最接近的等效方法是使用更高的限制进行查询，然后在事务中再次获取文档并再次检查所需的输入条件......

// this will do the OP sets/updates, but *doesn't* assume the input
// doc meets the required condition (isAvailable==true)
// return a bool indicating success
function reserveDoc(lottoDoc) {
  return firestore.runTransaction(transaction => {
    return transaction.get(lottoDoc.ref).then(ticketDoc => {
      // important, check the original condition again here
      if (ticketDoc.data().isAvailable) {
        // OP code to set/update goes here
        return true
      } else {
        return false
      }
    })
  })
}

// try a transaction on the first doc in the array.  return if successful
// otherwise, try recursively on the remainder of the array
// return a bool indicating success
function reserveOneOf(lottoDocs) {
  if (lottoDocs.length === 0) return false
  return reserveDoc(lottoDocs[0]).then(success => {
    // did it work?  if not, try another doc
    return success ? success : reserveOneOf(lottoDocs.slice(1))
  })
}

function originalOPFunction() {
  const query = firestore
    .collection("Lottories")
    .doc("R3DYubrqFbbVfQNjYXfh")
    .collection("sets")
    .where("isAvailable", "==", true)  // note, I assume the OP query had a typo, checking for false
    .limit(10);
  return query.get().then(snapshot => {
    return reserveOneOf(snapshot.docs)
  }).then(success => {
    // if !success here, the app must deal with NO docs currently meeting the criterion
    // the OP needed to handle this circumstance anyway (if the limit(1) query returned no docs
  })
}

事务get的第一个参数实际上是一个文档引用，而不是查询。这是令人困惑的，因为documentReference.get()、transaction.get(documentReference)和query.get()看起来和听起来都差不多，但事务只是单个文档上的原子，而不是查询中的集合上的原子，即使是限制为1的集合。

transaction.get(query)query必须是一个文档引用。因此，您需要类似以下的内容：

db.collection("Lottories/R3DYubrqFbbVfQNjYXfh/sets")
    .where("isAvailable", "==", false)
    .limit(1)
    .get()
    .then((docs) => {
      db.runTransaction((transaction) => transaction.get(docs[0]).then((doc) => {
        if (doc.exists) //do something
      }));
    });