Question

我想帮助用java编写相当于下面curl命令的内容。

curl -X POST --header 'Content-Type: application/json' --header 'Accept: application/json' --header  'cookie: [APP COOKIES];' -d 'sampleFile.json' 'https://url.com'

Answer 1

您尝试使用RESTful Web服务，因此您应该使用类似jax-rs或spring REST的库，这是example of consuming a RESTful web service，您可以找到许多具有更多详细信息的示例。

Answer 2

我认为您正在寻找的答案是哪个库提供Java来创建HTTP请求。 What is the best Java library to use for HTTP POST, GET etc.? 这个问题为您提供了所需的所有答案

Answer 3

使用here apache，您可以执行此操作：

...
import org.apache.http.client.methods.HttpPost;
import org.apache.http.NameValuePair;
import org.apache.http.client.entity.UrlEncodedFormEntity
import org.apache.http.Consts;
import org.apache.http.client.methods.CloseableHttpResponse;
import org.apache.http.HttpEntity;
....

public String sendPost(String url, Map<String, String> postParams, 
        Map<String, String> header, String cookies) {

    HttpPost httpPost = new HttpPost(url);
    List<NameValuePair> formParams = new ArrayList<NameValuePair>();
    HttpClientBuilder clientBuilder = HttpClients.createDefault();
    CloseableHttpClient httpclient = clientBuilder.build();

    if (header != null) {
        Iterator<Entry<String, String>> itCabecera = header.entrySet().iterator();
        while (itCabecera.hasNext()) {
            Entry<String, String> entry = itCabecera.next();

            httpPost.addHeader(entry.getKey(), entry.getValue());
        }
    }

    httpPost.setHeader("Cookie", cookies);

    if (postParams != null) {
        Iterator<Entry<String, String>> itParms = postParams.entrySet().iterator();
        while (itParms.hasNext()) {
            Entry<String, String> entry = itParms.next();

            formParams.add(new BasicNameValuePair(entry.getKey(), entry.getValue()));
        }
    }

    UrlEncodedFormEntity formEntity = new UrlEncodedFormEntity(formParams, Consts.UTF_8);
    httpPost.setEntity(formEntity);

    CloseableHttpResponse httpResponse = httpclient.execute(httpPost);

    HttpEntity entity = httpResponse.getEntity();
    if (entity != null) {
        pageContent = EntityUtils.toString(entity);
    }

    return pageContent;
}

我希望这对你有所帮助。

编辑：

我添加了发送文件的方式。我将代码分开放入不混合代码（这是一个基于HttpClient的近似值）：

File file = new File("path/to/file");
String message = "This is a multipart post";
MultipartEntityBuilder builder = MultipartEntityBuilder.create();
builder.setMode(HttpMultipartMode.BROWSER_COMPATIBLE);

if (postParams != null) {
    Iterator<Entry<String, String>> itParms = postParams.entrySet().iterator();
    while (itParms.hasNext()) {
        Entry<String, String> entry = itParms.next();

        builder.addTextBody(entry.getKey(), entry.getValue(), ContentType.DEFAULT_BINARY);
    }
}

builder.addTextBody("text", message, ContentType.DEFAULT_BINARY);

HttpEntity entity = builder.build();
httpPost.setEntity(entity);

相当于java中的curl命令

3 个答案: