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Java HTTP Post Multipart

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我正在尝试将HTTP Post Multipart请求发送到JAVA中的本地服务器 . 我正在尝试发送以下内容：

{
 "content-disposition": "form-data; name=\"metadata\"",
"content-type": "application/x-dmas+json",
 "body":         JSON.stringify(client_req)
},
{
"content-disposition": "attachment; filename=\"" + file + "\"; name=\"file\"",
"content-type": "application/octet-stream",
 "body":         [file content]
}

我查看了Apache HTTP组件，但它不允许我指定每个部分的内容类型和处置 . 这是我使用Apache HTTP API在JAVA中编写的内容：

`CloseableHttpClient httpclient = HttpClients.createDefault（）;

try {
        HttpPost httppost = new HttpPost("IP");

        FileBody bin = new FileBody(new File(args[0]), "application/octet-stream");
        StringBody hash = new StringBody("{\"hash\": \"\", \"policy\": {\"retention_permitted\": true, \"distribution\": \"global\"}}", ContentType.create("application/x-dmas+json"));

        HttpEntity reqEntity = MultipartEntityBuilder.create()
                .addPart("metadata", hash)
                .addPart("file", bin)
                .build();


        httppost.setEntity(reqEntity);

`

java http multipartform-data

2 回答

  • 0

    FilePart和StringPart的构造函数参数和方法，您使用它们组成构成多部分请求的Part []，提供此信息 .

    回复于
  • 0

    可能为时已晚，但作为对同一问题的答案的任何人的参考，MultipartEntityBuilder类中有几种方法允许您为每个部分设置内容类型和内容处置 . 例如，

    addBinaryBody（String name，File file，ContentType contentType，String filename）addTextBody（String name，String text，ContentType contentType）

    如果我们在您的示例中使用上述方法，

    CloseableHttpClient httpClient = HttpClients.createDefault();
HttpPost uploadFile = new HttpPost("http://url-to-post/");

MultipartEntityBuilder builder = MultipartEntityBuilder.create();
String jsonStr = "{\"hash\": \"\", \"policy\": {\"retention_permitted\": true, \"distribution\": \"global\"}}";
builder.addTextBody("metadata", jsonStr, ContentType.create("application/x-dmas+json"));
builder.addBinaryBody("file", new File("/path/to/file"),
    ContentType.APPLICATION_OCTET_STREAM, "filename");

HttpEntity multipart = builder.build();
uploadFile.setEntity(multipart);
HttpResponse response = httpClient.execute(uploadFile);
    回复于

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