使用 Jackson Fasterxml 将 Cucumber DataTable 转换为 POJO 时找不到字段

Question

我正在尝试使用 Jackson Faster XML 将以下数据table 转换为 POJO class，但是当我的测试运行时出现以下错误。我不确定为什么 table 没有映射。 customerId 出现在 POJO class.

中

错误：

cucumber.runtime.CucumberException: No such field learning.pojo.customerId

步骤特征：

  And the user imported the following product file
      | customerId   | ....
      |customer1     | ....

步骤 Java:

@When("^the user imported the following product file$")
public void uploadFile(DataTable table) throws IOException {
     Example productImportFileModel = table.asList(Example.class).get(0);

根 POJO：

import com.fasterxml.jackson.annotation.JsonInclude;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.annotation.JsonPropertyOrder;

@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonPropertyOrder({
                           "products"
                   })
public class Example{
    
    @JsonProperty("products")
    private List<Products> products = null;

    @JsonProperty("products")
    public List<Products> getProducts() {
        return products;
    }

    @JsonProperty("products")
    public void setProducts(List<Products> products) {
        this.products = products;
    }
}

POJO 产品：

import com.fasterxml.jackson.annotation.JsonInclude;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.annotation.JsonPropertyOrder;

@JsonInclude(JsonInclude.Include.NON_NULL)
@JsonPropertyOrder({
                           "customerId",
                           .....
                           .....
                           .....
                           .....
                           .....

                   })
public class Products {
    
    @JsonProperty("customerId")
    private String customerId;

    @JsonProperty("customerId")
    public String getCustomerId() {
        return customerId;
    }
    
    @JsonProperty("customerId")
    public void setCustomerId(String customerId) {
        this.customerId = customerId;
    }
                           .....
                           .....
                           .....
                           .....
                           .....
}

Answer 1

Example的json表示为：

{
  "products": [ ..... ]
}

Products的json表示为：

{
  "customerId": "customer1"
   ....
}

table 作为列表的 json 表示是：

[ 
  {
    "customerId": "customer1"
     ....
  }
]

所以考虑使用：

Product productImportFileModel = table.asList(Product.class).get(0);

或者更好：

@When("^the user imported the following product file$")
public void uploadFile(List<Product> products) throws IOException {
     Product productImportFileModel = products.get(0);

您可以通过向 Jackson 询问 json 表示来调试它：

JsonNode json = table.asList(JsonNode.class).get(0);
System.out.println(json);

使用 Jackson Fasterxml 将 Cucumber DataTable 转换为 POJO 时找不到字段

Cant find field when converting Cucumber DataTable to POJO using Jackson Fasterxml

java

cucumber

jackson

fasterxml