How to make this code to take multiple yaml input files and convert them to .json and save all of them to specified path

Question

def convert():             #function to convert .yaml to .json 

    in_file = filepath     #assigning input file from GUI to variable
    out_file = savepath    #assigning output path from GUI to variable 
    yaml = ruamel.yaml.YAML(typ='safe')
    with open(in_file ) as i:       #opening yaml file 
        data = yaml.load(i)
    with open(out_file, 'w') as o:  #writing json file
        json.dump(data, o, indent=2)

Answer 1

Your code is not formatted properly. But let's say your code looks like:

def convert():
    in_file = filepath     #assigning input file from GUI to variable
    out_file = savepath    #assigning output path from GUI to variable 
    yaml = ruamel.yaml.YAML(typ='safe')
    with open(in_file ) as i:       #opening yaml file 
        data = yaml.load(i)
    with open(out_file, 'w') as o:  #writing json file
        json.dump(data, o, indent=2)

You must change some logic here:

1. You must get file names information instead of file name. It can be a list of paths which can be got from glob. Then you can get individual file name looping in files.

2. Since you have multiple files you cannot provide 1 file name and save all your files with same name. The best I can think of is (Let's say you are converting yaml files to json files) you can replace the extension of the file and save it with the same name.

Getting file names:

The glob module finds all the pathnames matching a specified pattern according to the rules used by the Unix shell, although results are returned in arbitrary order.

see glob

Now let's use it:

from glob import glob
files = glob("/your/path/*.yaml")

for file in files:
    print(files)

Here file is path for each yaml file. Now you can open/use that file name inside the loop and do whatever you want.

Saving files

To save the file as json you can:

1. Add .json to the file name

Basically use + or fstrings to add .json.

json_file_name = file + ".json"

or

json_file_name = f"{file}.json"

Be aware that your file names might look weird. Because you will have my_file.yaml changes to my_file.yaml.json.

2. Replace yaml with json

Since file paths are strings you can use replace method and replace the file name.

json_file_name = file.replace("yaml", "json")

Here your file name would look better but it will change all yamls in the name. So if your file name has yaml in it it will be changed as well.

Like my_yaml_file.yaml changes to my_json_file.json.