许你可以去掉条件 block if (payloadMessage)，因为try - catch确保你的变量有一个值，但idk肯定是这样。

此处的规范:https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/async_function