MBJ*_*BJH 10 html javascript java xmlhttprequest
我想从一个文件(图像或视频)
<input type='file' id='file_i'/>
// Not this <input type='submit'/>
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使用XMLHttpRequest这样的
function img() {
var fd = new FormData();
fd.append('file', document.getElementById("file_i").files[0]);
var req;
if (window.ActiveXObject) {
req=new ActiveXObject();
} else {
req=new XMLHttpRequest();
}
req.open("post", "Image", true);
req.send(fd);
}
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例如.
然后在servlet中执行此操作:
new FileInputStream(new File(request.getParameter("file")))
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我该如何检索该文件?提前致谢.
我修好了它.这里是:
var fd = new FormData();
fd.append('file', document.getElementById("file_i").files[0]);
var req;
if (window.ActiveXObject) {
req=new ActiveXObject();
} else {
req=new XMLHttpRequest();
}
req.open("post", "Image", true);
req.send(fd);
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@MultipartConfig
public class addImage extends HttpServlet {
protected void doPost(HttpServletRequest request, HttpServletResponse response) throws IOException, ServletException {
Part filePart = request.getPart("file");
InputStream fileContent = filePart.getInputStream();
}
}
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<servlet>
<servlet-name>Add Image</servlet-name>
<servlet-class>servlets.addImage</servlet-class>
</servlet>
<servlet-mapping>
<servlet-name>Add Image</servlet-name>
<url-pattern>/Image</url-pattern>
</servlet-mapping>
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