FlutterFire - 如果 Firestore 查询 'result.exist' 在有结果时返回 true，那么如果没有结果，返回值是什么？

Question

In my database there is a document that has a field bizCode: '123456'. Running the below returns a TRUE and prints 'code exists'.

However, when I change isEqualTo: 'abcdef' (which does not exist in my database), nothing is returned or printed out.

How should I code it that when the query has no results, it prints returns a false or null or blank and prints 'no such code'?

                  await FirebaseFirestore.instance
                      .collection("users")
                      .where('bizCode', isEqualTo: '123456')
                      .get()
                      .then((querySnapshot) {
                    querySnapshot.docs.forEach((result) {
                      print(result.exists);
                      if (result.exists) {
                        print('code exists');
                      } else {
                        print('no such code');
                      }
                    });

Answer 1

I think you want to check querySnapshot.hasError , and querySnapshot.size > 0 .

Answer 2

I could give you an idea. You can check the length of the result.
Do something like this.

await FirebaseFirestore.instance
  .collection("users")
  .where('bizCode', isEqualTo: '123456')
  .get()
  .then((querySnapshot) {
    
    if(querySnapshot.size > 0){
      var results = querySnapshot.docs;
      results.forEach((result) {
        if (result.exists) {
          print('code exists');
        } else {
          print('no such code');
        }
      }else{
        print("No Results Found!");
      }
});

Hope this suits your case.