Question

我有要在Python中解析的JSON，看起来像这样

{
  "New Slim Testing":{
    "name":"New Slim Testing",
    "id":"6496",
    "type":1,
    "fullpath":"\\New Slim Testing",
    "children": {
        "sf_account":{
          "name":"sf_account",
          "type":1,
          "fullpath":"",
          "id":"6516"
        },
        "sf_case":{
          "name":"sf_case",
          "type":1,
          "fullpath":"",
          "id":"6517",
          "children": {
              "sf_case_delete":{
                "name":"sf_case_delete",
                "type":1,
                "fullpath":"",
                "id":"6518"
              }
            }
        },
        "sync_incr_sfdc_b2b_to_rds_sf_case":{
          "name":"sync_incr_sfdc_b2b_to_rds_sf_case",
          "type":2,
          "fullpath":"",
          "command":"&lt;T4I_DNA_SCRIPTS.23&gt;run_infacloud_task.ksh",
          "parameters":"0 &lt;JobName&gt; DSS",
          "id":"6520"
      }
    }
  }
}

此JSON将用于进行API调用。尽管id目前是硬编码的，但它应该是动态的，并由API调用返回。迭代时如何更改此字典中的某些值？

例如：将创建父作业，将为其分配一个ID，创建子作业时，我将使用其父ID等。

对于全路径字段也是如此。我想在运行时派生它（迭代字典），而不是对其进行硬编码。

这就是我想要做的

def jsonTraverser(json,parentJson):

    templateXML = settings.JOBXMLTEMPLATE

    for key, val in json.items():
        if isinstance(val, dict):
            pp = pprint.PrettyPrinter(indent=2)
            if(key!="children"):
                parentId = parentJson["id"]
                fullpath = '{}\\{}'.format(parentJson["fullpath"],json[key]["name"])
                print('Fullpath is {}'.format(fullpath))
                json['fullpath'] = fullpath
                if val['type']==1:
                    print('Creating job group {}'.format(val["name"]))
                    createJob(val,parentId)
                else:
                    print('Creating job {}'.format(val["name"]))
                    createJob(val,parentId)
            jsonTraverser(val,json)

我遇到了错误：字典在迭代过程中更改了大小

Answer 1

通过list(dct.keys())使用json数据的独立键（列表适用于python3 +）：

In [5]: dct = {1: 2, 2: 3}

In [6]: for k in list(dct.keys()):
   ...:     dct[str(k)] = dct[k]
   ...:

In [7]: dct
Out[7]: {1: 2, 2: 3, '1': 2, '2': 3}

迭代时更改JSON结构

1 个答案: