只写一个回复，为什么？

Question

尝试获取每个响应以保存。 到目前为止，它只打印所有响应并保存一个。 请帮忙。 编辑：需要帮助保存所有回复。

import requests
import json

url = 'https://www.virustotal.com/vtapi/v2/file/report'
resource = []
with open("/PATH/hashes.txt", "r") as f:
    for resource in f:
        print(resource)
        params = {'apikey': 'API KEY HERE', 'resource': resource}
        response = requests.get(url, params=params)
        response1 = response.json()
        print(response.json())
        with open('data.json', 'w') as J:
            json.dump(response1, J, indent=6)

更新！！感谢所有帮助！

import requests
import json

url = 'https://www.virustotal.com/vtapi/v2/file/report'
params = {'apikey': 'API KEY HERE'}
resources = []
with open("/PATH/hashes.txt", "r") as f:
    for resource in f:
        print(resource)
        params['resource'] = resource
        response = requests.get(url, params=params)
        json_data = response.json()['sha256']
        print(json_data)
        resources.append(json_data)

with open('data.json', 'w') as J:
    json.dump(resources, J, indent=6)
J.close()
f.close()

Answer 1

import requests
import json

url = 'https://www.virustotal.com/vtapi/v2/file/report'
params = {'apikey': 'API KEY HERE'}
resources = []
with open("/PATH/hashes.txt", "r") as f:
    for resource in f:
        print(resource)
        params['resource'] = resource
        response = requests.get(url, params=params)
        json_data = response.json()
        print(json_data)
        resources.append(json_data)

with open('data.json', 'w') as f:
    json.dump(resources, f, indent=6)

在您的原始代码中，您在循环的每次迭代中覆盖data.json。 将每个 JSON 响应附加到 resources 列表（注意名称中的复数 - 在您的代码中，您只需通过在 for 循环中重用名称 resource 来覆盖空列表）。然后将列表（它将是 JSON 数组）转储到 JSON 文件中。

Answer 2

在循环的每次迭代中，您都会覆盖data.json。相反，您可以通过使用a 模式（“append”）而不是w 模式（“write”）打开它来追加它：

with open('data.json', 'a') as J:
    # Here -------------^