如何在JAVA中放心api的请求正文中传递参数？

Question

In below example, if i run this it is taking z as string, not as parameter, 

String z=   "1234";
        System.out.println("converted"+z);
        try {
            RestAssured.baseURI = "http://192.189.11.51:05/oceans/fssdata/";
            String requestBody ="{\"CompanyID\":z"
                    + ""
                    + ""
                    + "}";`

如何在公司ID中传递参数Z，现在是以公司ID为z，我希望它应该取z的值。

Answer 1

直接使用字符串时需要小心转义引号。您可以尝试使用下面的代码。

String z = "\"1234\"";

String requestBody: "{\"name\":"+z+"}";

Answer 2

你可以用这个：

• 马文 com.github.cliftonlabs

   import com.github.cliftonlabs.json_simple.JsonObject;
       public void createAnAddress() {
       RestAssured.baseURI = "http://192.189.11.51:05/oceans/fssdata/";
       JsonObject requestParams = new JsonObject();
       requestParams.put("CompanyID", "Monroe Company");
  
  
       Response response = given().
               .header("Content-Type", "application/json").body(requestParams)
               .post("/oceans/fssdata/");
       System.out.println("Response Body create Add is =>  " + response.getBody().asString());
   }
  

Answer 3

可以使用JSONObject来传递参数

 import org.json.simple.JSONObject;

    String z = "1234";
    JSONObject requestParam = new JSONObject();
    requestParams.put("CompanyID", z);

    given()
            .body(requestParam.toJSONString())
            .post("/your_endpoint");   
}

或者您可以使用 String 以您的方式传递

"{\"CompanyID\" : \""+z+"\",\n" +
 "\"property\": \"value\"\n" +
"}"