如何使用Java 11 HTTP Client为POST请求定义多个参数

Question

我有一个对特定端点发出POST请求的代码。这段代码是使用Apache的HttpClient，我想开始使用HttpClientJava（JDK11）的本机代码。但是我不明白如何指定请求的参数。

这是我使用Apache Httpclient的代码：

var path = Path.of("file.txt");
var entity = MultipartEntityBuilder.create()
            .addPart("file", new FileBody(path.toFile()))
            .addTextBody("token", "<any-token>")
            .build();

和代码使用HttpClient：

var client = HttpClient.newHttpClient();
var request = HttpRequest.newBuilder()
                         .uri(URI.create("https://myendpoint.com/"))
                         .POST( /* How can I set the parameters here? */ );

如何设置file和token参数？

Answer 1

不幸的是，Java 11 HTTP 客户端没有为多部分类型的正文提供任何方便的支持。但是我们可以在它之上构建自定义实现：

Map<Object, Object> data = new LinkedHashMap<>();
data.put("token", "some-token-value";);
data.put("file", File.createTempFile("temp", "txt").toPath(););

// add extra parameters if needed

// Random 256 length string is used as multipart boundary
String boundary = new BigInteger(256, new Random()).toString();

HttpRequest.newBuilder()
              .uri(URI.create("http://example.com"))
              .header("Content-Type", "multipart/form-data;boundary=" + boundary)
              .POST(ofMimeMultipartData(data, boundary))
              .build();

public HttpRequest.BodyPublisher ofMimeMultipartData(Map<Object, Object> data,
                                                     String boundary) throws IOException {
        // Result request body
        List<byte[]> byteArrays = new ArrayList<>();

        // Separator with boundary
        byte[] separator = ("--" + boundary + "\r\nContent-Disposition: form-data; name=").getBytes(StandardCharsets.UTF_8);

        // Iterating over data parts
        for (Map.Entry<Object, Object> entry : data.entrySet()) {

            // Opening boundary
            byteArrays.add(separator);

            // If value is type of Path (file) append content type with file name and file binaries, otherwise simply append key=value
            if (entry.getValue() instanceof Path) {
                var path = (Path) entry.getValue();
                String mimeType = Files.probeContentType(path);
                byteArrays.add(("\"" + entry.getKey() + "\"; filename=\"" + path.getFileName()
                        + "\"\r\nContent-Type: " + mimeType + "\r\n\r\n").getBytes(StandardCharsets.UTF_8));
                byteArrays.add(Files.readAllBytes(path));
                byteArrays.add("\r\n".getBytes(StandardCharsets.UTF_8));
            } else {
                byteArrays.add(("\"" + entry.getKey() + "\"\r\n\r\n" + entry.getValue() + "\r\n")
                        .getBytes(StandardCharsets.UTF_8));
            }
        }

        // Closing boundary
        byteArrays.add(("--" + boundary + "--").getBytes(StandardCharsets.UTF_8));

        // Serializing as byte array
        return HttpRequest.BodyPublishers.ofByteArrays(byteArrays);
    }

这是Github 上的工作示例（您需要更改 VirusTotal API 密钥）