I have an API that i'm testing. The API receive POST Request and read it like this StringBuffer jb = new StringBuffer(); String line = null; try { BufferedReader reader = request.getReader(); while ((line = reader.readLine()) != null) jb.append(line); System.out.println("jb: "+jb); System.out.println("request.getHeader('content-
我有一个API，我正在测试。 API接收POST请求并像这样读取它 StringBuffer jb = new StringBuffer(); String line = null; try { BufferedReader reader = request.getReader(); while ((line = reader.readLine()) != null) jb.append(line); System.out.println("jb: "+jb); System.out.println("request.getHeader('content-type'): "+request.getHeader("co
This question already has an answer here: How to upload files to server using JSP/Servlet? 12 answers While using enctype="multipart/form-data" form fields aren't available as parameter of the request, you will always get request.getParameter(name); as null. They are included in the stream. You can use Apache Commons Fileupload to upload file with other form fields data. In
这个问题在这里已经有了答案： 如何使用JSP / Servlet将文件上传到服务器？ 12个答案 虽然使用enctype="multipart/form-data"表单字段不可用作请求的参数，但您始终会获得request.getParameter(name); 为空。 它们包含在流中。 您可以使用Apache Commons Fileupload与其他表单字段数据上传文件。 为了使用Apache Commons FileUpload，您需要在Web应用程序的/ WEB-INF / lib中包含以下文件： 1.）commons-fi
I'm trying to setup an Apache Camel route in a Java application where the consumer endpoint is a restlet component that handles an HTTP file upload as a POST of multipart form data and then the producer endpoint forwards the request to a rest service that also accepts multipart form data. I'm new to Camel and can't quite figure out how to wire it up properly. Below is how my route l
我试图在Java应用程序中设置一个Apache Camel路由，其中消费者端点是一个restlet组件，它将HTTP文件上载作为多部分表单数据的POST进行处理，然后生产者端点将该请求转发给还接受多部分表单数据。 我是骆驼新手，无法完全弄清楚如何正确连线。 以下是我的路线看起来如何。 我是否需要对正文进行任何转换，或者将多部分表单数据按原样转发？ 有人能为我提供一些正确的方法来做到这一点，或指向我正确的文件？ <route
I am trying to upload a 260k image file as part of multipart form using Apache HttpAsyncClient library. I create my form this way: val multipartEntityBuilder = new MultipartEntityBuilder multipartEntityBuilder.addBinaryBody("file", file) val multipartEntity = multipartEntityBuilder.build() And then I receive a ContentTooLongException when performing request basically because of this line in t
我试图使用Apache HttpAsyncClient库作为multipart表单的一部分上载一个260k图像文件。 我通过这种方式创建我的表单： val multipartEntityBuilder = new MultipartEntityBuilder multipartEntityBuilder.addBinaryBody("file", file) val multipartEntity = multipartEntityBuilder.build() 然后，我在执行请求时收到一个ContentTooLongException，这基本上是由于图书馆源代码中的这一行：https：//github.com/apache/httpc
In Talend Data Integration I need to upload a file (with multipart/form-data) to http server. Server in response returns http code other than 200 OK - for example 201 Created , 202 Accepted . I tried to use tFileFetch but this component only accepts responses with a code 200. Is in Talend any component that during file upload can handle http status code other than 200? I resolved my proble
在Talend数据集成中，我需要上传一个文件（带有multipart / form-data）到http服务器。 响应中的服务器返回200以外的http代码 - 例如201创建 ， 202接受 。 我试图使用tFileFetch，但该组件只接受代码为200的响应。 在Talend中是否有文件上传过程中的任何组件可以处理200以外的http状态码？ 我通过编写自定义组件解决了我的问题。 我从tFileFetch中复制源代码 ，然后修改它以满足我的要求。 我遵循以下教程：http://ww
