我是servlet的新手。我試圖做的是使用servlet將文件上傳到我的服務器,同時發送一個文本值,這是文件名在服務器端更改。問題是我使用ajax post將表單數據提交給servlet,但在我的servlet中,我從request.getParameter中獲得了空值。在Servlet中獲取表單值
這是我的html;
$("#fileUp").html( "<form enctype='multipart/form-data' id='uploadForm' action='" + url + "/PrjHRService/FileUpload'>"+
"<input name='file' id='file' type='file' size='50'>"+
"<input name='fname' type='text' >"+
"<input id='btnUpload' value='Upload' type='submit'>"+
//"<div id='imgLink'></div>"+
"</form>");
這是我的jquery ajax調用服務器;
$("#uploadForm").submit(function() {
var formData = new FormData($(this)[0]);
$.ajax({
type: "POST",
url: $("#uploadForm").attr("action"),
data: formData,
async: false,
success: function(data)
{
if(data.res === "true"){
jsi.showMsg("Uploaded Successfully");
$("#imgLink").html("Uploaded Successfully");
}else{
jsi.showMsg("Error Uploading");
}
},
contentType: false,
processData: false
});
return false; // avoid to execute the actual submit of the form.
});
這就是我如何從我的servlet中獲取表單的值;在那裏你會看到一行String fileExt = request.getParameter(「fname」);我發現fileExt爲空,這是我的問題。
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, java.io.IOException {
// Check that we have a file upload request
isMultipart = ServletFileUpload.isMultipartContent(request);
response.setContentType("application/json");
String fileExt=request.getParameter("fname");
java.io.PrintWriter out = response.getWriter();
if(!isMultipart){
JSONObject o=new JSONObject();
try {
o.put("res", "false");
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
out.println(o.toString());
return;
}
DiskFileItemFactory factory = new DiskFileItemFactory();
// maximum size that will be stored in memory
factory.setSizeThreshold(maxMemSize);
// Location to save data that is larger than maxMemSize.
factory.setRepository(new File("c:\\temp"));
// Create a new file upload handler
ServletFileUpload upload = new ServletFileUpload(factory);
// maximum file size to be uploaded.
upload.setSizeMax(maxFileSize);
try{
// Parse the request to get file items.
List fileItems = upload.parseRequest(request);
// Process the uploaded file items
Iterator i = fileItems.iterator();
while (i.hasNext())
{
FileItem fi = (FileItem)i.next();
if (!fi.isFormField())
{
// Get the uploaded file parameters
String fieldName = fi.getFieldName();
String fileName = fi.getName();
//String[] parts = fileName.split(".");
//fileName = parts[0] + "." + parts[1];
String contentType = fi.getContentType();
boolean isInMemory = fi.isInMemory();
long sizeInBytes = fi.getSize();
// Write the file
if(fileName.lastIndexOf("\\") >= 0){
file = new File(filePath +
fileName.substring(fileName.lastIndexOf("\\"))) ;
}else{
file = new File(filePath +
fileName.substring(fileName.lastIndexOf("\\")+1)) ;
}
fi.write(file) ;
JSONObject o=new JSONObject();
o.put("res", "true");
//out.println("Uploaded Filename: " + fileName + "<br>");
out.println(o.toString());
//out.println("Uploaded Successfully");
}
}
/* out.println("</body>");
out.println("</html>");*/
}catch(Exception ex) {
System.out.println(ex);
}
}
兩者都是POST。我可以上傳文件,但我無法獲得文本輸入值「fname」。 – ksh
你可以提供處理它的servlet的'doPost'方法嗎?另請查看我剛纔在答案中提供的文檔鏈接。 – Cromax
我剛加了doPost方法。謝謝。 – ksh