json - select object without quotation using jq tool

I have the following json file:

{
  "aNo": 66,
  "name": "Fella"
}
{
  "aNo": 77,
  "name": "Bella"
}

I am trying to select the name of entry which has aNo = 66 without success. However the vice versa is working, I am using this example :

cat file.json | jq '.[] | select(.name=="Fella")

It works, I get the whole key :

{
  "aNo": 66,
  "name": "Fella"
}

Now what I am asking about, is :

1. How to get the key based on aNo?
2. How to get only the name without any other values in the result.
3. Why I can't pipeline jq results ?? I get error messages when I do.

Thank you all for your great help !

to select the name of entry which has aNo = 66

This response assumes file.json contains the array:

[
{
  "aNo": 66,
  "name": "Fella"
},
{
  "aNo": 77,
  "name": "Bella"
}
]

jq approach:

jq '.[] | select(.aNo == 66).name' file.json

The output:

"Fella"

First convert the unstructured json into proper one using --slurp/-s option

--slurp/-s: Instead of running the filter for each JSON object in the input, read the entire input stream into a large array and run the filter just once.

cat file.json
{
  "aNo": 66,
  "name": "Fella"
}
{
  "aNo": 77,
  "name": "Bella"
}

cat file.json | jq -s
[
  {
    "aNo": 66,
    "name": "Fella"
  },
  {
    "aNo": 77,
    "name": "Bella"
  }
]

cat file.json | jq -s | jq -r '.[]|select(.aNo == 66).name'
Fella