如何查看 MultipartForm 请求的内容？

发布于 2024-10-12 15:44:01 字数 565 浏览 8 评论 0原文

我正在使用 Apache HTTPClient 4。我正在做非常正常的多部分工作，如下所示：

val entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
entity.addPart("filename", new FileBody(new File(fileName), "application/zip").asInstanceOf[ContentBody])
entity.addPart("shared", new StringBody(sharedValue, "text/plain", Charset.forName("UTF-8")));

val post = new HttpPost(uploadUrl);
post.setEntity(entity);

我想在发送实体（或帖子，无论什么）之前查看它的内容。但是，该特定方法并未实现：

entity.getContent() // not defined for MultipartEntity

我怎样才能看到我发布的内容？

原文

I am using Apache HTTPClient 4. I am doing very normal multipart stuff like this:

val entity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
entity.addPart("filename", new FileBody(new File(fileName), "application/zip").asInstanceOf[ContentBody])
entity.addPart("shared", new StringBody(sharedValue, "text/plain", Charset.forName("UTF-8")));

val post = new HttpPost(uploadUrl);
post.setEntity(entity);

I want to see the contents of the entity (or post, whatever) before I send it. However, that specific method is not implemented:

entity.getContent() // not defined for MultipartEntity

How can I see what I am posting?

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远昼 2024-10-19 15:44:01

使用 org.apache.http.entity.mime.MultipartEntity writeTo(java.io.OutputStream) 方法将内容写入 java.io. OutputStream，然后将该流转换为 String 或 byte[]：

// import java.io.ByteArrayOutputStream;
// import org.apache.http.entity.mime.MultipartEntity;
// ...
// MultipartEntity entity = ...;
// ...

ByteArrayOutputStream out = new ByteArrayOutputStream(entity.getContentLength());

// write content to stream
entity.writeTo(out);

// either convert stream to string
String string = out.toString();

// or convert stream to bytes
byte[] bytes = out.toByteArray();

注意：这仅适用于小到足以读入内存的多部分实体，并且小于 Java 中字节数组的最大大小 2Gb。

Use the org.apache.http.entity.mime.MultipartEntity writeTo(java.io.OutputStream) method to write the content to an java.io.OutputStream, and then convert that stream to a String or byte[]:

// import java.io.ByteArrayOutputStream;
// import org.apache.http.entity.mime.MultipartEntity;
// ...
// MultipartEntity entity = ...;
// ...

ByteArrayOutputStream out = new ByteArrayOutputStream(entity.getContentLength());

// write content to stream
entity.writeTo(out);

// either convert stream to string
String string = out.toString();

// or convert stream to bytes
byte[] bytes = out.toByteArray();

Note: this only works for multipart entities small enough to be read into memory, and smaller than 2Gb which is the maximum size of a byte array in Java.

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临风闻羌笛 2024-10-19 15:44:01

以下肯定会有帮助：

ByteArrayOutputStream content = new ByteArrayOutputStream();
httpEntity.writeTo(content);
logger.info("Calling "+url+" with data: "+content.toString());

与第一个答案相比，上面有一个修复，不需要将任何参数传递给 ByteArrayOutputStream 构造函数。

Following would help for sure:

ByteArrayOutputStream content = new ByteArrayOutputStream();
httpEntity.writeTo(content);
logger.info("Calling "+url+" with data: "+content.toString());

The above has a fix in comparison to the first answer, there is no need to pass any parameter to ByteArrayOutputStream constructor.

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财迷小姐 2024-10-19 15:44:01

你不知道内容吗？不过，您是通过提供 sharedValue 来构建 StringBody 的。那么，它与 sharedValue 有何不同。

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栩栩如生 2024-10-19 15:44:01

我已经通过以下代码打印了多部分请求，您可以尝试像

ByteArrayOutputStream bytes = new ByteArrayOutputStream();

entity.writeTo(bytes);

String content = bytes.toString();

Log.e("MultiPartEntityRequest:",content);

I have printed the Multipart request by following code, You can try like

ByteArrayOutputStream bytes = new ByteArrayOutputStream();

entity.writeTo(bytes);

String content = bytes.toString();

Log.e("MultiPartEntityRequest:",content);

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