spring 如何使用inputStream获取文件名和文件类型？
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How to get file name and file Type using inputStream?
提问by Arunprasad
guys i'm using inputStream for file Download . now i want to pass fileName and fileType into DefaultStreamedContent .now how can i find fileName and FileType using inputStream .
伙计们，我正在使用 inputStream 进行文件下载。现在我想将 fileName 和 fileType 传递到 DefaultStreamedContent 中。现在我如何使用 inputStream 找到 fileName 和 FileType 。
InputStream inputStream = new BufferedInputStream(new FileInputStream(filePath));
fileDownload = new DefaultStreamedContent(inputStream,**fileType,fileName**);
回答by BalusC
This information can't be extracted from InputStream. This information can only be extracted based on the filePath(and with little help of java.io.Fileto easily get the filename).
无法从 中提取此信息InputStream。此信息只能根据filePath（并且几乎没有帮助java.io.File来轻松获取文件名）提取。
File file = new File(filePath);
InputStream inputStream = new FileInputStream(file);
String fileName = file.getName();
String fileType = FacesContext.getCurrentInstance().getEexternalContext().getMimeType(fileName);
fileDownload = new DefaultStreamedContent(inputStream, fileType, fileName);
The ExternalContext#getMimeType()is determined based on <mime-mapping>entries in web.xml. The servletcontainer has already a whole bunch definied by itself (in Tomcat, check /conf/web.xml) but you can extend and override it by (re)defining them in webapp's own /WEB-INF/web.xmllike as follows for an XLSX type:
在ExternalContext#getMimeType()基于确定<mime-mapping>的条目web.xml。servletcontainer 已经自己定义了一大堆（在 Tomcat 中， check /conf/web.xml），但是您可以通过（重新）在 webapp 自己中定义它们来扩展和覆盖它，/WEB-INF/web.xml如下所示，对于 XLSX 类型：
<mime-mapping>
<extension>xlsx</extension>
<mime-type>application/vnd.openxmlformats-officedocument.spreadsheetml.sheet</mime-type>
</mime-mapping>
