如何基于子级标记python dict父级失败或通过

我有一个测试json报告，如果有任何失败的子节点，如何标记父节点失败，如果子节点全部通过，如何标记父节点通过

起初的

---

我必须将dict的原始dict转换为dict列表，因为没有空间添加新的

结果

键

期待

---

您可以创建两个函数：一个用于前瞻并确定用户是否失败，另一个用于遍历整个结构的主函数：

def lookahead(d):
  if isinstance(d, str):
    return d == 'passed'
  if all('outcome' in b for b in d.values()):
     return all(b["outcome"] == "passed" for b in d.values())
  return all(lookahead(b) for b in d.values())

def new_struct(d):
  return [{'node_name':a, 'outcome':['failed', 'pass'][lookahead(b)] if isinstance(b, dict) and 'outcome' not in b else b['outcome'], 'children':[] if not isinstance(b, dict) or 'outcome' in b else new_struct(b)} for a, b in d.items()]

import json
d = {'scenario': {'server': {'testsuit_1': {'test_get': {'outcome': 'passed'}, 'test_set': {'outcome': 'failed'}}, 'testsui_2': {'test_get': {'outcome': 'passed'}, 'test_set': {'outcome': 'passed'}}}, 'client': {'test_receive': {'outcome': 'pass'}, 'test_send': {'outcome': 'fail'}}}}
print(json.dumps(new_struct(d), indent=4))

输出：

 [
    {
    "node_name": "scenario",
    "outcome": "failed",
    "children": [
        {
            "node_name": "server",
            "outcome": "failed",
            "children": [
                {
                    "node_name": "test_1",
                    "outcome": "failed",
                    "children": [
                        {
                            "node_name": "get",
                            "outcome": "passed",
                            "children": []
                        },
                        {
                            "node_name": "set",
                            "outcome": "failed",
                            "children": []
                        }
                    ]
                },
                {
                    "node_name": "test_2",
                    "outcome": "pass",
                    "children": [
                        {
                            "node_name": "get",
                            "outcome": "passed",
                            "children": []
                        },
                        {
                            "node_name": "set",
                            "outcome": "passed",
                            "children": []
                        }
                    ]
                }
            ]
        },
        {
            "node_name": "client",
            "outcome": "failed",
            "children": [
                {
                    "node_name": "receive",
                    "outcome": "pass",
                    "children": []
                },
                {
                    "node_name": "send",
                    "outcome": "fail",
                    "children": []
                }
            ]
        }
    ]
  }
]

---

您可以创建两个函数：一个用于前瞻并确定用户是否失败，另一个用于遍历整个结构的主函数：

def lookahead(d):
  if isinstance(d, str):
    return d == 'passed'
  if all('outcome' in b for b in d.values()):
     return all(b["outcome"] == "passed" for b in d.values())
  return all(lookahead(b) for b in d.values())

def new_struct(d):
  return [{'node_name':a, 'outcome':['failed', 'pass'][lookahead(b)] if isinstance(b, dict) and 'outcome' not in b else b['outcome'], 'children':[] if not isinstance(b, dict) or 'outcome' in b else new_struct(b)} for a, b in d.items()]

import json
d = {'scenario': {'server': {'testsuit_1': {'test_get': {'outcome': 'passed'}, 'test_set': {'outcome': 'failed'}}, 'testsui_2': {'test_get': {'outcome': 'passed'}, 'test_set': {'outcome': 'passed'}}}, 'client': {'test_receive': {'outcome': 'pass'}, 'test_send': {'outcome': 'fail'}}}}
print(json.dumps(new_struct(d), indent=4))

输出：

 [
    {
    "node_name": "scenario",
    "outcome": "failed",
    "children": [
        {
            "node_name": "server",
            "outcome": "failed",
            "children": [
                {
                    "node_name": "test_1",
                    "outcome": "failed",
                    "children": [
                        {
                            "node_name": "get",
                            "outcome": "passed",
                            "children": []
                        },
                        {
                            "node_name": "set",
                            "outcome": "failed",
                            "children": []
                        }
                    ]
                },
                {
                    "node_name": "test_2",
                    "outcome": "pass",
                    "children": [
                        {
                            "node_name": "get",
                            "outcome": "passed",
                            "children": []
                        },
                        {
                            "node_name": "set",
                            "outcome": "passed",
                            "children": []
                        }
                    ]
                }
            ]
        },
        {
            "node_name": "client",
            "outcome": "failed",
            "children": [
                {
                    "node_name": "receive",
                    "outcome": "pass",
                    "children": []
                },
                {
                    "node_name": "send",
                    "outcome": "fail",
                    "children": []
                }
            ]
        }
    ]
  }
]

---

很抱歉，我已经修复了这些错误，并且在你的新结构函数中有一个语法错误。感谢你help@ChwengMega一切都准备好了。很乐意帮忙！很抱歉，我已经修复了这些错误，并且在你的新结构函数中有一个语法错误。感谢你help@ChwengMega一切都准备好了。很乐意帮忙！