Data output in Flask?

Question

Python

Good evening. I ask for help, because I did not find the answer on the Internet. There is data - dictionaries saved in files. I can open them by iterating through all the files in the folder. But passing normally to Flask doesn't work. Namely, when iterating over dictionaries, I want to display each dictionary (data in it) on the page. At the output I have only the output of the data of the last dictionary. I think I need to learn the mother part of Flask, but I hope for help ...
My code in python

@app.route("/monitoring")
def index(): 
    title = "START:"
    # Открываем файлы
    WATCH_DIRECTORY = 'data'
    for filename in os.listdir(WATCH_DIRECTORY):
        file_data = os.path.join(WATCH_DIRECTORY, filename)
        try:
            with open(file_data, 'rb') as file:
                data_local = pickle.load(file)
                file.close()
                print(data_local) # Здесь я нормально получаю вывод всех данных словарей.
                g.data_local = data_local
                return g.data_local().all() # Это нахимичил сам...

        except Exception as e:
            print(e)
    return render_template("monitoring.html", thetitle=title, temps=temp, peopls=g.data_local, year=datetime.date.today().year)

<table>
  {% for peopl in peopls %}
    <tr><td>{{ peopl.temp1 }}</td>
    <td>{{ peopl.znac1 }}</td></tr>
     {% endfor %}</table></div>
    <div></div>
На выходе после этого я ничего не получаю... пустые блоки.

    {% if peopls%}
    {{ peopls.znac1 }}
    {% endif %}
Здесь удается получить только последнего словаря данные..

Answer 1

Create before the loop, for example, immediately after WATCH_DIRECTORY = 'data' the variable people = []
replace this line g.data_local = data_local with people.append(data_local) and in render, replace this peopls=g.data_local with peopls=people

Answer 2

for filename in os.listdir(WATCH_DIRECTORY):
        file_data = os.path.join(WATCH_DIRECTORY, filename)
        try:
            with open(file_data, 'rb') as file:
                data_local = pickle.load(file)
                file.close()
                print(data_local) # здесь вывод данных с одного файла, открытого в данный момент
                g.data_local = data_local # здесь значение g.data_local затирается последним (на каждой итерации)
                return g.data_local().all()

At each iteration of the loop, the value of g.data_local is overwritten
and as a result, only data from the last file gets into the view template

Answer 3

1. I didn't really understand why you used g. Just for the accumulation of data during processing, an ordinary variable would suffice, but it is not applicable for storing data between requests (See note on this subject).
2. If you yourself catch an exception, then it is better to handle it completely, indicating the template and response code. It doesn't affect your problem, but it's just a good habit to do nothing, or finish it all the way.
3. What does return mean inside the with block? If you declare a variable accumulator of values ​​from files above the loop through files, and instead of g.data_local ... and return ... write the code for sending the read value to the accumulator, for example, like this, and then pass the contents of the accumulator to the template, under the name peopls, you won’t get exactly the effect you wanted?
A couple of thoughts besides your code:
What is the fate of the files that you have already processed, will you read them again, at the next iteration? If files are added often enough, then each next request will take longer than the previous one. Maybe you should think again?
Depending on the volume of files and / or their number, in especially severe cases, you may run into a response time limit. It's easy to fix, but it's worth remembering.