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Send a file as multipart through XMLHttpRequest

Can I send a file as multipart by XMLHttpRequest to a servlet?

I am making a form and submitting it as multipart, but somehow I am not getting a response for successfully uploading it. I do not want the page to be refreshed, so it has to take place by Ajax.

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Tejasva Dhyani Avatar asked Feb 22 '12 13:02

Tejasva Dhyani


People also ask

How do I set the multipart form data in XMLHttpRequest?

It's not possible to send multipart/form-data with XMLHttpRequest (though it is possible in modern browsers, with XHR2 . See BalusC's answer). A common way to achieve what you want is to use a regular form , but in an iframe instead. This way, only the iframe is refreshed on upload.

How do I upload a file using XMLHttpRequest?

You have to use the FormData object to wrap the file into a multipart/form-data post data object: var formData = new FormData(); formData. append("thefile", file); xhr. send(formData);

How do I upload a file to multipart?

Follow this rules when creating a multipart form: Specify enctype="multipart/form-data" attribute on a form tag. Add a name attribute to a single input type="file" tag. DO NOT add a name attribute to any other input, select or textarea tags.


2 Answers

That's only possible with the XHR FormData API (previously known being part of as "XHR2" or "XHR Level 2", currently known as "XHR Advanced Features").

Given this HTML,

<input type="file" id="myFileField" name="myFile" /> 

you can upload it as below:

var formData = new FormData(); formData.append("myFile", document.getElementById("myFileField").files[0]);  var xhr = new XMLHttpRequest(); xhr.open("POST", "myServletUrl"); xhr.send(formData); 

XHR will take care about proper headers and request body encoding and the file will in this example be available on the server side as form-data part with the name myFile.

You need to keep in mind that FormData API is not supported in older browsers. At caniuse.com you can see that it's currently implemented in Chrome 7+, Firefox 3.5+, Safari 5+, Internet Explorer 10+ and Opera 12+.

In case you're using jQuery, then you might be tempted to use its $.val() function as below:

formData.append("myFile", $("#myFileField").val()); 

But this is incorrect as it doesn't return the whole File object, but merely the file name as String which is utterly useless as it doesn't contain the file contents.

If you don't want to use document.getElementById() for some reason, then use one of the following instead:

formData.append("myFile", $("#myFileField").prop("files")[0]); 
formData.append("myFile", $("#myFileField")[0].files[0]); 

An alternative is to use the jQuery Form plugin. Your entire form, when written and functioning properly without any line of JavaScript code, will then instantly be ajaxified with just the following line:

$("#formId").ajaxForm(function(response) {     // Handle Ajax response here. }); 

It also supports file uploads as well by a hidden iframe trick. See also this jQuery Form documentation for an in-depth explanation. You may only need to change the servlet code to be able to intercept on both normal (synchronous) and Ajax (asynchronous) requests. See also this answer for a concrete example: Simple calculator with JSP/Servlet and Ajax

Either way, the uploaded file should then be available in the doPost() method of a @MultipartConfig servlet as follows:

Part myFile = request.getPart("myFile"); 

Or if you're still on Servlet 2.5 or older, use Apache Commons FileUpload the usual way. See also this answer for a concrete example: How can I upload files to a server using JSP/Servlet?

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BalusC Avatar answered Sep 21 '22 06:09

BalusC


It's not possible to send multipart/form-data with XMLHttpRequest (though it is possible in modern browsers, with XHR2. See BalusC's answer).

A common way to achieve what you want is to use a regular form, but in an iframe instead. This way, only the iframe is refreshed on upload.

like image 25
Linus Thiel Avatar answered Sep 21 '22 06:09

Linus Thiel