如何中断功能并从循环中返回最后一个值。返回值为无

Question

我试图在if语句中使用return破坏函数。我认为我不会停止该循环，而是使另一个循环不再返回。但我找不到解决该问题的方法。

这是输入obj的示例

svg{border:1px solid}

这是功能代码

<svg id="theSVG" viewBox="0 0 30 30" width="300">
  <g id="theG">
<rect x="20" y="0" width="100" height="20" transform="rotate(45)" fill="black" />
  </g>
</svg>

此函数应该获取一个对象并将路径返回到所需值。

Answer 1

首先，您输入的对象是错误的，您在最外层有一个字典，但是缺少键，所以我假设您的对象是一个像这样的列表

b =  [
       {
           "id": "160407",
           "created": "2017-10-30T09:41:37.960+0000",
           "items": [
               {
                   "field": "status",
                   "fieldtype": "test",
                   "from": "10407",
                   "fromString": "Analysis",
                   "to": "4",
                   "toString": "To Do"
               }
           ]
       },
       {
           "id": "160407",
           "created": "2019-10-30T09:41:37.960+0000",
           "items": [
               {
                   "field": "status",
                   "fieldtype": "test",
                   "from": "10407",
                   "fromString": "Analysis",
                   "to": "3",
                   "toString": "In Progress"
               }
           ]
       }
   ]

要忽略返回值的复杂性，使用全局变量要容易得多，在达到条件并分配该变量并将其打印出来时

result = []
def recursive(obj, path=None):

    global result
    if path is None:
        path = []

    # Check for object type and unpacked
    if isinstance(obj, dict):
        for key, value in obj.items():
            # Write path for this key
            new_path = list(path)
            new_path.append(key)

            # Check Status = in progress
            condition = 'to' in obj and '3' == obj['to']
            if condition:
                result = new_path
                return new_path #None

            recursive(value, path=new_path)

    # Check for list type and unpacked
    elif isinstance(obj, list):
        for i, item in enumerate(obj):
            new_path = list(path)
            new_path.append(i)
            recursive(item, path=new_path)

recursive(b) #None
print(result)
#[1, 'items', 0, 'field']