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优化转存结果的查询逻辑和回显
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@ -11,6 +11,7 @@ import os
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import re
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import sys
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import json
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import time
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import random
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import requests
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from datetime import datetime
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@ -533,22 +534,37 @@ class Quark:
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return False
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def query_task(self, task_id):
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url = "https://drive-m.quark.cn/1/clouddrive/task"
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querystring = {
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"pr": "ucpro",
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"fr": "pc",
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"uc_param_str": "",
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"task_id": task_id,
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"retry_index": "1",
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"__dt": int(random.uniform(1, 5) * 60 * 1000),
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"__t": datetime.now().timestamp(),
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}
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headers = self.common_headers()
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response = requests.request(
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"GET", url, headers=headers, params=querystring
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).json()
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if response["code"] == 32003:
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response["message"] = "容量超限"
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retry_index = 0
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while True:
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url = "https://drive-m.quark.cn/1/clouddrive/task"
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querystring = {
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"pr": "ucpro",
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"fr": "pc",
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"uc_param_str": "",
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"task_id": task_id,
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"retry_index": retry_index,
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"__dt": int(random.uniform(1, 5) * 60 * 1000),
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"__t": datetime.now().timestamp(),
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}
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headers = self.common_headers()
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response = requests.request(
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"GET", url, headers=headers, params=querystring
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).json()
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if response["data"]["status"] != 0:
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if retry_index > 0:
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print()
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break
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else:
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if retry_index == 0:
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print(
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f"正在等待[{response['data']['task_title']}]执行结果",
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end="",
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flush=True,
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)
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else:
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print(".", end="", flush=True)
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retry_index += 1
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time.sleep(0.500)
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return response
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def do_rename_task(self, task):
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