无法在java

时间:2018-01-26 18:37:21

标签: java api curl get httpclient

我在我的java类中运行下面的代码,理想情况下它应该返回给我一个持票人令牌....但是当我运行代码时我得到了异常

以下是我正在运行的代码

String[] command = {"curl","-v","-
X","POST","https://<Domain name>/oauth2/token","-H","cache-control: 
no-cache","-H","content-type: application/x-www-form-urlencoded","-
H","postman-token: a1f5e569-cbc7-63cc-8667-45d85d74784b","-d","client_id=XXXXXXX","&","client_secret=XXXXXXXXXX","&","grant_type=client_credentials"};
for (String s: command)
            System.out.print(s); 
        ProcessBuilder process = new ProcessBuilder(command);
         Process p;
try {

                p = process.start();
     }

我得到的例外是

java.io.IOException: Cannot run program "curl": CreateProcess error=2, The system cannot find the file specified

有人可以让我知道问题是什么或如何从JAVA运行curl命令

提前致谢

1 个答案:

答案 0 :(得分:0)

您是否尝试过Rest Assured Framework?我在某个时候有相同的要求,可以使用Rest Assured来实现它。

RestAssured.baseURI = CURLURL;
Response r = (Response) given().relaxedHTTPSValidation().contentType("application/json").body(curlRequest)
.when().post("");
return r.getBody().asString();