java实现文件上传,代码尽量简洁~~~·

发布网友发布时间：2022-05-05 10:41

共3个回答

热心网友时间：2022-06-27 09:36

你说的2种方法都是很简单的，参考网上的资料都不难做出，用io流做更是基础中的基础，我说下smartupload好了，有的人是直接写在jsp上面，感觉比较乱，我一般都是写在action里面，打好jar包和配置后

SmartUpload mySmartUpload = new SmartUpload();

//如果是struts2.0或者webwork 则是mySmartUpload.initialize(ServletActionContext.getServletConfig(),ServletActionContext.getRequest(),ServletActionContext.getResponse());

mySmartUpload.initialize(servlet.getServletConfig(), request,response);
mySmartUpload.setTotalMaxFileSize(500000);
//如果上传任意文件不设置mySmartUpload.setAllowedFilesList（文件后缀名）就可以了
mySmartUpload.upload();
for (int i = 0; i < mySmartUpload.getFiles().getCount(); i++) {
com.jspsmart.upload.File file = mySmartUpload.getFiles().getFile(i);
if (file.isMissing()) continue;
file.saveAs(保存的地址 + file.getFileName(),
su.SAVE_PHYSICAL);

热心网友时间：2022-06-27 09:37

一个SpringMVC的文件上传示例
1、jsp页面代码
<%@ page pageEncoding="utf-8"%>

<!DOCTYPE html>

<html>

<head>

<meta charset="utf-8">

<title>上传图片</title>

</head>

<body>

<form action="message.do">

<input type="text" name="test" value="Submit" /> <input type="submit"

value="Submit" />

</form>

<form action="upload.do" method="post" enctype="multipart/form-data">

<input type="file" name="file" /> <input type="submit" value="Submit" />

</form>

</body>

</html>

2、action代码
@RequestMapping(value = "/upload.do")

public String upload(@RequestParam(value = "file", required = false) MultipartFile file, HttpServletRequest request,

ModelMap model) {

String path = "/Users/xieyuhai/Desktop";//更换一下路径

String fileName = file.getOriginalFilename();

// String fileName = new Date().getTime()+".jpg";

System.out.println(path);

File targetFile = new File(path, fileName);

if (!targetFile.exists()) {

targetFile.mkdirs();

}

// 保存

try {

file.transferTo(targetFile);

} catch (Exception e) {

e.printStackTrace();

}

model.addAttribute("resultUrl", request.getContextPath() + "/upload/" + fileName);

return "result";

}

热心网友时间：2022-06-27 09:37

fis = new FileInputStream(getUpload());