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如何从嵌套列表中提取项目 - Rest Assured

[英]How to extract item from nested list - Rest Assured

I've got this json object 我有这个json对象

{
   "type" : "employee",
   "columns" :
      [
         {
            "id" : 1,
            "human" :
               [
                  {
                     "name" : "ANA",
                     "age" : "23"
                  },
                  {
                     "name" : "IULIA",
                     "age" : "22"
                  }
               ]
         },
         {
            "id" : 2,
            "human" :
               [
                  {
                     "name" : "ADI",
                     "age" : "21"
                  },
                  {
                     "name" : "GELU",
                     "age" : "18"
                  }
               ]
         }
      ]
}

and I need to extract the first name from each human list. 我需要从每个人名单中提取名字。 I've tried .body("columns.human.name[0]", everyItem(containsString("A"))) but it doesn't work. 我试过.body("columns.human.name[0]", everyItem(containsString("A")))但它不起作用。 Any ideas? 有任何想法吗?

Using JsonPath you can get all columns and all humans. 使用JsonPath您可以获得所有列和所有人类。

Each of JSON Object is represented as HashMap<> . 每个JSON Object都表示为HashMap<> If it contains only fields it's HashMap<String, String> but if contains arrays or nested JSON Objects then it is HashMap<String, Object> where Object is either another JSON Object or array. 如果它只包含字段,那么它是HashMap<String, String>但是如果包含数组或嵌套的JSON对象,则它是HashMap<String, Object> ,其中Object是另一个JSON对象或数组。

Given the above you can use following code to get all columns and name of first human in each column: 鉴于上述情况,您可以使用以下代码获取每列中第一个人的所有列和名称:

JsonPath path = response.jsonPath();
List<HashMap<String, Object>> columns = path.getList("columns");
for (HashMap<String, Object> singleColumn : columns) {
    List<HashMap<String, Object>> humans = (List<HashMap<String, Object>>) singleColumn.get("human");
    System.out.println(humans.get(0).get("name"));
}

The above code will print ANA and ADI in the console. 上面的代码将在控制台中打印ANAADI You can store the results in List<String> for further processing 您可以将结果存储在List<String>以进行进一步处理

you can get all "name" from humans with jsonPath : $.columns[*].human[*].name it will give below result : 你可以通过jsonPath获取人类的所有“名字”: $.columns[*].human[*].name它将给出以下结果:

[
  "ANA",
  "IULIA",
  "ADI",
  "GELU"
]

And if you want only first "name" then you need to use josn : $.columns[*].human[0].name this will give you below result: 如果你只想要第一个“名字”,那么你需要使用josn: $.columns[*].human[0].name这将给你以下结果:

[
  "ANA",
  "ADI"
]

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