如何从以列表为值的字典中获取结果

Question

I have a JSON file with n number of dictionaries as listed below in the snippet. I am trying to fetch the value against the key but it fails in my code when the value is defined as a list like in the below example for key affected_packages . I tried to check why my code fails, so it looks like it pulls no data out of it this fails. I just see two brackets [] as output instead of "thunderbird-0:78.9.1-1.el8_1","thunderbird-0:78.9.1-1.el8_2","thunderbird-0:78.9.1-1.el8_3","thunderbird-0:78.9.1-1.el7_9"

{"bugzilla_description":"CVE-2021-23992 Mozilla: A crafted OpenPGP key with an invalid user ID could be used to confuse the user","cvss_score":null,"cvss_scoring_vector":null,"CWE":"CWE-347","affected_packages":["thunderbird-0:78.9.1-1.el8_1","thunderbird-0:78.9.1-1.el8_2","thunderbird-0:78.9.1-1.el8_3","thunderbird-0:78.9.1-1.el7_9"],"resource_url":"https://access.redhat.com/hydra/rest/securitydata/cve/CVE-2021-23992.json","cvss3_scoring_vector":"CVSS:3.1/AV:N/AC:L/PR:N/UI:R/S:U/C:N/I:N/A:L","cvss3_score":"4.3"}

I am doing like below in my code as I need to prepare a worksheet. Sample snippet:

for i in range(offs):
    ws.cell(row=r+1+i,column=2).value = v['current'][i]
    if 'affected_packages' in list(tmp1.keys()):
          ws.cell(row=r+1+index1,column=11).value = tmp1['affected_packages']
          print("affected_packages done")
    if 'advisories' in list(tmp1.keys()):
          ws.cell(row=r+1+index2,column=13).value = tmp1['advisories']
          print("advisories done")

Is there a way I can pull the value correctly for those defined as a list in the dictionary? I need a way so that it won't hamper my existing logic to pull value for normal key: value since while looking up into my JSON file.

So need something which can fulfil both whether my value in the dictionary is as a list or not as a list and I can get value against the keys in my json file.

Answer 1

As mentioned in the other answers, you can test the type of a variable using

if type(some_variable) == list: 
    # do what you need to do

You do mention that your code breaks, and I guess it's because inserting into a cell expects a String, not the list you pass in the line

          ws.cell(row=r+1+index1,column=11).value = tmp1['affected_packages']

So how do we get a string out of a list of strings? It's pretty easy using the join method.

my_list = ["thunderbird-0:78.9.1-1.el8_1","thunderbird-0:78.9.1-1.el8_2","thunderbird-0:78.9.1-1.el8_3","thunderbird-0:78.9.1-1.el7_9"]

as_one_string = ", ".join(my_list)
print(as_one_string)
# Prints out 'thunderbird-0:78.9.1-1.el8_1, thunderbird-0:78.9.1-1.el8_2, thunderbird-0:78.9.1-1.el8_3, thunderbird-0:78.9.1-1.el7_9'

So combining the two ideas:

    if 'affected_packages' in list(tmp1.keys()):
          ws.cell(row=r+1+index1,column=11).value = tmp1['affected_packages'] if type(tmp1['affected_packages']) != list else ", ".join(tmp1['affected_packages'])
          print("affected_packages done")

Quick feedback because I can't comment yet: Please always include an error message and/or the output you get when running your code when you ask a question

Answer 2

Regarding your second problem, when you don't know if it is a list or something else, you can just check the type, maybe like this:

if type(tmp1['affected_packages']) == list:
    # process the list
else:
    # process other types

Since you don't know the data type, having this explicit type check seems necessary.

Answer 3

If I understand it correctly, you just need to determine if a value in dict is list. You can do that as below:

for i in d.items(): # gets key, value as a tuple. 
   if isinstance(i[1],list):
       print('its a list, process it accordingly')
   else:
       print('Not a list')