我试图使用apache HttpClient api将“myfile.txt”文件上传到apache服务器（WAMP）中作为“uploaded.txt”，但我得到了“404未找到”状态。 public class Httpupload1 { public static void main(String[] args) throws Exception { String url = "http://username:password@localhost/uploaded.txt"; DefaultHttpClient httpclient = new DefaultHttpClient(); HttpPost httppost = new HttpPos
我试图使用apache HttpClient api将“myfile.txt”文件上传到apache服务器（WAMP）中作为“uploaded.txt”，但我得到了“404未找到”状态。 public class Httpupload1 { public static void main(String[] args) throws Exception { String url = "http://username:password@localhost/uploaded.txt"; DefaultHttpClient httpclient = new DefaultHttpClient(); HttpPost httppost = new HttpPos
A web server expects a file to be uploaded via an html form. This is the way how i construct the MultipartEntity, that already works: FileBody filePart = new FileBody(new File("emptyFile.txt"), "text/plain"); FormBodyPart fbp = new FormBodyPart("UploadService", filePart); MultipartEntity mpe = new MultipartEntity(); mpe.addPart(fbp); The fact is that i have my data in memory, so i don't
Web服务器希望通过HTML表单上传文件。 这就是我构建MultipartEntity的方式，它已经起作用了： FileBody filePart = new FileBody(new File("emptyFile.txt"), "text/plain"); FormBodyPart fbp = new FormBodyPart("UploadService", filePart); MultipartEntity mpe = new MultipartEntity(); mpe.addPart(fbp); 事实是，我有我的数据在内存中，所以我不喜欢把它保存到磁盘的想法，所以我试图替换 FileBody filePart = new
I am writing a simple file download servlet and I can't get correct filenames. Tried URLEncoding and MimeEncoding the filename as seen in existing answers, but none of them worked. The fileData object in the following snippet contains the mime type, the byte[] content and the filename, that needs at least ISO-8859-2 charset, ISO-8859-1 is not enough. How can I get my browser to display t
我正在写一个简单的文件下载servlet，我无法得到正确的文件名。 尝试URLEncoding和MimeEncoding在现有的答案中看到的文件名，但没有一个工作。 以下片段中的fileData对象包含需要至少ISO-8859-2字符集的MIME类型，字节[]内容和文件名，ISO-8859-1是不够的。 我如何让浏览器正确显示下载的文件名？ 以下是文件名的一个例子：árvíztűrőtükörfúrógép.xls，结果如下：árvíztqrptükörfúrógép.xls protected void renderMerge
I have an image that is Base64 encoded. What is the best way to decode that in Java? Hopefully using only the libraries included with Sun Java 6. As of v6, Java SE ships with JAXB. javax.xml.bind.DatatypeConverter has static methods that make this easy. See parseBase64Binary() and printBase64Binary() . As of Java 8 , there is an officially supported API for Base64 encoding and decoding.
我有一个Base64编码的图像。 在Java中解码的最佳方式是什么？ 希望仅使用Sun Java 6附带的库。 从第6版开始，Java SE随JAXB一起提供。 javax.xml.bind.DatatypeConverter有一些简单的静态方法。 请参阅parseBase64Binary()和printBase64Binary() 。 从Java 8开始 ，有一个官方支持的Base64编码和解码API。 及时这可能会成为默认选择。 该API包含类java.util.Base64及其嵌套类。 它支持三种不同的风格：基本，URL安全
This is sort of a contrived example, but I'm trying to get at a general principle here. Given phrases written in English using this list-like form: I have a cat I have a cat and a dog I have a cat, a dog, and a guinea pig I have a cat, a dog, a guinea pig, and a snake Can I use a regular expression to get all of the items, regardless of how many there are? Note that the items may contain
这是一个人为的例子，但我试图在这里得到一个一般原则。 鉴于使用这种列表形式的英文用英文书写： I have a cat I have a cat and a dog I have a cat, a dog, and a guinea pig I have a cat, a dog, a guinea pig, and a snake 我可以使用正则表达式来获取所有项目，无论有多少项？ 请注意，这些项目可能包含多个单词。 显然，如果我只有一个，那么我可以使用I have a (.+) ，如果有两个， I have a (.+) and a (.+)作