将响应类型存储到数据帧

Question

我正在尝试哟给下载结果到一个数据框下载部分工作，但在最后的数据框是空白。

df = pd.DataFrame()

url = 'https://www.cms.gov/files/zip/monthly-contract-summary-report-april-2020.zip'
FolderYear = '2020'
FolderName = 'ContractSummary'
FileName = 'monthly-contract-summary.zip'
FileDirectory = rootpath+FolderYear+"/"+FolderName+"/"
FullWritePath = rootpath+FolderYear+"/"+FolderName+"/"+FileName
if not os.path.exists(FileDirectory):
    os.makedirs(FileDirectory)

r = requests.get(url)

with open(FullWritePath, 'wb') as f:
    f.write(r.content)

# Retrieve HTTP meta-data
print(r.status_code)
print(r.headers['content-type'])
print(r.encoding)

df['Status'] = r.status_code
df['headers'] = r.status_code
df['FileName'] = FileName
df['FullWritePath'] = FullWritePath
df['ZipFileDowlondLink'] = url

Answer 1

用这句话代替你的最后几行。

rows = []    
columns = ['Status', 'headers', 'FileName', 'FullWritePath', 'ZipFileDowlondLink']
rows.append([r.status_code, r.status_code, FileName, FullWritePath, url])
df = pd.DataFrame(rows, columns=columns)

print(df)

Status  headers                      FileName                                      FullWritePath                                 ZipFileDowlondLink
0     200      200  monthly-contract-summary.zip  .2020/ContractSummary/monthly-contract-summary...  https://www.cms.gov/files/zip/monthly-contract...